Computing $int_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}}dx$ using residue calculus












21












$begingroup$


I need to find $displaystyleint_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}} dx$ where $a > 0$. To do this, I set $f(z) = displaystylefrac{cos z}{z^{2} + a^{2}}$ and integrate along the semi-circle of radius $R$. For the residue at $ia$ I get $displaystylefrac{cos(ia)}{2ia}$. Then letting $R rightarrow infty$, the integral over the arc is zero, so I get $displaystyleint_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}} dx = 2 pi i frac{cos(ia)}{2ia} = frac{pi cos(ia)}{a}$. But this is supposed to be $displaystylefrac{pi e^{-a}}{a}$, so I am doing something wrong.



In a similar problem, I have to evaluate $displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}} dx$ and get $pi i sin(ia)$ whereas this is supposed to be $pi e^{-a}$. I think I am getting some detail wrong in both cases. Can anyone enlighten me?










share|cite|improve this question











$endgroup$












  • $begingroup$
    The integral over the arc is NOT zero if you are integrating $cos(x)$ (cosines become hyperbolic cosines for imaginary $x$ which grow exponentially!). It IS, if you replace $cos x$ with $e^{ix}$ and then take the real part at the very end of the evaluation.
    $endgroup$
    – Alex R.
    May 3 '12 at 20:51












  • $begingroup$
    I was able to derive the result correctly by replacing $cos(x)$ with $e^{ix}$ as you said, but I do not understand why the integral would not be zero. The length of the curve is $pi R$. The supremum would be at most $frac{1}{R^{2} + a^{2}}$. How come this isn't zero?
    $endgroup$
    – Pedro
    May 3 '12 at 21:03










  • $begingroup$
    Again, you are thinking that $cos(x)$ is between -1 and 1, but this is only true for real $x$. Otherwise you have $cos(x)=frac{e^{ix}+e^{-ix}}{2}$ so try to plug in $x=a+bi$ and you will see that you are gurunteed to have exponential blowup as $|x|rightarrow infty$ along any path away from the real axis
    $endgroup$
    – Alex R.
    May 3 '12 at 21:52










  • $begingroup$
    You really need either to use $cos(x)=frac{exp(ix)+exp(-ix)}{2}$ or $cos(x)=Re(exp(ix))$ so that we work with $exp(ix)$ over the upper half-plane. $cos(ix)$ blows up exponentially as the imaginary part of $x$ gets large, so we can't use it in the contour integration.
    $endgroup$
    – robjohn
    May 4 '12 at 2:23


















21












$begingroup$


I need to find $displaystyleint_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}} dx$ where $a > 0$. To do this, I set $f(z) = displaystylefrac{cos z}{z^{2} + a^{2}}$ and integrate along the semi-circle of radius $R$. For the residue at $ia$ I get $displaystylefrac{cos(ia)}{2ia}$. Then letting $R rightarrow infty$, the integral over the arc is zero, so I get $displaystyleint_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}} dx = 2 pi i frac{cos(ia)}{2ia} = frac{pi cos(ia)}{a}$. But this is supposed to be $displaystylefrac{pi e^{-a}}{a}$, so I am doing something wrong.



In a similar problem, I have to evaluate $displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}} dx$ and get $pi i sin(ia)$ whereas this is supposed to be $pi e^{-a}$. I think I am getting some detail wrong in both cases. Can anyone enlighten me?










share|cite|improve this question











$endgroup$












  • $begingroup$
    The integral over the arc is NOT zero if you are integrating $cos(x)$ (cosines become hyperbolic cosines for imaginary $x$ which grow exponentially!). It IS, if you replace $cos x$ with $e^{ix}$ and then take the real part at the very end of the evaluation.
    $endgroup$
    – Alex R.
    May 3 '12 at 20:51












  • $begingroup$
    I was able to derive the result correctly by replacing $cos(x)$ with $e^{ix}$ as you said, but I do not understand why the integral would not be zero. The length of the curve is $pi R$. The supremum would be at most $frac{1}{R^{2} + a^{2}}$. How come this isn't zero?
    $endgroup$
    – Pedro
    May 3 '12 at 21:03










  • $begingroup$
    Again, you are thinking that $cos(x)$ is between -1 and 1, but this is only true for real $x$. Otherwise you have $cos(x)=frac{e^{ix}+e^{-ix}}{2}$ so try to plug in $x=a+bi$ and you will see that you are gurunteed to have exponential blowup as $|x|rightarrow infty$ along any path away from the real axis
    $endgroup$
    – Alex R.
    May 3 '12 at 21:52










  • $begingroup$
    You really need either to use $cos(x)=frac{exp(ix)+exp(-ix)}{2}$ or $cos(x)=Re(exp(ix))$ so that we work with $exp(ix)$ over the upper half-plane. $cos(ix)$ blows up exponentially as the imaginary part of $x$ gets large, so we can't use it in the contour integration.
    $endgroup$
    – robjohn
    May 4 '12 at 2:23
















21












21








21


19



$begingroup$


I need to find $displaystyleint_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}} dx$ where $a > 0$. To do this, I set $f(z) = displaystylefrac{cos z}{z^{2} + a^{2}}$ and integrate along the semi-circle of radius $R$. For the residue at $ia$ I get $displaystylefrac{cos(ia)}{2ia}$. Then letting $R rightarrow infty$, the integral over the arc is zero, so I get $displaystyleint_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}} dx = 2 pi i frac{cos(ia)}{2ia} = frac{pi cos(ia)}{a}$. But this is supposed to be $displaystylefrac{pi e^{-a}}{a}$, so I am doing something wrong.



In a similar problem, I have to evaluate $displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}} dx$ and get $pi i sin(ia)$ whereas this is supposed to be $pi e^{-a}$. I think I am getting some detail wrong in both cases. Can anyone enlighten me?










share|cite|improve this question











$endgroup$




I need to find $displaystyleint_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}} dx$ where $a > 0$. To do this, I set $f(z) = displaystylefrac{cos z}{z^{2} + a^{2}}$ and integrate along the semi-circle of radius $R$. For the residue at $ia$ I get $displaystylefrac{cos(ia)}{2ia}$. Then letting $R rightarrow infty$, the integral over the arc is zero, so I get $displaystyleint_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}} dx = 2 pi i frac{cos(ia)}{2ia} = frac{pi cos(ia)}{a}$. But this is supposed to be $displaystylefrac{pi e^{-a}}{a}$, so I am doing something wrong.



In a similar problem, I have to evaluate $displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}} dx$ and get $pi i sin(ia)$ whereas this is supposed to be $pi e^{-a}$. I think I am getting some detail wrong in both cases. Can anyone enlighten me?







integration complex-analysis definite-integrals residue-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '13 at 3:01









user85798

1




1










asked May 3 '12 at 20:46









PedroPedro

2,79131834




2,79131834












  • $begingroup$
    The integral over the arc is NOT zero if you are integrating $cos(x)$ (cosines become hyperbolic cosines for imaginary $x$ which grow exponentially!). It IS, if you replace $cos x$ with $e^{ix}$ and then take the real part at the very end of the evaluation.
    $endgroup$
    – Alex R.
    May 3 '12 at 20:51












  • $begingroup$
    I was able to derive the result correctly by replacing $cos(x)$ with $e^{ix}$ as you said, but I do not understand why the integral would not be zero. The length of the curve is $pi R$. The supremum would be at most $frac{1}{R^{2} + a^{2}}$. How come this isn't zero?
    $endgroup$
    – Pedro
    May 3 '12 at 21:03










  • $begingroup$
    Again, you are thinking that $cos(x)$ is between -1 and 1, but this is only true for real $x$. Otherwise you have $cos(x)=frac{e^{ix}+e^{-ix}}{2}$ so try to plug in $x=a+bi$ and you will see that you are gurunteed to have exponential blowup as $|x|rightarrow infty$ along any path away from the real axis
    $endgroup$
    – Alex R.
    May 3 '12 at 21:52










