Solving Riemann-Stieltjes integral:$int_{- pi/4}^{pi/4} f(x)dg(x)$












8












$begingroup$


I'm having trouble solving this Riemann-Stieltjes integral:



$$int_{- pi/4}^{pi/4} f(x)dg(x),$$ where $$f(x):= begin{cases} frac{sin^4x}{cos^2x}{} &text{if }xge0, \{}\ frac1{cos^3x} &text{if }x<0,end{cases}$$



and $$g(x)=begin{cases} phantom{-} 1+sin(x) &text{if }-pi/4 <x<pi/4, \ -1 &text{otherwise}.end{cases}$$



I believe the only jump discontinuities are at $-pi/4$ and $pi/4$. Which $g=-1$ at both of those points. I'm struggling with the rest. What formula should I be using to compute the integral and what should my answer look like? Thanks for any help!






real-analysis calculus integration trigonometry stieltjes-integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    Check this theorem.
    $endgroup$
    – Mhenni Benghorbal
    May 3 '13 at 1:08










  • $begingroup$
    Can you elaborate how to solve the last integral in that theorem (related to my problem). I can see why I would use it since both f and g have no common discontinuities.
    $endgroup$
    – Drake
    May 6 '13 at 13:51










  • $begingroup$
    CAS says:$1-{frac {7,sqrt {2}}{12}}+ln left( 1+sqrt {2} right) $
    $endgroup$
    – Mariusz Iwaniuk
    Oct 1 '18 at 22:14
















8












$begingroup$


I'm having trouble solving this Riemann-Stieltjes integral:



$$int_{- pi/4}^{pi/4} f(x)dg(x),$$ where $$f(x):= begin{cases} frac{sin^4x}{cos^2x}{} &text{if }xge0, \{}\ frac1{cos^3x} &text{if }x<0,end{cases}$$



and $$g(x)=begin{cases} phantom{-} 1+sin(x) &text{if }-pi/4 <x<pi/4, \ -1 &text{otherwise}.end{cases}$$



I believe the only jump discontinuities are at $-pi/4$ and $pi/4$. Which $g=-1$ at both of those points. I'm struggling with the rest. What formula should I be using to compute the integral and what should my answer look like? Thanks for any help!






real-analysis calculus integration trigonometry stieltjes-integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    Check this theorem.
    $endgroup$
    – Mhenni Benghorbal
    May 3 '13 at 1:08










  • $begingroup$
    Can you elaborate how to solve the last integral in that theorem (related to my problem). I can see why I would use it since both f and g have no common discontinuities.
    $endgroup$
    – Drake
    May 6 '13 at 13:51










  • $begingroup$
    CAS says:$1-{frac {7,sqrt {2}}{12}}+ln left( 1+sqrt {2} right) $
    $endgroup$
    – Mariusz Iwaniuk
    Oct 1 '18 at 22:14














8












8








8


4



$begingroup$


I'm having trouble solving this Riemann-Stieltjes integral:



$$int_{- pi/4}^{pi/4} f(x)dg(x),$$ where $$f(x):= begin{cases} frac{sin^4x}{cos^2x}{} &text{if }xge0, \{}\ frac1{cos^3x} &text{if }x<0,end{cases}$$



and $$g(x)=begin{cases} phantom{-} 1+sin(x) &text{if }-pi/4 <x<pi/4, \ -1 &text{otherwise}.end{cases}$$



I believe the only jump discontinuities are at $-pi/4$ and $pi/4$. Which $g=-1$ at both of those points. I'm struggling with the rest. What formula should I be using to compute the integral and what should my answer look like? Thanks for any help!






real-analysis calculus integration trigonometry stieltjes-integral







share|cite|improve this question











$endgroup$




I'm having trouble solving this Riemann-Stieltjes integral:



$$int_{- pi/4}^{pi/4} f(x)dg(x),$$ where $$f(x):= begin{cases} frac{sin^4x}{cos^2x}{} &text{if }xge0, \{}\ frac1{cos^3x} &text{if }x<0,end{cases}$$



and $$g(x)=begin{cases} phantom{-} 1+sin(x) &text{if }-pi/4 <x<pi/4, \ -1 &text{otherwise}.end{cases}$$



I believe the only jump discontinuities are at $-pi/4$ and $pi/4$. Which $g=-1$ at both of those points. I'm struggling with the rest. What formula should I be using to compute the integral and what should my answer look like? Thanks for any help!






