Dtyjlui

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?

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 $x=ia$.

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!

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:
$$
displaystyleint_{-infty}^{infty} frac{x sin x}{x^{2} + a^{2}}, dx=
lim_{Rtoinfty}oint_{Gamma} frac{x sin x}{x^{2} + a^{2}}, dz=
operatorname{Im},left(displaystyleint_{-infty}^{infty} frac{ze^{iz}}{(z+ia)(z-ia)}, dzright)=
operatorname{Im},(2pi ib)=
operatorname{Im},(2pi i frac{e^{-a}}{2})=
operatorname{Im},(pi e^{-a}i)=
pi e^{-a}
$$