  • $begingroup$
    You really need either to use $cos(x)=frac{exp(ix)+exp(-ix)}{2}$ or $cos(x)=Re(exp(ix))$ so that we work with $exp(ix)$ over the upper half-plane. $cos(ix)$ blows up exponentially as the imaginary part of $x$ gets large, so we can't use it in the contour integration.
    $endgroup$
    – robjohn
    May 4 '12 at 2:23




















  • $begingroup$
    The integral over the arc is NOT zero if you are integrating $cos(x)$ (cosines become hyperbolic cosines for imaginary $x$ which grow exponentially!). It IS, if you replace $cos x$ with $e^{ix}$ and then take the real part at the very end of the evaluation.
    $endgroup$
    – Alex R.
    May 3 '12 at 20:51












  • $begingroup$
    I was able to derive the result correctly by replacing $cos(x)$ with $e^{ix}$ as you said, but I do not understand why the integral would not be zero. The length of the curve is $pi R$. The supremum would be at most $frac{1}{R^{2} + a^{2}}$. How come this isn't zero?
    $endgroup$
    – Pedro
    May 3 '12 at 21:03










  • $begingroup$
    Again, you are thinking that $cos(x)$ is between -1 and 1, but this is only true for real $x$. Otherwise you have $cos(x)=frac{e^{ix}+e^{-ix}}{2}$ so try to plug in $x=a+bi$ and you will see that you are gurunteed to have exponential blowup as $|x|rightarrow infty$ along any path away from the real axis
    $endgroup$
    – Alex R.
    May 3 '12 at 21:52










  • $begingroup$
    You really need either to use $cos(x)=frac{exp(ix)+exp(-ix)}{2}$ or $cos(x)=Re(exp(ix))$ so that we work with $exp(ix)$ over the upper half-plane. $cos(ix)$ blows up exponentially as the imaginary part of $x$ gets large, so we can't use it in the contour integration.
    $endgroup$
    – robjohn
    May 4 '12 at 2:23


















$begingroup$
The integral over the arc is NOT zero if you are integrating $cos(x)$ (cosines become hyperbolic cosines for imaginary $x$ which grow exponentially!). It IS, if you replace $cos x$ with $e^{ix}$ and then take the real part at the very end of the evaluation.
$endgroup$
– Alex R.
May 3 '12 at 20:51






$begingroup$
The integral over the arc is NOT zero if you are integrating $cos(x)$ (cosines become hyperbolic cosines for imaginary $x$ which grow exponentially!). It IS, if you replace $cos x$ with $e^{ix}$ and then take the real part at the very end of the evaluation.
$endgroup$
– Alex R.
May 3 '12 at 20:51














$begingroup$
I was able to derive the result correctly by replacing $cos(x)$ with $e^{ix}$ as you said, but I do not understand why the integral would not be zero. The length of the curve is $pi R$. The supremum would be at most $frac{1}{R^{2} + a^{2}}$. How come this isn't zero?
$endgroup$
– Pedro
May 3 '12 at 21:03




$begingroup$
I was able to derive the result correctly by replacing $cos(x)$ with $e^{ix}$ as you said, but I do not understand why the integral would not be zero. The length of the curve is $pi R$. The supremum would be at most $frac{1}{R^{2} + a^{2}}$. How come this isn't zero?
$endgroup$
– Pedro
May 3 '12 at 21:03












$begingroup$
Again, you are thinking that $cos(x)$ is between -1 and 1, but this is only true for real $x$. Otherwise you have $cos(x)=frac{e^{ix}+e^{-ix}}{2}$ so try to plug in $x=a+bi$ and you will see that you are gurunteed to have exponential blowup as $|x|rightarrow infty$ along any path away from the real axis
$endgroup$
– Alex R.
May 3 '12 at 21:52




$begingroup$
Again, you are thinking that $cos(x)$ is between -1 and 1, but this is only true for real $x$. Otherwise you have $cos(x)=frac{e^{ix}+e^{-ix}}{2}$ so try to plug in $x=a+bi$ and you will see that you are gurunteed to have exponential blowup as $|x|rightarrow infty$ along any path away from the real axis
$endgroup$
– Alex R.
May 3 '12 at 21:52












$begingroup$
You really need either to use $cos(x)=frac{exp(ix)+exp(-ix)}{2}$ or $cos(x)=Re(exp(ix))$ so that we work with $exp(ix)$ over the upper half-plane. $cos(ix)$ blows up exponentially as the imaginary part of $x$ gets large, so we can't use it in the contour integration.
$endgroup$
– robjohn
May 4 '12 at 2:23






$begingroup$
You really need either to use $cos(x)=frac{exp(ix)+exp(-ix)}{2}$ or $cos(x)=Re(exp(ix))$ so that we work with $exp(ix)$ over the upper half-plane. $cos(ix)$ blows up exponentially as the imaginary part of $x$ gets large, so we can't use it in the contour integration.
$endgroup$
– robjohn
May 4 '12 at 2:23












3 Answers
3






active

oldest

votes


















17












$begingroup$

Let $gamma$ be the path along the real axis then circling back counter-clockwise through the upper half-plane, letting the circle get infinitely big.
$$
begin{align}
int_{-infty}^inftyfrac{cos(x)}{x^2+a^2}mathrm{d}xtag{1}
&=Releft(int_{-infty}^inftyfrac{exp(ix)}{x^2+a^2}mathrm{d}xright)\tag{2}
&=Releft(int_{gamma}frac{exp(ix)}{x^2+a^2}mathrm{d}xright)\tag{3}
&=Releft(2pi i,mathrm{Res}left(frac{exp(ix)}{x^2+a^2},iaright)right)\tag{4}
&=Releft(2pi i,lim_{zto ia}frac{exp(ix)}{x+ia}right)\tag{5}
&=Releft(2pi i,frac{exp(-a)}{2ia}right)\
&=frac{pi exp(-a)}{a}
end{align}
$$
$(1)$: $Re(exp(ix))=cos(x)$



$(2)$: The integral along the circle back through the upper half-plane vanishes as the circle gets bigger.



$(3)$: There is only one singularity of $dfrac{exp(ix)}{x^2+a^2}$ in the upper half-plane, at $x=ia$. The integral along $gamma$ is the residue of $dfrac{exp(ix)}{x^2+a^2}$ at $x=ia$.



$(4)$: The singularity of $dfrac{exp(ix)}{x^2+a^2}$ at $x=ia$ is a simple pole. We can compute the residue as $displaystylelim_{xto ia}(x-ia)frac{exp(ix)}{x^2+a^2}=lim_{xto ia}frac{exp(ix)}{x+ia}$.



$(5)$: plug in $x=ia$.






Following the same strategy,
$$
begin{align}
int_{-infty}^inftyfrac{xsin(x)}{x^2+a^2}mathrm{d}x
&=Imleft(int_{-infty}^inftyfrac{xexp(ix)}{x^2+a^2}mathrm{d}xright)\
&=Imleft(int_{gamma}frac{xexp(ix)}{x^2+a^2}mathrm{d}xright)\
&=Imleft(2pi i,mathrm{Res}left(frac{xexp(ix)}{x^2+a^2},iaright)right)\
&=Imleft(2pi i,lim_{zto ia}frac{xexp(ix)}{x+ia}right)\
&=Imleft(2pi i,frac{iaexp(-a)}{2ia}right)\
&=pi exp(-a)
end{align}
$$




share|cite|improve this answer











$endgroup$





















    9












    $begingroup$


    Although the OP is searching for a way forward using contour integration and the residue theorem, I thought it might be instructive to present an approach that uses real analysis only. To that end, we proceed.