real-analysis calculus integration trigonometry stieltjes-integral




real-analysis calculus integration trigonometry stieltjes-integral






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 22:25









Batominovski

33.1k33293




33.1k33293










asked May 3 '13 at 0:53









DrakeDrake

4681515




4681515












  • $begingroup$
    Check this theorem.
    $endgroup$
    – Mhenni Benghorbal
    May 3 '13 at 1:08










  • $begingroup$
    Can you elaborate how to solve the last integral in that theorem (related to my problem). I can see why I would use it since both f and g have no common discontinuities.
    $endgroup$
    – Drake
    May 6 '13 at 13:51










  • $begingroup$
    CAS says:$1-{frac {7,sqrt {2}}{12}}+ln left( 1+sqrt {2} right) $
    $endgroup$
    – Mariusz Iwaniuk
    Oct 1 '18 at 22:14


















  • $begingroup$
    Check this theorem.
    $endgroup$
    – Mhenni Benghorbal
    May 3 '13 at 1:08










  • $begingroup$
    Can you elaborate how to solve the last integral in that theorem (related to my problem). I can see why I would use it since both f and g have no common discontinuities.
    $endgroup$
    – Drake
    May 6 '13 at 13:51










  • $begingroup$
    CAS says:$1-{frac {7,sqrt {2}}{12}}+ln left( 1+sqrt {2} right) $
    $endgroup$
    – Mariusz Iwaniuk
    Oct 1 '18 at 22:14
















$begingroup$
Check this theorem.
$endgroup$
– Mhenni Benghorbal
May 3 '13 at 1:08




$begingroup$
Check this theorem.
$endgroup$
– Mhenni Benghorbal
May 3 '13 at 1:08












$begingroup$
Can you elaborate how to solve the last integral in that theorem (related to my problem). I can see why I would use it since both f and g have no common discontinuities.
$endgroup$
– Drake
May 6 '13 at 13:51




$begingroup$
Can you elaborate how to solve the last integral in that theorem (related to my problem). I can see why I would use it since both f and g have no common discontinuities.
$endgroup$
– Drake
May 6 '13 at 13:51












$begingroup$
CAS says:$1-{frac {7,sqrt {2}}{12}}+ln left( 1+sqrt {2} right) $
$endgroup$
– Mariusz Iwaniuk
Oct 1 '18 at 22:14




$begingroup$
CAS says:$1-{frac {7,sqrt {2}}{12}}+ln left( 1+sqrt {2} right) $
$endgroup$
– Mariusz Iwaniuk
Oct 1 '18 at 22:14










3 Answers
3






active

oldest

votes


















0












$begingroup$

Since $g$ is differentiable on $-pi/4 <x<pi/4$ your integral changes to Riemann integral simply by the following theorem: $$int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx$$ so you will have $int_{- pi/4}^{pi/4} f(x)dg(x)= int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}d(1+sin x)=int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}cos x dx$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
    $endgroup$
    – New day rising
    Jan 7 '17 at 3:50












  • $begingroup$
    Psst... both of you need to take a closer look to what happens at the endpoints.
    $endgroup$
    – user361424
    Jun 28 '17 at 0:55



















0












$begingroup$


Since $g$ is differentiable there is no trouble with sum this boils down to Riemann integral of $g'cdot f$




Namely from here we have $$color{blue}{int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx}$$
Therefore,
$$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)\=int_{-pi/4}^0 frac 1{cos^2 x} dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x)$$



But $$int_{-pi/4}^0 frac 1{cos^2 x} dx =int^{pi/4}_0 (tan x)' dx =1$$



and
$$int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x) = int_0^{pi/4} frac{sin^4 x}{1-sin^2 x} d(sin x) =int_0^{sqrt{2}/2} frac{t^4 }{1-t^2 } dt\=int_0^{sqrt{2}/2} -t^2+frac{t^2 }{1-t^2 } dt$$
$$
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The first formula isn't applicable when $g$ has jump discontinuities.
    $endgroup$
    – Maxim
    May 28 '18 at 21:15










  • $begingroup$
    @Maxim It is possible the discontinuities are of first order.
    $endgroup$
    – Guy Fsone
    May 30 '18 at 7:14










  • $begingroup$
    There are zero order discontinuities at the endpoints. See my answer.
    $endgroup$
    – Maxim
    May 30 '18 at 9:42










