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Real integral using residue theorem - why doesn't this work?



The Next CEO of Stack OverflowMistake with using residue theory for calculating $int_-infty^inftyfracsin(x)xdx$Evaluating Integral with Residue TheoremReal Pole Residue theoremSolving this complicated integral using the Residue TheoremIntegrating secans over the imaginary axis using the residue theoremWhy doesn't this residue method work for calculating $sum_k=1^k=infty fraccos(k x)k^2$Compute integral using residue theoremEvaluating a real definite integral using residue theoremCalculating this integral using Residue TheoremCalculating integrals using the residue theoremsolving integral with real exponent and real pole with residue theorem










2












$begingroup$


Consider the following definite real integral:
$$I = int_0^infty dx frace^-ix - e^ixx$$



Using the $textSi(x)$ function, I can solve it easily,
$$I = -2i int_0^infty dx frace^-ix - e^ix-2ix = -2i int_0^infty dx fracsinxx = -2i lim_x to infty textSi(x) = -2i left(fracpi2right) = - i pi,$$
simply because I happen to know that $mathrmSi(x)$ asymptotically approaches $pi/2$.



However, if I try to calculate it using the residue theorem, I get the wrong answer, off by a factor of $2$ and I'm not sure if I understand why. Here's the procedure:
$$I= int_0^infty dx frace^-ixx - int_0^infty dx frac e^ixx = colorred-int_-infty^0 dx frace^ixx - int_0^infty dx frac e^ixx
= -int_-infty^infty dx frace^ixx $$

Then I define $$I_epsilon := -int_-infty^infty dx frace^ixx-ivarepsilon$$ for $varepsilon > 0$ so that$$I=lim_varepsilon to 0^+ I_varepsilon.$$
Then I complexify the integration variable and integrate over a D-shaped contour over the upper half of the complex plane. I choose that contour because
$$lim_x to +iinfty frace^ixx-ivarepsilon = 0$$ and it contains the simple pole at $x_0 = i varepsilon$. Using the residue theorem with the contour enclosing $x_0$ $$I_varepsilon = -2 pi i , textRes_x_0 left( frace^ixx-ivarepsilonright) = -2 pi i left( frace^ix1 right)Biggrvert_x=x_0=ivarepsilon=-2 pi i , e^-varepsilon.$$
Therefore,
$$I=lim_varepsilon to 0^+ left( -2 pi i , e^-varepsilon right) = -2pi i.$$



However, that is obviously wrong. Where exactly is the mistake?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    math.stackexchange.com/a/2270510/155436
    $endgroup$
    – Count Iblis
    5 hours ago










  • $begingroup$
    @CountIblis Didn't catch that one before, thank you!
    $endgroup$
    – Ivan V.
    5 hours ago















2












$begingroup$


Consider the following definite real integral:
$$I = int_0^infty dx frace^-ix - e^ixx$$



Using the $textSi(x)$ function, I can solve it easily,
$$I = -2i int_0^infty dx frace^-ix - e^ix-2ix = -2i int_0^infty dx fracsinxx = -2i lim_x to infty textSi(x) = -2i left(fracpi2right) = - i pi,$$
simply because I happen to know that $mathrmSi(x)$ asymptotically approaches $pi/2$.



However, if I try to calculate it using the residue theorem, I get the wrong answer, off by a factor of $2$ and I'm not sure if I understand why. Here's the procedure:
$$I= int_0^infty dx frace^-ixx - int_0^infty dx frac e^ixx = colorred-int_-infty^0 dx frace^ixx - int_0^infty dx frac e^ixx
= -int_-infty^infty dx frace^ixx $$

Then I define $$I_epsilon := -int_-infty^infty dx frace^ixx-ivarepsilon$$ for $varepsilon > 0$ so that$$I=lim_varepsilon to 0^+ I_varepsilon.$$
Then I complexify the integration variable and integrate over a D-shaped contour over the upper half of the complex plane. I choose that contour because
$$lim_x to +iinfty frace^ixx-ivarepsilon = 0$$ and it contains the simple pole at $x_0 = i varepsilon$. Using the residue theorem with the contour enclosing $x_0$ $$I_varepsilon = -2 pi i , textRes_x_0 left( frace^ixx-ivarepsilonright) = -2 pi i left( frace^ix1 right)Biggrvert_x=x_0=ivarepsilon=-2 pi i , e^-varepsilon.$$
Therefore,
$$I=lim_varepsilon to 0^+ left( -2 pi i , e^-varepsilon right) = -2pi i.$$