    Let $g(a)$ be given by the convergent improper integral



    $$g(a)=int_{-infty}^infty frac{cos(x)}{x^2+a^2},dx tag1$$



    Exploiting the even symmetry of the integrand and enforcing the substitution $xto ax$ reveals



    $$g(a)=frac2aint_0^infty frac{cos(ax)}{x^2+1},dx$$



    Now let $f(a)=frac a2 g(a)=int_0^infty frac{cos(ax)}{x^2+1},dx$



    Since the integral $int_0^infty frac{xsin(ax)}{x^2+1},dx$ is uniformly convergent for $|a|ge delta>0$, we may differentiate under the integral in $(3)$ for $|a|>delta>0$ to obtain



    $$begin{align}
    f'(a)&=-int_0^infty frac{xsin(ax)}{x^2+1},dx\\
    &=-int_0^infty frac{(x^2+1-1)sin(ax)}{x(x^2+1)},dx\\
    &=-int_0^infty frac{sin(ax)}{x},dx+int_0^infty frac{sin(ax)}{x(x^2+1)},dx\\
    &=-frac{pi}{2}+int_0^infty frac{sin(ax)}{x(x^2+1)},dxtag4
    end{align}$$



    Again, since the integral $int_0^infty frac{cos(ax)}{x^2+1},dx$ converges uniformly for all $a$, we may differentiate under the integral in $(4)$ to obtain



    $$f''(a)=int_0^infty frac{cos(ax)}{x^2+1},dx=f(a)tag 5$$



    Solving the second-order ODE in $(5)$ reveals



    $$f(a)=C_1 e^{a}+C_2 e^{-a}$$



    Using $f(0)=pi/2$ and $f'(0)=-pi/2$, we find that $C_1=0$ and $C_2=frac{pi}{2}$ and hence $f(a)=frac{pi e^{-a}}{2}$. Multiplying by $2/a$ yields the coveted result



    $$bbox[5px,border:2px solid #C0A000]{int_{-infty}^infty frac{cos(x)}{x^2+a^2},dx=frac{pi}{ae^a}}$$



    as expected!






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I just corrected I typo, I hope you don't mind.
      $endgroup$
      – Shashi
      Dec 26 '17 at 10:01






    • 1




      $begingroup$
      @shashi Not at all. Much appreciative.
      $endgroup$
      – Mark Viola
      Dec 26 '17 at 16:14










    • $begingroup$
      @Fawad My answer is correct and DOES agree with both Rob's answer and Argon's answer. So, I'm not sure what on earth you are talking about.
      $endgroup$
      – Mark Viola
      Jan 3 at 5:09












    • $begingroup$
      I am sorry,I saw their end part which is for sine function.
      $endgroup$
      – Fawad
      Jan 3 at 16:06










    • $begingroup$
      @Fawad Thank you for the reply. And feel free to up vote an answer if you find it useful. ;-)
      $endgroup$
      – Mark Viola
      Jan 3 at 16:25



















    4












    $begingroup$

    Let us integrate along the contour $Gamma$, which is the semicircle of radius $R$ described above. Let $C_R$ be the arc of the contour with radius $R$. We may say that



    $$oint_{Gamma}frac{cos z}{z^2+a^2}, dz=int_{-R}^{R}frac{cos x}{x^2+a^2}, dx+int_{C_R}frac{cos z}{z^2+a^2}, dz$$



    Note that $cos x = operatorname{Re},(e^{ix})$. Thus



    $$
    f(z)=frac{operatorname{Re},(e^{iz})}{z^2+a^2} = operatorname{Re},left(frac{e^{iz}}{(z+ia)(z-ia)}right)
    $$



    Because $f(z)$ can be written ias $g(z)e^{iaz}$ and suffices all of the conditions of Jordan's lemma, we see that the integral along the arc tends to zero when $Rto infty$. Thus



    $$
    lim_{Rtoinfty}oint_{Gamma}frac{cos z}{z^2+a^2}, dz=lim_{Rtoinfty}
    int_{-R}^{R}frac{cos x}{x^2+a^2}, dx+0
    $$



    To solve the LHS, we find the residues. The residue, similar to the one you found is
    $$
    b=frac{e^{i^2a}}{ia+ia}=frac{e^{-a}}{2ia}
    $$



    Thus



    $$
    displaystyleint_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}}, dx=
    lim_{Rtoinfty}oint_{Gamma}frac{cos z}{z^2+a^2}, dz=operatorname{Re},left(displaystyleint_{-infty}^{infty} frac{e^{iz}}{z^{2} + a^{2}}, dzright)
    =operatorname{Re},(2pi ib)=operatorname{Re},(2pi ifrac{e^{-a}}{2ia}) = frac{pi e^{-a}}{a}
    $$





    Similarly, with $displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}} dx$ we change $sin x$ to $operatorname{Im},(e^{ix})$ and make the function complex-valued.



    $$g(z)=frac{zsin z}{z^2+a^2}=operatorname{Im},left(frac{ze^{iz}}{(z+ia)(z-ia)}right)$$



    Jordan's lemma again can be used to show that the integral around the arc tends to zero as $Rtoinfty$. We then proceed to find the residues.
    The residue of pole $z_1=ia$ is



    $$b=frac{iae^{-a}}{2ia}=frac{e^{-a}}{2}$$



    $z_1$ is the unique pole in the contour, so upon multiplying its residue, $b$, with $2pi i$ we find:
    $$
    displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}}, dx=
    lim_{Rtoinfty}oint_{Gamma} frac{x sin x}{x^{2} + a^{2}}, dz=
    operatorname{Im},left(displaystyleint_{-infty}^{infty} frac{ze^{iz}}{(z+ia)(z-ia)}, dzright)=
    operatorname{Im},(2pi ib)=
    operatorname{Im},(2pi i frac{e^{-a}}{2})=
    operatorname{Im},(pi e^{-a}i)=
    pi e^{-a}
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It would be good to describe the contour along which you are integrating.
      $endgroup$
      – robjohn
      May 4 '12 at 0:34










    • $begingroup$
      @robjohn I am using the one described above: "integrate along the semi-circle of radius R"...
      $endgroup$
      – Argon
      May 4 '12 at 1:09










    • $begingroup$
      @Argon Maybe also mention one needs to use Jordan's lemma to estimate the contribution from the arc for the $int frac{xsin x}{x^2+a^2} dx $ integral, just the usual Length * Max estimate fails there.
      $endgroup$
      – Ragib Zaman
      May 4 '12 at 1:19










    • $begingroup$
      @RagibZaman Very true, I will include this.
      $endgroup$
      – Argon
      May 4 '12 at 1:49










    • $begingroup$
      @robjohn the contour he uses is clearly stated the first sentence, re-read the solution. Very nice solution Argon, thanks a lot!
      $endgroup$
      – Jeff Faraci
      Dec 17 '13 at 5:59











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f140580%2fcomputing-int-infty-infty-frac-cos-xx2-a2dx-using-residue%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    17












    $begingroup$

    Let $gamma$ be the path along the real axis then circling back counter-clockwise through the upper half-plane, letting the circle get infinitely big.
    $$
    begin{align}
    int_{-infty}^inftyfrac{cos(x)}{x^2+a^2}mathrm{d}xtag{1}
    &=Releft(int_{-infty}^inftyfrac{exp(ix)}{x^2+a^2}mathrm{d}xright)\tag{2}
    &=Releft(int_{gamma}frac{exp(ix)}{x^2+a^2}mathrm{d}xright)\tag{3}
    &=Releft(2pi i,mathrm{Res}left(frac{exp(ix)}{x^2+a^2},iaright)right)\tag{4}
    &=Releft(2pi i,lim_{zto ia}frac{exp(ix)}{x+ia}right)\tag{5}
    &=Releft(2pi i,frac{exp(-a)}{2ia}right)\
    &=frac{pi exp(-a)}{a}
    end{align}
    $$
    $(1)$: $Re(exp(ix))=cos(x)$



    $(2)$: The integral along the circle back through the upper half-plane vanishes as the circle gets bigger.



    $(3)$: There is only one singularity of $dfrac{exp(ix)}{x^2+a^2}$ in the upper half-plane, at $x=ia$. The integral along $gamma$ is the residue of $dfrac{exp(ix)}{x^2+a^2}$ at $x=ia$.



    $(4)$: The singularity of $dfrac{exp(ix)}{x^2+a^2}$ at $x=ia$ is a simple pole. We can compute the residue as $displaystylelim_{xto ia}(x-ia)frac{exp(ix)}{x^2+a^2}=lim_{xto ia}frac{exp(ix)}{x+ia}$.



    $(5)$: plug in $x=ia$.