  • $begingroup$
    @Maxim what do you mean by zero order discontnuity?
    $endgroup$
    – Guy Fsone
    May 31 '18 at 1:52










  • $begingroup$
    I mean that the function itself (the derivative of zero order) is discontinuous.
    $endgroup$
    – Maxim
    May 31 '18 at 9:45



















0












$begingroup$

Let $T = pi/4$. The first term in the Riemann-Stieltjes sum, $f(xi_1)(g(x_1) - g(-T))$, does not tend to zero when $x_1 downarrow -T$ and $xi_1 downarrow -T$, and similarly for the last term. The integral is
$$I = int_{-T}^T f(x) dg(x) = \
lim_{epsilon downarrow 0} {
f(-T) (g(-T + epsilon)- g(-T)) + \
f(T) (g(T)- g(T - epsilon)) } + \
int_{-T}^T f(x) cos x ,dx = \
2 operatorname{arctanh}(sqrt 2 - 1) + frac {19 sqrt 2} 6 - 2.$$
It can be verified that
$$I = f(T) g(T) - f(-T) g(-T) - int_{-T}^T g(x) df(x) = \
f(T) g(T) - f(-T) g(-T) - \
lim_{epsilon downarrow 0} g(0) (f(epsilon) - f(-epsilon)) - \
int_{-T}^0 g(x) left( frac 1 {cos^3 x} right)' dx -
int_0^T g(x) left( frac {sin^4 x} {cos^2 x} right)' dx.$$






share|cite|improve this answer











$endgroup$














    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%2f379818%2fsolving-riemann-stieltjes-integral-int-pi-4-pi-4-fxdgx%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









    0












    $begingroup$

    Since $g$ is differentiable on $-pi/4 <x<pi/4$ your integral changes to Riemann integral simply by the following theorem: $$int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx$$ so you will have $int_{- pi/4}^{pi/4} f(x)dg(x)= int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}d(1+sin x)=int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}cos x dx$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
      $endgroup$
      – New day rising
      Jan 7 '17 at 3:50












    • $begingroup$
      Psst... both of you need to take a closer look to what happens at the endpoints.
      $endgroup$
      – user361424
      Jun 28 '17 at 0:55
















    0












    $begingroup$

    Since $g$ is differentiable on $-pi/4 <x<pi/4$ your integral changes to Riemann integral simply by the following theorem: $$int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx$$ so you will have $int_{- pi/4}^{pi/4} f(x)dg(x)= int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}d(1+sin x)=int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}cos x dx$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
      $endgroup$
      – New day rising
      Jan 7 '17 at 3:50












    • $begingroup$
      Psst... both of you need to take a closer look to what happens at the endpoints.
      $endgroup$
      – user361424
      Jun 28 '17 at 0:55














    0












    0








    0





    $begingroup$

    Since $g$ is differentiable on $-pi/4 <x<pi/4$ your integral changes to Riemann integral simply by the following theorem: $$int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx$$ so you will have $int_{- pi/4}^{pi/4} f(x)dg(x)= int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}d(1+sin x)=int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}cos x dx$






    share|cite|improve this answer









    $endgroup$



    Since $g$ is differentiable on $-pi/4 <x<pi/4$ your integral changes to Riemann integral simply by the following theorem: $$int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx$$ so you will have $int_{- pi/4}^{pi/4} f(x)dg(x)= int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}d(1+sin x)=int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}cos x dx$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 4 '15 at 10:01









    AliAli

    1,9812520




    1,9812520












    • $begingroup$
      The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
      $endgroup$
      – New day rising
      Jan 7 '17 at 3:50












    • $begingroup$
      Psst... both of you need to take a closer look to what happens at the endpoints.
      $endgroup$
      – user361424
      Jun 28 '17 at 0:55


















    • $begingroup$
      The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
      $endgroup$
      – New day rising
      Jan 7 '17 at 3:50












    • $begingroup$
      Psst... both of you need to take a closer look to what happens at the endpoints.
      $endgroup$
      – user361424
      Jun 28 '17 at 0:55
















    $begingroup$
    The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
    $endgroup$
    – New day rising
    Jan 7 '17 at 3:50






    $begingroup$
    The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
    $endgroup$
    – New day rising
    Jan 7 '17 at 3:50














    $begingroup$
    Psst... both of you need to take a closer look to what happens at the endpoints.
    $endgroup$
    – user361424
    Jun 28 '17 at 0:55