However, that is obviously wrong. Where exactly is the mistake?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    math.stackexchange.com/a/2270510/155436
    $endgroup$
    – Count Iblis
    5 hours ago










  • $begingroup$
    @CountIblis Didn't catch that one before, thank you!
    $endgroup$
    – Ivan V.
    5 hours ago













2












2








2





$begingroup$


Consider the following definite real integral:
$$I = int_0^infty dx frace^-ix - e^ixx$$



Using the $textSi(x)$ function, I can solve it easily,
$$I = -2i int_0^infty dx frace^-ix - e^ix-2ix = -2i int_0^infty dx fracsinxx = -2i lim_x to infty textSi(x) = -2i left(fracpi2right) = - i pi,$$
simply because I happen to know that $mathrmSi(x)$ asymptotically approaches $pi/2$.



However, if I try to calculate it using the residue theorem, I get the wrong answer, off by a factor of $2$ and I'm not sure if I understand why. Here's the procedure:
$$I= int_0^infty dx frace^-ixx - int_0^infty dx frac e^ixx = colorred-int_-infty^0 dx frace^ixx - int_0^infty dx frac e^ixx
= -int_-infty^infty dx frace^ixx $$

Then I define $$I_epsilon := -int_-infty^infty dx frace^ixx-ivarepsilon$$ for $varepsilon > 0$ so that$$I=lim_varepsilon to 0^+ I_varepsilon.$$
Then I complexify the integration variable and integrate over a D-shaped contour over the upper half of the complex plane. I choose that contour because
$$lim_x to +iinfty frace^ixx-ivarepsilon = 0$$ and it contains the simple pole at $x_0 = i varepsilon$. Using the residue theorem with the contour enclosing $x_0$ $$I_varepsilon = -2 pi i , textRes_x_0 left( frace^ixx-ivarepsilonright) = -2 pi i left( frace^ix1 right)Biggrvert_x=x_0=ivarepsilon=-2 pi i , e^-varepsilon.$$
Therefore,
$$I=lim_varepsilon to 0^+ left( -2 pi i , e^-varepsilon right) = -2pi i.$$



However, that is obviously wrong. Where exactly is the mistake?










share|cite|improve this question









$endgroup$




Consider the following definite real integral:
$$I = int_0^infty dx frace^-ix - e^ixx$$



Using the $textSi(x)$ function, I can solve it easily,
$$I = -2i int_0^infty dx frace^-ix - e^ix-2ix = -2i int_0^infty dx fracsinxx = -2i lim_x to infty textSi(x) = -2i left(fracpi2right) = - i pi,$$
simply because I happen to know that $mathrmSi(x)$ asymptotically approaches $pi/2$.



However, if I try to calculate it using the residue theorem, I get the wrong answer, off by a factor of $2$ and I'm not sure if I understand why. Here's the procedure:
$$I= int_0^infty dx frace^-ixx - int_0^infty dx frac e^ixx = colorred-int_-infty^0 dx frace^ixx - int_0^infty dx frac e^ixx
= -int_-infty^infty dx frace^ixx $$

Then I define $$I_epsilon := -int_-infty^infty dx frace^ixx-ivarepsilon$$ for $varepsilon > 0$ so that$$I=lim_varepsilon to 0^+ I_varepsilon.$$
Then I complexify the integration variable and integrate over a D-shaped contour over the upper half of the complex plane. I choose that contour because
$$lim_x to +iinfty frace^ixx-ivarepsilon = 0$$ and it contains the simple pole at $x_0 = i varepsilon$. Using the residue theorem with the contour enclosing $x_0$ $$I_varepsilon = -2 pi i , textRes_x_0 left( frace^ixx-ivarepsilonright) = -2 pi i left( frace^ix1 right)Biggrvert_x=x_0=ivarepsilon=-2 pi i , e^-varepsilon.$$
Therefore,
$$I=lim_varepsilon to 0^+ left( -2 pi i , e^-varepsilon right) = -2pi i.$$



However, that is obviously wrong. Where exactly is the mistake?







integration residue-calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









Ivan V.Ivan V.