    Following the same strategy,
    $$
    begin{align}
    int_{-infty}^inftyfrac{xsin(x)}{x^2+a^2}mathrm{d}x
    &=Imleft(int_{-infty}^inftyfrac{xexp(ix)}{x^2+a^2}mathrm{d}xright)\
    &=Imleft(int_{gamma}frac{xexp(ix)}{x^2+a^2}mathrm{d}xright)\
    &=Imleft(2pi i,mathrm{Res}left(frac{xexp(ix)}{x^2+a^2},iaright)right)\
    &=Imleft(2pi i,lim_{zto ia}frac{xexp(ix)}{x+ia}right)\
    &=Imleft(2pi i,frac{iaexp(-a)}{2ia}right)\
    &=pi exp(-a)
    end{align}
    $$




    share|cite|improve this answer











    $endgroup$


















      17












      $begingroup$

      Let $gamma$ be the path along the real axis then circling back counter-clockwise through the upper half-plane, letting the circle get infinitely big.
      $$
      begin{align}
      int_{-infty}^inftyfrac{cos(x)}{x^2+a^2}mathrm{d}xtag{1}
      &=Releft(int_{-infty}^inftyfrac{exp(ix)}{x^2+a^2}mathrm{d}xright)\tag{2}
      &=Releft(int_{gamma}frac{exp(ix)}{x^2+a^2}mathrm{d}xright)\tag{3}
      &=Releft(2pi i,mathrm{Res}left(frac{exp(ix)}{x^2+a^2},iaright)right)\tag{4}
      &=Releft(2pi i,lim_{zto ia}frac{exp(ix)}{x+ia}right)\tag{5}
      &=Releft(2pi i,frac{exp(-a)}{2ia}right)\
      &=frac{pi exp(-a)}{a}
      end{align}
      $$
      $(1)$: $Re(exp(ix))=cos(x)$



      $(2)$: The integral along the circle back through the upper half-plane vanishes as the circle gets bigger.



      $(3)$: There is only one singularity of $dfrac{exp(ix)}{x^2+a^2}$ in the upper half-plane, at $x=ia$. The integral along $gamma$ is the residue of $dfrac{exp(ix)}{x^2+a^2}$ at $x=ia$.



      $(4)$: The singularity of $dfrac{exp(ix)}{x^2+a^2}$ at $x=ia$ is a simple pole. We can compute the residue as $displaystylelim_{xto ia}(x-ia)frac{exp(ix)}{x^2+a^2}=lim_{xto ia}frac{exp(ix)}{x+ia}$.



      $(5)$: plug in $x=ia$.






      Following the same strategy,
      $$
      begin{align}
      int_{-infty}^inftyfrac{xsin(x)}{x^2+a^2}mathrm{d}x
      &=Imleft(int_{-infty}^inftyfrac{xexp(ix)}{x^2+a^2}mathrm{d}xright)\
      &=Imleft(int_{gamma}frac{xexp(ix)}{x^2+a^2}mathrm{d}xright)\
      &=Imleft(2pi i,mathrm{Res}left(frac{xexp(ix)}{x^2+a^2},iaright)right)\
      &=Imleft(2pi i,lim_{zto ia}frac{xexp(ix)}{x+ia}right)\
      &=Imleft(2pi i,frac{iaexp(-a)}{2ia}right)\
      &=pi exp(-a)
      end{align}
      $$




      share|cite|improve this answer











      $endgroup$
















        17












        17








        17





        $begingroup$

        Let $gamma$ be the path along the real axis then circling back counter-clockwise through the upper half-plane, letting the circle get infinitely big.
        $$
        begin{align}
        int_{-infty}^inftyfrac{cos(x)}{x^2+a^2}mathrm{d}xtag{1}
        &=Releft(int_{-infty}^inftyfrac{exp(ix)}{x^2+a^2}mathrm{d}xright)\tag{2}
        &=Releft(int_{gamma}frac{exp(ix)}{x^2+a^2}mathrm{d}xright)\tag{3}
        &=Releft(2pi i,mathrm{Res}left(frac{exp(ix)}{x^2+a^2},iaright)right)\tag{4}
        &=Releft(2pi i,lim_{zto ia}frac{exp(ix)}{x+ia}right)\tag{5}
        &=Releft(2pi i,frac{exp(-a)}{2ia}right)\
        &=frac{pi exp(-a)}{a}
        end{align}
        $$
        $(1)$: $Re(exp(ix))=cos(x)$



        $(2)$: The integral along the circle back through the upper half-plane vanishes as the circle gets bigger.



        $(3)$: There is only one singularity of $dfrac{exp(ix)}{x^2+a^2}$ in the upper half-plane, at $x=ia$. The integral along $gamma$ is the residue of $dfrac{exp(ix)}{x^2+a^2}$ at $x=ia$.



        $(4)$: The singularity of $dfrac{exp(ix)}{x^2+a^2}$ at $x=ia$ is a simple pole. We can compute the residue as $displaystylelim_{xto ia}(x-ia)frac{exp(ix)}{x^2+a^2}=lim_{xto ia}frac{exp(ix)}{x+ia}$.



        $(5)$: plug in $x=ia$.






        Following the same strategy,
        $$
        begin{align}
        int_{-infty}^inftyfrac{xsin(x)}{x^2+a^2}mathrm{d}x
        &=Imleft(int_{-infty}^inftyfrac{xexp(ix)}{x^2+a^2}mathrm{d}xright)\
        &=Imleft(int_{gamma}frac{xexp(ix)}{x^2+a^2}mathrm{d}xright)\
        &=Imleft(2pi i,mathrm{Res}left(frac{xexp(ix)}{x^2+a^2},iaright)right)\
        &=Imleft(2pi i,lim_{zto ia}frac{xexp(ix)}{x+ia}right)\
        &=Imleft(2pi i,frac{iaexp(-a)}{2ia}right)\
        &=pi exp(-a)
        end{align}
        $$




        share|cite|improve this answer











        $endgroup$



        Let $gamma$ be the path along the real axis then circling back counter-clockwise through the upper half-plane, letting the circle get infinitely big.
        $$
        begin{align}
        int_{-infty}^inftyfrac{cos(x)}{x^2+a^2}mathrm{d}xtag{1}
        &=Releft(int_{-infty}^inftyfrac{exp(ix)}{x^2+a^2}mathrm{d}xright)\tag{2}
        &=Releft(int_{gamma}frac{exp(ix)}{x^2+a^2}mathrm{d}xright)\tag{3}
        &=Releft(2pi i,mathrm{Res}left(frac{exp(ix)}{x^2+a^2},iaright)right)\tag{4}
        &=Releft(2pi i,lim_{zto ia}frac{exp(ix)}{x+ia}right)\tag{5}
        &=Releft(2pi i,frac{exp(-a)}{2ia}right)\
        &=frac{pi exp(-a)}{a}
        end{align}
        $$
        $(1)$: $Re(exp(ix))=cos(x)$



        $(2)$: The integral along the circle back through the upper half-plane vanishes as the circle gets bigger.



        $(3)$: There is only one singularity of $dfrac{exp(ix)}{x^2+a^2}$ in the upper half-plane, at $x=ia$. The integral along $gamma$ is the residue of $dfrac{exp(ix)}{x^2+a^2}$ at $x=ia$.



        $(4)$: The singularity of $dfrac{exp(ix)}{x^2+a^2}$ at $x=ia$ is a simple pole. We can compute the residue as $displaystylelim_{xto ia}(x-ia)frac{exp(ix)}{x^2+a^2}=lim_{xto ia}frac{exp(ix)}{x+ia}$.



        $(5)$: plug in $x=ia$.






        Following the same strategy,
        $$
        begin{align}
        int_{-infty}^inftyfrac{xsin(x)}{x^2+a^2}mathrm{d}x
        &=Imleft(int_{-infty}^inftyfrac{xexp(ix)}{x^2+a^2}mathrm{d}xright)\
        &=Imleft(int_{gamma}frac{xexp(ix)}{x^2+a^2}mathrm{d}xright)\
        &=Imleft(2pi i,mathrm{Res}left(frac{xexp(ix)}{x^2+a^2},iaright)right)\
        &=Imleft(2pi i,lim_{zto ia}frac{xexp(ix)}{x+ia}right)\
        &=Imleft(2pi i,frac{iaexp(-a)}{2ia}right)\
        &=pi exp(-a)
        end{align}
        $$





        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited May 3 '12 at 21:56

























        answered May 3 '12 at 21:24









        robjohnrobjohn

        266k27306630




        266k27306630























            9












            $begingroup$


            Although the OP is searching for a way forward using contour integration and the residue theorem, I thought it might be instructive to present an approach that uses real analysis only. To that end, we proceed.