    $begingroup$
    Psst... both of you need to take a closer look to what happens at the endpoints.
    $endgroup$
    – user361424
    Jun 28 '17 at 0:55











    0












    $begingroup$


    Since $g$ is differentiable there is no trouble with sum this boils down to Riemann integral of $g'cdot f$




    Namely from here we have $$color{blue}{int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx}$$
    Therefore,
    $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)\=int_{-pi/4}^0 frac 1{cos^2 x} dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x)$$



    But $$int_{-pi/4}^0 frac 1{cos^2 x} dx =int^{pi/4}_0 (tan x)' dx =1$$



    and
    $$int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x) = int_0^{pi/4} frac{sin^4 x}{1-sin^2 x} d(sin x) =int_0^{sqrt{2}/2} frac{t^4 }{1-t^2 } dt\=int_0^{sqrt{2}/2} -t^2+frac{t^2 }{1-t^2 } dt$$
    $$
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The first formula isn't applicable when $g$ has jump discontinuities.
      $endgroup$
      – Maxim
      May 28 '18 at 21:15










    • $begingroup$
      @Maxim It is possible the discontinuities are of first order.
      $endgroup$
      – Guy Fsone
      May 30 '18 at 7:14










    • $begingroup$
      There are zero order discontinuities at the endpoints. See my answer.
      $endgroup$
      – Maxim
      May 30 '18 at 9:42










    • $begingroup$
      @Maxim what do you mean by zero order discontnuity?
      $endgroup$
      – Guy Fsone
      May 31 '18 at 1:52










    • $begingroup$
      I mean that the function itself (the derivative of zero order) is discontinuous.
      $endgroup$
      – Maxim
      May 31 '18 at 9:45
















    0












    $begingroup$


    Since $g$ is differentiable there is no trouble with sum this boils down to Riemann integral of $g'cdot f$




    Namely from here we have $$color{blue}{int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx}$$
    Therefore,
    $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)\=int_{-pi/4}^0 frac 1{cos^2 x} dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x)$$



    But $$int_{-pi/4}^0 frac 1{cos^2 x} dx =int^{pi/4}_0 (tan x)' dx =1$$



    and
    $$int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x) = int_0^{pi/4} frac{sin^4 x}{1-sin^2 x} d(sin x) =int_0^{sqrt{2}/2} frac{t^4 }{1-t^2 } dt\=int_0^{sqrt{2}/2} -t^2+frac{t^2 }{1-t^2 } dt$$
    $$
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The first formula isn't applicable when $g$ has jump discontinuities.
      $endgroup$
      – Maxim
      May 28 '18 at 21:15










    • $begingroup$
      @Maxim It is possible the discontinuities are of first order.
      $endgroup$
      – Guy Fsone
      May 30 '18 at 7:14










    • $begingroup$
      There are zero order discontinuities at the endpoints. See my answer.
      $endgroup$
      – Maxim
      May 30 '18 at 9:42










    • $begingroup$
      @Maxim what do you mean by zero order discontnuity?
      $endgroup$
      – Guy Fsone
      May 31 '18 at 1:52










    • $begingroup$
      I mean that the function itself (the derivative of zero order) is discontinuous.
      $endgroup$
      – Maxim
      May 31 '18 at 9:45














    0












    0








    0





    $begingroup$


    Since $g$ is differentiable there is no trouble with sum this boils down to Riemann integral of $g'cdot f$




    Namely from here we have $$color{blue}{int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx}$$
    Therefore,
    $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)\=int_{-pi/4}^0 frac 1{cos^2 x} dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x)$$



    But $$int_{-pi/4}^0 frac 1{cos^2 x} dx =int^{pi/4}_0 (tan x)' dx =1$$



    and
    $$int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x) = int_0^{pi/4} frac{sin^4 x}{1-sin^2 x} d(sin x) =int_0^{sqrt{2}/2} frac{t^4 }{1-t^2 } dt\=int_0^{sqrt{2}/2} -t^2+frac{t^2 }{1-t^2 } dt$$
    $$
    $$






    share|cite|improve this answer











    $endgroup$




    Since $g$ is differentiable there is no trouble with sum this boils down to Riemann integral of $g'cdot f$




    Namely from here we have $$color{blue}{int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx}$$
    Therefore,
    $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)\=int_{-pi/4}^0 frac 1{cos^2 x} dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x)$$