            Let $g(a)$ be given by the convergent improper integral



            $$g(a)=int_{-infty}^infty frac{cos(x)}{x^2+a^2},dx tag1$$



            Exploiting the even symmetry of the integrand and enforcing the substitution $xto ax$ reveals



            $$g(a)=frac2aint_0^infty frac{cos(ax)}{x^2+1},dx$$



            Now let $f(a)=frac a2 g(a)=int_0^infty frac{cos(ax)}{x^2+1},dx$



            Since the integral $int_0^infty frac{xsin(ax)}{x^2+1},dx$ is uniformly convergent for $|a|ge delta>0$, we may differentiate under the integral in $(3)$ for $|a|>delta>0$ to obtain



            $$begin{align}
            f'(a)&=-int_0^infty frac{xsin(ax)}{x^2+1},dx\\
            &=-int_0^infty frac{(x^2+1-1)sin(ax)}{x(x^2+1)},dx\\
            &=-int_0^infty frac{sin(ax)}{x},dx+int_0^infty frac{sin(ax)}{x(x^2+1)},dx\\
            &=-frac{pi}{2}+int_0^infty frac{sin(ax)}{x(x^2+1)},dxtag4
            end{align}$$



            Again, since the integral $int_0^infty frac{cos(ax)}{x^2+1},dx$ converges uniformly for all $a$, we may differentiate under the integral in $(4)$ to obtain



            $$f''(a)=int_0^infty frac{cos(ax)}{x^2+1},dx=f(a)tag 5$$



            Solving the second-order ODE in $(5)$ reveals



            $$f(a)=C_1 e^{a}+C_2 e^{-a}$$



            Using $f(0)=pi/2$ and $f'(0)=-pi/2$, we find that $C_1=0$ and $C_2=frac{pi}{2}$ and hence $f(a)=frac{pi e^{-a}}{2}$. Multiplying by $2/a$ yields the coveted result



            $$bbox[5px,border:2px solid #C0A000]{int_{-infty}^infty frac{cos(x)}{x^2+a^2},dx=frac{pi}{ae^a}}$$



            as expected!






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I just corrected I typo, I hope you don't mind.
              $endgroup$
              – Shashi
              Dec 26 '17 at 10:01






            • 1




              $begingroup$
              @shashi Not at all. Much appreciative.
              $endgroup$
              – Mark Viola
              Dec 26 '17 at 16:14










            • $begingroup$
              @Fawad My answer is correct and DOES agree with both Rob's answer and Argon's answer. So, I'm not sure what on earth you are talking about.
              $endgroup$
              – Mark Viola
              Jan 3 at 5:09












            • $begingroup$
              I am sorry,I saw their end part which is for sine function.
              $endgroup$
              – Fawad
              Jan 3 at 16:06










            • $begingroup$
              @Fawad Thank you for the reply. And feel free to up vote an answer if you find it useful. ;-)
              $endgroup$
              – Mark Viola
              Jan 3 at 16:25
















            9












            $begingroup$


            Although the OP is searching for a way forward using contour integration and the residue theorem, I thought it might be instructive to present an approach that uses real analysis only. To that end, we proceed.






            Let $g(a)$ be given by the convergent improper integral



            $$g(a)=int_{-infty}^infty frac{cos(x)}{x^2+a^2},dx tag1$$



            Exploiting the even symmetry of the integrand and enforcing the substitution $xto ax$ reveals



            $$g(a)=frac2aint_0^infty frac{cos(ax)}{x^2+1},dx$$



            Now let $f(a)=frac a2 g(a)=int_0^infty frac{cos(ax)}{x^2+1},dx$



            Since the integral $int_0^infty frac{xsin(ax)}{x^2+1},dx$ is uniformly convergent for $|a|ge delta>0$, we may differentiate under the integral in $(3)$ for $|a|>delta>0$ to obtain



            $$begin{align}
            f'(a)&=-int_0^infty frac{xsin(ax)}{x^2+1},dx\\
            &=-int_0^infty frac{(x^2+1-1)sin(ax)}{x(x^2+1)},dx\\
            &=-int_0^infty frac{sin(ax)}{x},dx+int_0^infty frac{sin(ax)}{x(x^2+1)},dx\\
            &=-frac{pi}{2}+int_0^infty frac{sin(ax)}{x(x^2+1)},dxtag4
            end{align}$$



            Again, since the integral $int_0^infty frac{cos(ax)}{x^2+1},dx$ converges uniformly for all $a$, we may differentiate under the integral in $(4)$ to obtain



            $$f''(a)=int_0^infty frac{cos(ax)}{x^2+1},dx=f(a)tag 5$$



            Solving the second-order ODE in $(5)$ reveals



            $$f(a)=C_1 e^{a}+C_2 e^{-a}$$



            Using $f(0)=pi/2$ and $f'(0)=-pi/2$, we find that $C_1=0$ and $C_2=frac{pi}{2}$ and hence $f(a)=frac{pi e^{-a}}{2}$. Multiplying by $2/a$ yields the coveted result



            $$bbox[5px,border:2px solid #C0A000]{int_{-infty}^infty frac{cos(x)}{x^2+a^2},dx=frac{pi}{ae^a}}$$



            as expected!






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I just corrected I typo, I hope you don't mind.
              $endgroup$
              – Shashi
              Dec 26 '17 at 10:01






            • 1




              $begingroup$
              @shashi Not at all. Much appreciative.
              $endgroup$
              – Mark Viola
              Dec 26 '17 at 16:14










            • $begingroup$
              @Fawad My answer is correct and DOES agree with both Rob's answer and Argon's answer. So, I'm not sure what on earth you are talking about.
              $endgroup$
              – Mark Viola
              Jan 3 at 5:09












            • $begingroup$
              I am sorry,I saw their end part which is for sine function.
              $endgroup$
              – Fawad
              Jan 3 at 16:06










            • $begingroup$
              @Fawad Thank you for the reply. And feel free to up vote an answer if you find it useful. ;-)
              $endgroup$
              – Mark Viola
              Jan 3 at 16:25














            9












            9








            9





            $begingroup$


            Although the OP is searching for a way forward using contour integration and the residue theorem, I thought it might be instructive to present an approach that uses real analysis only. To that end, we proceed.






            Let $g(a)$ be given by the convergent improper integral



            $$g(a)=int_{-infty}^infty frac{cos(x)}{x^2+a^2},dx tag1$$



            Exploiting the even symmetry of the integrand and enforcing the substitution $xto ax$ reveals



            $$g(a)=frac2aint_0^infty frac{cos(ax)}{x^2+1},dx$$



            Now let $f(a)=frac a2 g(a)=int_0^infty frac{cos(ax)}{x^2+1},dx$



            Since the integral $int_0^infty frac{xsin(ax)}{x^2+1},dx$ is uniformly convergent for $|a|ge delta>0$, we may differentiate under the integral in $(3)$ for $|a|>delta>0$ to obtain



            $$begin{align}
            f'(a)&=-int_0^infty frac{xsin(ax)}{x^2+1},dx\\
            &=-int_0^infty frac{(x^2+1-1)sin(ax)}{x(x^2+1)},dx\\
            &=-int_0^infty frac{sin(ax)}{x},dx+int_0^infty frac{sin(ax)}{x(x^2+1)},dx\\
            &=-frac{pi}{2}+int_0^infty frac{sin(ax)}{x(x^2+1)},dxtag4
            end{align}$$



            Again, since the integral $int_0^infty frac{cos(ax)}{x^2+1},dx$ converges uniformly for all $a$, we may differentiate under the integral in $(4)$ to obtain



            $$f''(a)=int_0^infty frac{cos(ax)}{x^2+1},dx=f(a)tag 5$$



            Solving the second-order ODE in $(5)$ reveals



            $$f(a)=C_1 e^{a}+C_2 e^{-a}$$



            Using $f(0)=pi/2$ and $f'(0)=-pi/2$, we find that $C_1=0$ and $C_2=frac{pi}{2}$ and hence $f(a)=frac{pi e^{-a}}{2}$. Multiplying by $2/a$ yields the coveted result



            $$bbox[5px,border:2px solid #C0A000]{int_{-infty}^infty frac{cos(x)}{x^2+a^2},dx=frac{pi}{ae^a}}$$



            as expected!






            share|cite|improve this answer











            $endgroup$




            Although the OP is searching for a way forward using contour integration and the residue theorem, I thought it might be instructive to present an approach that uses real analysis only. To that end, we proceed.