    But $$int_{-pi/4}^0 frac 1{cos^2 x} dx =int^{pi/4}_0 (tan x)' dx =1$$



    and
    $$int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x) = int_0^{pi/4} frac{sin^4 x}{1-sin^2 x} d(sin x) =int_0^{sqrt{2}/2} frac{t^4 }{1-t^2 } dt\=int_0^{sqrt{2}/2} -t^2+frac{t^2 }{1-t^2 } dt$$
    $$
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 2 '18 at 9:02

























    answered Dec 15 '17 at 14:27









    Guy FsoneGuy Fsone

    17.2k43074




    17.2k43074












    • $begingroup$
      The first formula isn't applicable when $g$ has jump discontinuities.
      $endgroup$
      – Maxim
      May 28 '18 at 21:15










    • $begingroup$
      @Maxim It is possible the discontinuities are of first order.
      $endgroup$
      – Guy Fsone
      May 30 '18 at 7:14










    • $begingroup$
      There are zero order discontinuities at the endpoints. See my answer.
      $endgroup$
      – Maxim
      May 30 '18 at 9:42










    • $begingroup$
      @Maxim what do you mean by zero order discontnuity?
      $endgroup$
      – Guy Fsone
      May 31 '18 at 1:52










    • $begingroup$
      I mean that the function itself (the derivative of zero order) is discontinuous.
      $endgroup$
      – Maxim
      May 31 '18 at 9:45


















    • $begingroup$
      The first formula isn't applicable when $g$ has jump discontinuities.
      $endgroup$
      – Maxim
      May 28 '18 at 21:15










    • $begingroup$
      @Maxim It is possible the discontinuities are of first order.
      $endgroup$
      – Guy Fsone
      May 30 '18 at 7:14










    • $begingroup$
      There are zero order discontinuities at the endpoints. See my answer.
      $endgroup$
      – Maxim
      May 30 '18 at 9:42










    • $begingroup$
      @Maxim what do you mean by zero order discontnuity?
      $endgroup$
      – Guy Fsone
      May 31 '18 at 1:52










    • $begingroup$
      I mean that the function itself (the derivative of zero order) is discontinuous.
      $endgroup$
      – Maxim
      May 31 '18 at 9:45
















    $begingroup$
    The first formula isn't applicable when $g$ has jump discontinuities.
    $endgroup$
    – Maxim
    May 28 '18 at 21:15




    $begingroup$
    The first formula isn't applicable when $g$ has jump discontinuities.
    $endgroup$
    – Maxim
    May 28 '18 at 21:15












    $begingroup$
    @Maxim It is possible the discontinuities are of first order.
    $endgroup$
    – Guy Fsone
    May 30 '18 at 7:14




    $begingroup$
    @Maxim It is possible the discontinuities are of first order.
    $endgroup$
    – Guy Fsone
    May 30 '18 at 7:14












    $begingroup$
    There are zero order discontinuities at the endpoints. See my answer.
    $endgroup$
    – Maxim
    May 30 '18 at 9:42




    $begingroup$
    There are zero order discontinuities at the endpoints. See my answer.
    $endgroup$
    – Maxim
    May 30 '18 at 9:42












    $begingroup$
    @Maxim what do you mean by zero order discontnuity?
    $endgroup$
    – Guy Fsone
    May 31 '18 at 1:52




    $begingroup$
    @Maxim what do you mean by zero order discontnuity?
    $endgroup$
    – Guy Fsone
    May 31 '18 at 1:52












    $begingroup$
    I mean that the function itself (the derivative of zero order) is discontinuous.
    $endgroup$
    – Maxim
    May 31 '18 at 9:45




    $begingroup$
    I mean that the function itself (the derivative of zero order) is discontinuous.
    $endgroup$
    – Maxim
    May 31 '18 at 9:45