            Let $g(a)$ be given by the convergent improper integral



            $$g(a)=int_{-infty}^infty frac{cos(x)}{x^2+a^2},dx tag1$$



            Exploiting the even symmetry of the integrand and enforcing the substitution $xto ax$ reveals



            $$g(a)=frac2aint_0^infty frac{cos(ax)}{x^2+1},dx$$



            Now let $f(a)=frac a2 g(a)=int_0^infty frac{cos(ax)}{x^2+1},dx$



            Since the integral $int_0^infty frac{xsin(ax)}{x^2+1},dx$ is uniformly convergent for $|a|ge delta>0$, we may differentiate under the integral in $(3)$ for $|a|>delta>0$ to obtain



            $$begin{align}
            f'(a)&=-int_0^infty frac{xsin(ax)}{x^2+1},dx\\
            &=-int_0^infty frac{(x^2+1-1)sin(ax)}{x(x^2+1)},dx\\
            &=-int_0^infty frac{sin(ax)}{x},dx+int_0^infty frac{sin(ax)}{x(x^2+1)},dx\\
            &=-frac{pi}{2}+int_0^infty frac{sin(ax)}{x(x^2+1)},dxtag4
            end{align}$$



            Again, since the integral $int_0^infty frac{cos(ax)}{x^2+1},dx$ converges uniformly for all $a$, we may differentiate under the integral in $(4)$ to obtain



            $$f''(a)=int_0^infty frac{cos(ax)}{x^2+1},dx=f(a)tag 5$$



            Solving the second-order ODE in $(5)$ reveals



            $$f(a)=C_1 e^{a}+C_2 e^{-a}$$



            Using $f(0)=pi/2$ and $f'(0)=-pi/2$, we find that $C_1=0$ and $C_2=frac{pi}{2}$ and hence $f(a)=frac{pi e^{-a}}{2}$. Multiplying by $2/a$ yields the coveted result



            $$bbox[5px,border:2px solid #C0A000]{int_{-infty}^infty frac{cos(x)}{x^2+a^2},dx=frac{pi}{ae^a}}$$



            as expected!







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 3 at 5:07

























            answered Oct 19 '17 at 18:21









            Mark ViolaMark Viola

            131k1275171




            131k1275171












            • $begingroup$
              I just corrected I typo, I hope you don't mind.
              $endgroup$
              – Shashi
              Dec 26 '17 at 10:01






            • 1




              $begingroup$
              @shashi Not at all. Much appreciative.
              $endgroup$
              – Mark Viola
              Dec 26 '17 at 16:14










            • $begingroup$
              @Fawad My answer is correct and DOES agree with both Rob's answer and Argon's answer. So, I'm not sure what on earth you are talking about.
              $endgroup$
              – Mark Viola
              Jan 3 at 5:09












            • $begingroup$
              I am sorry,I saw their end part which is for sine function.
              $endgroup$
              – Fawad
              Jan 3 at 16:06










            • $begingroup$
              @Fawad Thank you for the reply. And feel free to up vote an answer if you find it useful. ;-)
              $endgroup$
              – Mark Viola
              Jan 3 at 16:25


















            • $begingroup$
              I just corrected I typo, I hope you don't mind.
              $endgroup$
              – Shashi
              Dec 26 '17 at 10:01






            • 1




              $begingroup$
              @shashi Not at all. Much appreciative.
              $endgroup$
              – Mark Viola
              Dec 26 '17 at 16:14










            • $begingroup$
              @Fawad My answer is correct and DOES agree with both Rob's answer and Argon's answer. So, I'm not sure what on earth you are talking about.
              $endgroup$
              – Mark Viola
              Jan 3 at 5:09












            • $begingroup$
              I am sorry,I saw their end part which is for sine function.
              $endgroup$
              – Fawad
              Jan 3 at 16:06










            • $begingroup$
              @Fawad Thank you for the reply. And feel free to up vote an answer if you find it useful. ;-)
              $endgroup$
              – Mark Viola
              Jan 3 at 16:25
















            $begingroup$
            I just corrected I typo, I hope you don't mind.
            $endgroup$
            – Shashi
            Dec 26 '17 at 10:01




            $begingroup$
            I just corrected I typo, I hope you don't mind.
            $endgroup$
            – Shashi
            Dec 26 '17 at 10:01




            1




            1




            $begingroup$
            @shashi Not at all. Much appreciative.
            $endgroup$
            – Mark Viola
            Dec 26 '17 at 16:14




            $begingroup$
            @shashi Not at all. Much appreciative.
            $endgroup$
            – Mark Viola
            Dec 26 '17 at 16:14












            $begingroup$
            @Fawad My answer is correct and DOES agree with both Rob's answer and Argon's answer. So, I'm not sure what on earth you are talking about.
            $endgroup$
            – Mark Viola
            Jan 3 at 5:09






            $begingroup$
            @Fawad My answer is correct and DOES agree with both Rob's answer and Argon's answer. So, I'm not sure what on earth you are talking about.
            $endgroup$
            – Mark Viola
            Jan 3 at 5:09














            $begingroup$
            I am sorry,I saw their end part which is for sine function.
            $endgroup$
            – Fawad
            Jan 3 at 16:06




            $begingroup$
            I am sorry,I saw their end part which is for sine function.
            $endgroup$
            – Fawad
            Jan 3 at 16:06












            $begingroup$
            @Fawad Thank you for the reply. And feel free to up vote an answer if you find it useful. ;-)
            $endgroup$
            – Mark Viola
            Jan 3 at 16:25




            $begingroup$
            @Fawad Thank you for the reply. And feel free to up vote an answer if you find it useful. ;-)
            $endgroup$
            – Mark Viola
            Jan 3 at 16:25











            4












            $begingroup$

            Let us integrate along the contour $Gamma$, which is the semicircle of radius $R$ described above. Let $C_R$ be the arc of the contour with radius $R$. We may say that



            $$oint_{Gamma}frac{cos z}{z^2+a^2}, dz=int_{-R}^{R}frac{cos x}{x^2+a^2}, dx+int_{C_R}frac{cos z}{z^2+a^2}, dz$$



            Note that $cos x = operatorname{Re},(e^{ix})$. Thus



            $$
            f(z)=frac{operatorname{Re},(e^{iz})}{z^2+a^2} = operatorname{Re},left(frac{e^{iz}}{(z+ia)(z-ia)}right)
            $$



            Because $f(z)$ can be written ias $g(z)e^{iaz}$ and suffices all of the conditions of Jordan's lemma, we see that the integral along the arc tends to zero when $Rto infty$. Thus



            $$
            lim_{Rtoinfty}oint_{Gamma}frac{cos z}{z^2+a^2}, dz=lim_{Rtoinfty}
            int_{-R}^{R}frac{cos x}{x^2+a^2}, dx+0
            $$



            To solve the LHS, we find the residues. The residue, similar to the one you found is
            $$
            b=frac{e^{i^2a}}{ia+ia}=frac{e^{-a}}{2ia}
            $$



            Thus



            $$
            displaystyleint_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}}, dx=
            lim_{Rtoinfty}oint_{Gamma}frac{cos z}{z^2+a^2}, dz=operatorname{Re},left(displaystyleint_{-infty}^{infty} frac{e^{iz}}{z^{2} + a^{2}}, dzright)
            =operatorname{Re},(2pi ib)=operatorname{Re},(2pi ifrac{e^{-a}}{2ia}) = frac{pi e^{-a}}{a}
            $$





            Similarly, with $displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}} dx$ we change $sin x$ to $operatorname{Im},(e^{ix})$ and make the function complex-valued.



            $$g(z)=frac{zsin z}{z^2+a^2}=operatorname{Im},left(frac{ze^{iz}}{(z+ia)(z-ia)}right)$$



            Jordan's lemma again can be used to show that the integral around the arc tends to zero as $Rtoinfty$. We then proceed to find the residues.
            The residue of pole $z_1=ia$ is



            $$b=frac{iae^{-a}}{2ia}=frac{e^{-a}}{2}$$



            $z_1$ is the unique pole in the contour, so upon multiplying its residue, $b$, with $2pi i$ we find:
            $$
            displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}}, dx=
            lim_{Rtoinfty}oint_{Gamma} frac{x sin x}{x^{2} + a^{2}}, dz=
            operatorname{Im},left(displaystyleint_{-infty}^{infty} frac{ze^{iz}}{(z+ia)(z-ia)}, dzright)=
            operatorname{Im},(2pi ib)=
            operatorname{Im},(2pi i frac{e^{-a}}{2})=
            operatorname{Im},(pi e^{-a}i)=
            pi e^{-a}
            $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It would be good to describe the contour along which you are integrating.
              $endgroup$
              – robjohn
              May 4 '12 at 0:34










            • $begingroup$
              @robjohn I am using the one described above: "integrate along the semi-circle of radius R"...
              $endgroup$
              – Argon
              May 4 '12 at 1:09










            • $begingroup$
              @Argon Maybe also mention one needs to use Jordan's lemma to estimate the contribution from the arc for the $int frac{xsin x}{x^2+a^2} dx $ integral, just the usual Length * Max estimate fails there.
              $endgroup$
              – Ragib Zaman
              May 4 '12 at 1:19










            • $begingroup$
              @RagibZaman Very true, I will include this.
              $endgroup$
              – Argon
              May 4 '12 at 1:49










            • $begingroup$
              @robjohn the contour he uses is clearly stated the first sentence, re-read the solution. Very nice solution Argon, thanks a lot!
              $endgroup$
              – Jeff Faraci
              Dec 17 '13 at 5:59
















            4












            $begingroup$

            Let us integrate along the contour $Gamma$, which is the semicircle of radius $R$ described above. Let $C_R$ be the arc of the contour with radius $R$. We may say that



            $$oint_{Gamma}frac{cos z}{z^2+a^2}, dz=int_{-R}^{R}frac{cos x}{x^2+a^2}, dx+int_{C_R}frac{cos z}{z^2+a^2}, dz$$



            Note that $cos x = operatorname{Re},(e^{ix})$. Thus



            $$
            f(z)=frac{operatorname{Re},(e^{iz})}{z^2+a^2} = operatorname{Re},left(frac{e^{iz}}{(z+ia)(z-ia)}right)
            $$



            Because $f(z)$ can be written ias $g(z)e^{iaz}$ and suffices all of the conditions of Jordan's lemma, we see that the integral along the arc tends to zero when $Rto infty$. Thus



            $$
            lim_{Rtoinfty}oint_{Gamma}frac{cos z}{z^2+a^2}, dz=lim_{Rtoinfty}
            int_{-R}^{R}frac{cos x}{x^2+a^2}, dx+0
            $$



            To solve the LHS, we find the residues. The residue, similar to the one you found is
            $$
            b=frac{e^{i^2a}}{ia+ia}=frac{e^{-a}}{2ia}
            $$



            Thus



            $$
            displaystyleint_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}}, dx=
            lim_{Rtoinfty}oint_{Gamma}frac{cos z}{z^2+a^2}, dz=operatorname{Re},left(displaystyleint_{-infty}^{infty} frac{e^{iz}}{z^{2} + a^{2}}, dzright)
            =operatorname{Re},(2pi ib)=operatorname{Re},(2pi ifrac{e^{-a}}{2ia}) = frac{pi e^{-a}}{a}
            $$





            Similarly, with $displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}} dx$ we change $sin x$ to $operatorname{Im},(e^{ix})$ and make the function complex-valued.



            $$g(z)=frac{zsin z}{z^2+a^2}=operatorname{Im},left(frac{ze^{iz}}{(z+ia)(z-ia)}right)$$



            Jordan's lemma again can be used to show that the integral around the arc tends to zero as $Rtoinfty$. We then proceed to find the residues.
            The residue of pole $z_1=ia$ is



            $$b=frac{iae^{-a}}{2ia}=frac{e^{-a}}{2}$$



            $z_1$ is the unique pole in the contour, so upon multiplying its residue, $b$, with $2pi i$ we find:
            $$
            displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}}, dx=
            lim_{Rtoinfty}oint_{Gamma} frac{x sin x}{x^{2} + a^{2}}, dz=
            operatorname{Im},left(displaystyleint_{-infty}^{infty} frac{ze^{iz}}{(z+ia)(z-ia)}, dzright)=
            operatorname{Im},(2pi ib)=
            operatorname{Im},(2pi i frac{e^{-a}}{2})=
            operatorname{Im},(pi e^{-a}i)=
            pi e^{-a}
            $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It would be good to describe the contour along which you are integrating.
              $endgroup$
              – robjohn
              May 4 '12 at 0:34










            • $begingroup$
              @robjohn I am using the one described above: "integrate along the semi-circle of radius R"...
              $endgroup$
              – Argon
              May 4 '12 at 1:09










            • $begingroup$
              @Argon Maybe also mention one needs to use Jordan's lemma to estimate the contribution from the arc for the $int frac{xsin x}{x^2+a^2} dx $ integral, just the usual Length * Max estimate fails there.
              $endgroup$
              – Ragib Zaman
              May 4 '12 at 1:19










            • $begingroup$
              @RagibZaman Very true, I will include this.
              $endgroup$
              – Argon
              May 4 '12 at 1:49










            • $begingroup$
              @robjohn the contour he uses is clearly stated the first sentence, re-read the solution. Very nice solution Argon, thanks a lot!
              $endgroup$
              – Jeff Faraci
              Dec 17 '13 at 5:59














            4












            4








            4





            $begingroup$

            Let us integrate along the contour $Gamma$, which is the semicircle of radius $R$ described above. Let $C_R$ be the arc of the contour with radius $R$. We may say that



            $$oint_{Gamma}frac{cos z}{z^2+a^2}, dz=int_{-R}^{R}frac{cos x}{x^2+a^2}, dx+int_{C_R}frac{cos z}{z^2+a^2}, dz$$



            Note that $cos x = operatorname{Re},(e^{ix})$. Thus



            $$
            f(z)=frac{operatorname{Re},(e^{iz})}{z^2+a^2} = operatorname{Re},left(frac{e^{iz}}{(z+ia)(z-ia)}right)
            $$



            Because $f(z)$ can be written ias $g(z)e^{iaz}$ and suffices all of the conditions of Jordan's lemma, we see that the integral along the arc tends to zero when $Rto infty$. Thus



            $$
            lim_{Rtoinfty}oint_{Gamma}frac{cos z}{z^2+a^2}, dz=lim_{Rtoinfty}
            int_{-R}^{R}frac{cos x}{x^2+a^2}, dx+0
            $$



            To solve the LHS, we find the residues. The residue, similar to the one you found is
            $$
            b=frac{e^{i^2a}}{ia+ia}=frac{e^{-a}}{2ia}
            $$



            Thus



            $$
            displaystyleint_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}}, dx=
            lim_{Rtoinfty}oint_{Gamma}frac{cos z}{z^2+a^2}, dz=operatorname{Re},left(displaystyleint_{-infty}^{infty} frac{e^{iz}}{z^{2} + a^{2}}, dzright)
            =operatorname{Re},(2pi ib)=operatorname{Re},(2pi ifrac{e^{-a}}{2ia}) = frac{pi e^{-a}}{a}
            $$





            Similarly, with $displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}} dx$ we change $sin x$ to $operatorname{Im},(e^{ix})$ and make the function complex-valued.



            $$g(z)=frac{zsin z}{z^2+a^2}=operatorname{Im},left(frac{ze^{iz}}{(z+ia)(z-ia)}right)$$