    0












    $begingroup$

    Let $T = pi/4$. The first term in the Riemann-Stieltjes sum, $f(xi_1)(g(x_1) - g(-T))$, does not tend to zero when $x_1 downarrow -T$ and $xi_1 downarrow -T$, and similarly for the last term. The integral is
    $$I = int_{-T}^T f(x) dg(x) = \
    lim_{epsilon downarrow 0} {
    f(-T) (g(-T + epsilon)- g(-T)) + \
    f(T) (g(T)- g(T - epsilon)) } + \
    int_{-T}^T f(x) cos x ,dx = \
    2 operatorname{arctanh}(sqrt 2 - 1) + frac {19 sqrt 2} 6 - 2.$$
    It can be verified that
    $$I = f(T) g(T) - f(-T) g(-T) - int_{-T}^T g(x) df(x) = \
    f(T) g(T) - f(-T) g(-T) - \
    lim_{epsilon downarrow 0} g(0) (f(epsilon) - f(-epsilon)) - \
    int_{-T}^0 g(x) left( frac 1 {cos^3 x} right)' dx -
    int_0^T g(x) left( frac {sin^4 x} {cos^2 x} right)' dx.$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Let $T = pi/4$. The first term in the Riemann-Stieltjes sum, $f(xi_1)(g(x_1) - g(-T))$, does not tend to zero when $x_1 downarrow -T$ and $xi_1 downarrow -T$, and similarly for the last term. The integral is
      $$I = int_{-T}^T f(x) dg(x) = \
      lim_{epsilon downarrow 0} {
      f(-T) (g(-T + epsilon)- g(-T)) + \
      f(T) (g(T)- g(T - epsilon)) } + \
      int_{-T}^T f(x) cos x ,dx = \
      2 operatorname{arctanh}(sqrt 2 - 1) + frac {19 sqrt 2} 6 - 2.$$
      It can be verified that
      $$I = f(T) g(T) - f(-T) g(-T) - int_{-T}^T g(x) df(x) = \
      f(T) g(T) - f(-T) g(-T) - \
      lim_{epsilon downarrow 0} g(0) (f(epsilon) - f(-epsilon)) - \
      int_{-T}^0 g(x) left( frac 1 {cos^3 x} right)' dx -
      int_0^T g(x) left( frac {sin^4 x} {cos^2 x} right)' dx.$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $T = pi/4$. The first term in the Riemann-Stieltjes sum, $f(xi_1)(g(x_1) - g(-T))$, does not tend to zero when $x_1 downarrow -T$ and $xi_1 downarrow -T$, and similarly for the last term. The integral is
        $$I = int_{-T}^T f(x) dg(x) = \
        lim_{epsilon downarrow 0} {
        f(-T) (g(-T + epsilon)- g(-T)) + \
        f(T) (g(T)- g(T - epsilon)) } + \
        int_{-T}^T f(x) cos x ,dx = \
        2 operatorname{arctanh}(sqrt 2 - 1) + frac {19 sqrt 2} 6 - 2.$$
        It can be verified that
        $$I = f(T) g(T) - f(-T) g(-T) - int_{-T}^T g(x) df(x) = \
        f(T) g(T) - f(-T) g(-T) - \
        lim_{epsilon downarrow 0} g(0) (f(epsilon) - f(-epsilon)) - \
        int_{-T}^0 g(x) left( frac 1 {cos^3 x} right)' dx -
        int_0^T g(x) left( frac {sin^4 x} {cos^2 x} right)' dx.$$






        share|cite|improve this answer











        $endgroup$



        Let $T = pi/4$. The first term in the Riemann-Stieltjes sum, $f(xi_1)(g(x_1) - g(-T))$, does not tend to zero when $x_1 downarrow -T$ and $xi_1 downarrow -T$, and similarly for the last term. The integral is
        $$I = int_{-T}^T f(x) dg(x) = \
        lim_{epsilon downarrow 0} {
        f(-T) (g(-T + epsilon)- g(-T)) + \
        f(T) (g(T)- g(T - epsilon)) } + \
        int_{-T}^T f(x) cos x ,dx = \
        2 operatorname{arctanh}(sqrt 2 - 1) + frac {19 sqrt 2} 6 - 2.$$
        It can be verified that
        $$I = f(T) g(T) - f(-T) g(-T) - int_{-T}^T g(x) df(x) = \
        f(T) g(T) - f(-T) g(-T) - \
        lim_{epsilon downarrow 0} g(0) (f(epsilon) - f(-epsilon)) - \
        int_{-T}^0 g(x) left( frac 1 {cos^3 x} right)' dx -
        int_0^T g(x) left( frac {sin^4 x} {cos^2 x} right)' dx.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited May 28 '18 at 21:35

























        answered May 28 '18 at 20:47









        MaximMaxim

        6,1681221




        6,1681221






























            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%2f379818%2fsolving-riemann-stieltjes-integral-int-pi-4-pi-4-fxdgx%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

            Image
            Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
            Read more

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

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
            Read more

            張江高科駅

            Image
            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2
            Read more