            Jordan's lemma again can be used to show that the integral around the arc tends to zero as $Rtoinfty$. We then proceed to find the residues.
            The residue of pole $z_1=ia$ is



            $$b=frac{iae^{-a}}{2ia}=frac{e^{-a}}{2}$$



            $z_1$ is the unique pole in the contour, so upon multiplying its residue, $b$, with $2pi i$ we find:
            $$
            displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}}, dx=
            lim_{Rtoinfty}oint_{Gamma} frac{x sin x}{x^{2} + a^{2}}, dz=
            operatorname{Im},left(displaystyleint_{-infty}^{infty} frac{ze^{iz}}{(z+ia)(z-ia)}, dzright)=
            operatorname{Im},(2pi ib)=
            operatorname{Im},(2pi i frac{e^{-a}}{2})=
            operatorname{Im},(pi e^{-a}i)=
            pi e^{-a}
            $$






            share|cite|improve this answer











            $endgroup$



            Let us integrate along the contour $Gamma$, which is the semicircle of radius $R$ described above. Let $C_R$ be the arc of the contour with radius $R$. We may say that



            $$oint_{Gamma}frac{cos z}{z^2+a^2}, dz=int_{-R}^{R}frac{cos x}{x^2+a^2}, dx+int_{C_R}frac{cos z}{z^2+a^2}, dz$$



            Note that $cos x = operatorname{Re},(e^{ix})$. Thus



            $$
            f(z)=frac{operatorname{Re},(e^{iz})}{z^2+a^2} = operatorname{Re},left(frac{e^{iz}}{(z+ia)(z-ia)}right)
            $$



            Because $f(z)$ can be written ias $g(z)e^{iaz}$ and suffices all of the conditions of Jordan's lemma, we see that the integral along the arc tends to zero when $Rto infty$. Thus



            $$
            lim_{Rtoinfty}oint_{Gamma}frac{cos z}{z^2+a^2}, dz=lim_{Rtoinfty}
            int_{-R}^{R}frac{cos x}{x^2+a^2}, dx+0
            $$



            To solve the LHS, we find the residues. The residue, similar to the one you found is
            $$
            b=frac{e^{i^2a}}{ia+ia}=frac{e^{-a}}{2ia}
            $$



            Thus



            $$
            displaystyleint_{-infty}^{infty} frac{cos x}{x^{2} + a^{2}}, dx=
            lim_{Rtoinfty}oint_{Gamma}frac{cos z}{z^2+a^2}, dz=operatorname{Re},left(displaystyleint_{-infty}^{infty} frac{e^{iz}}{z^{2} + a^{2}}, dzright)
            =operatorname{Re},(2pi ib)=operatorname{Re},(2pi ifrac{e^{-a}}{2ia}) = frac{pi e^{-a}}{a}
            $$





            Similarly, with $displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}} dx$ we change $sin x$ to $operatorname{Im},(e^{ix})$ and make the function complex-valued.



            $$g(z)=frac{zsin z}{z^2+a^2}=operatorname{Im},left(frac{ze^{iz}}{(z+ia)(z-ia)}right)$$



            Jordan's lemma again can be used to show that the integral around the arc tends to zero as $Rtoinfty$. We then proceed to find the residues.
            The residue of pole $z_1=ia$ is



            $$b=frac{iae^{-a}}{2ia}=frac{e^{-a}}{2}$$



            $z_1$ is the unique pole in the contour, so upon multiplying its residue, $b$, with $2pi i$ we find:
            $$
            displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}}, dx=
            lim_{Rtoinfty}oint_{Gamma} frac{x sin x}{x^{2} + a^{2}}, dz=
            operatorname{Im},left(displaystyleint_{-infty}^{infty} frac{ze^{iz}}{(z+ia)(z-ia)}, dzright)=
            operatorname{Im},(2pi ib)=
            operatorname{Im},(2pi i frac{e^{-a}}{2})=
            operatorname{Im},(pi e^{-a}i)=
            pi e^{-a}
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited May 4 '12 at 2:01

























            answered May 3 '12 at 21:20









            ArgonArgon

            16.3k673122




            16.3k673122












            • $begingroup$
              It would be good to describe the contour along which you are integrating.
              $endgroup$
              – robjohn
              May 4 '12 at 0:34










            • $begingroup$
              @robjohn I am using the one described above: "integrate along the semi-circle of radius R"...
              $endgroup$
              – Argon
              May 4 '12 at 1:09










            • $begingroup$
              @Argon Maybe also mention one needs to use Jordan's lemma to estimate the contribution from the arc for the $int frac{xsin x}{x^2+a^2} dx $ integral, just the usual Length * Max estimate fails there.
              $endgroup$
              – Ragib Zaman
              May 4 '12 at 1:19










            • $begingroup$
              @RagibZaman Very true, I will include this.
              $endgroup$
              – Argon
              May 4 '12 at 1:49










            • $begingroup$
              @robjohn the contour he uses is clearly stated the first sentence, re-read the solution. Very nice solution Argon, thanks a lot!
              $endgroup$
              – Jeff Faraci
              Dec 17 '13 at 5:59


















            • $begingroup$
              It would be good to describe the contour along which you are integrating.
              $endgroup$
              – robjohn
              May 4 '12 at 0:34










            • $begingroup$
              @robjohn I am using the one described above: "integrate along the semi-circle of radius R"...
              $endgroup$
              – Argon
              May 4 '12 at 1:09










            • $begingroup$
              @Argon Maybe also mention one needs to use Jordan's lemma to estimate the contribution from the arc for the $int frac{xsin x}{x^2+a^2} dx $ integral, just the usual Length * Max estimate fails there.
              $endgroup$
              – Ragib Zaman
              May 4 '12 at 1:19










            • $begingroup$
              @RagibZaman Very true, I will include this.
              $endgroup$
              – Argon
              May 4 '12 at 1:49










            • $begingroup$
              @robjohn the contour he uses is clearly stated the first sentence, re-read the solution. Very nice solution Argon, thanks a lot!
              $endgroup$
              – Jeff Faraci
              Dec 17 '13 at 5:59
















            $begingroup$
            It would be good to describe the contour along which you are integrating.
            $endgroup$
            – robjohn
            May 4 '12 at 0:34




            $begingroup$
            It would be good to describe the contour along which you are integrating.
            $endgroup$
            – robjohn
            May 4 '12 at 0:34












            $begingroup$
            @robjohn I am using the one described above: "integrate along the semi-circle of radius R"...
            $endgroup$
            – Argon
            May 4 '12 at 1:09




            $begingroup$
            @robjohn I am using the one described above: "integrate along the semi-circle of radius R"...
            $endgroup$
            – Argon
            May 4 '12 at 1:09












            $begingroup$
            @Argon Maybe also mention one needs to use Jordan's lemma to estimate the contribution from the arc for the $int frac{xsin x}{x^2+a^2} dx $ integral, just the usual Length * Max estimate fails there.
            $endgroup$
            – Ragib Zaman
            May 4 '12 at 1:19




            $begingroup$
            @Argon Maybe also mention one needs to use Jordan's lemma to estimate the contribution from the arc for the $int frac{xsin x}{x^2+a^2} dx $ integral, just the usual Length * Max estimate fails there.
            $endgroup$
            – Ragib Zaman
            May 4 '12 at 1:19












            $begingroup$
            @RagibZaman Very true, I will include this.
            $endgroup$
            – Argon
            May 4 '12 at 1:49




            $begingroup$
            @RagibZaman Very true, I will include this.
            $endgroup$
            – Argon
            May 4 '12 at 1:49












            $begingroup$
            @robjohn the contour he uses is clearly stated the first sentence, re-read the solution. Very nice solution Argon, thanks a lot!
            $endgroup$
            – Jeff Faraci
            Dec 17 '13 at 5:59




            $begingroup$
            @robjohn the contour he uses is clearly stated the first sentence, re-read the solution. Very nice solution Argon, thanks a lot!
            $endgroup$
            – Jeff Faraci
            Dec 17 '13 at 5:59


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f140580%2fcomputing-int-infty-infty-frac-cos-xx2-a2dx-using-residue%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Human spaceflight

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            張江高科駅