811216




811216







  • 1




    $begingroup$
    math.stackexchange.com/a/2270510/155436
    $endgroup$
    – Count Iblis
    5 hours ago










  • $begingroup$
    @CountIblis Didn't catch that one before, thank you!
    $endgroup$
    – Ivan V.
    5 hours ago












  • 1




    $begingroup$
    math.stackexchange.com/a/2270510/155436
    $endgroup$
    – Count Iblis
    5 hours ago










  • $begingroup$
    @CountIblis Didn't catch that one before, thank you!
    $endgroup$
    – Ivan V.
    5 hours ago







1




1




$begingroup$
math.stackexchange.com/a/2270510/155436
$endgroup$
– Count Iblis
5 hours ago




$begingroup$
math.stackexchange.com/a/2270510/155436
$endgroup$
– Count Iblis
5 hours ago












$begingroup$
@CountIblis Didn't catch that one before, thank you!
$endgroup$
– Ivan V.
5 hours ago




$begingroup$
@CountIblis Didn't catch that one before, thank you!
$endgroup$
– Ivan V.
5 hours ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

You've replaced the converging integral $int_0^infty fracmathrme^-mathrmi x - mathrme^mathrmi xx ,mathrmdx$ with two divergent integrals, $int_0^infty fracmathrme^-mathrmi xx ,mathrmdx$ and $int_0^infty fracmathrme^mathrmi xx ,mathrmdx$. (That something divergent has been introduced is evident in your need to sneak up on a singularity at $0$ that was not in the original integral.)



Also, notice that your D-shaped contour does not go around your freshly minted singularity at $x = 0$. The singularity lands on your contour. See the Sokhotski–Plemelj theorem to find that the multiplier for the residue of the pole is $pm pi mathrmi$, not $pm 2 pi mathrmi$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Ah, of course! And thank you for the additional info, very useful.
    $endgroup$
    – Ivan V.
    5 hours ago


















2












$begingroup$

You cannot shift the pole from the integration contour at will. Imagine that you shift it in the lower complex half-plane. Then instead of $-2pi i$ you would obtain for the integral the value $0$!



The correct way to handle the pole is to take the half of its residue value, which is equivalent to bypassing the pole along a tiny semicircle around it (observe that the result does not depend on the choice between upper and lower semicircle).






share|cite|improve this answer











$endgroup$




















    0












    $begingroup$

    There is a problem at the very first step. You cannot split the integral because both integrals are divergent.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      You've replaced the converging integral $int_0^infty fracmathrme^-mathrmi x - mathrme^mathrmi xx ,mathrmdx$ with two divergent integrals, $int_0^infty fracmathrme^-mathrmi xx ,mathrmdx$ and $int_0^infty fracmathrme^mathrmi xx ,mathrmdx$. (That something divergent has been introduced is evident in your need to sneak up on a singularity at $0$ that was not in the original integral.)



      Also, notice that your D-shaped contour does not go around your freshly minted singularity at $x = 0$. The singularity lands on your contour. See the Sokhotski–Plemelj theorem to find that the multiplier for the residue of the pole is $pm pi mathrmi$, not $pm 2 pi mathrmi$.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        Ah, of course! And thank you for the additional info, very useful.
        $endgroup$
        – Ivan V.
        5 hours ago















      3












      $begingroup$

      You've replaced the converging integral $int_0^infty fracmathrme^-mathrmi x - mathrme^mathrmi xx ,mathrmdx$ with two divergent integrals, $int_0^infty fracmathrme^-mathrmi xx ,mathrmdx$ and $int_0^infty fracmathrme^mathrmi xx ,mathrmdx$. (That something divergent has been introduced is evident in your need to sneak up on a singularity at $0$ that was not in the original integral.)



      Also, notice that your D-shaped contour does not go around your freshly minted singularity at $x = 0$. The singularity lands on your contour. See the Sokhotski–Plemelj theorem to find that the multiplier for the residue of the pole is $pm pi mathrmi$, not $pm 2 pi mathrmi$.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        Ah, of course! And thank you for the additional info, very useful.
        $endgroup$
        – Ivan V.
        5 hours ago













      3












      3








      3





      $begingroup$

      You've replaced the converging integral $int_0^infty fracmathrme^-mathrmi x - mathrme^mathrmi xx ,mathrmdx$ with two divergent integrals, $int_0^infty fracmathrme^-mathrmi xx ,mathrmdx$ and $int_0^infty fracmathrme^mathrmi xx ,mathrmdx$. (That something divergent has been introduced is evident in your need to sneak up on a singularity at $0$ that was not in the original integral.)



      Also, notice that your D-shaped contour does not go around your freshly minted singularity at $x = 0$. The singularity lands on your contour. See the Sokhotski–Plemelj theorem to find that the multiplier for the residue of the pole is $pm pi mathrmi$, not $pm 2 pi mathrmi$.






      share|cite|improve this answer









      $endgroup$



      You've replaced the converging integral $int_0^infty fracmathrme^-mathrmi x - mathrme^mathrmi xx ,mathrmdx$ with two divergent integrals, $int_0^infty fracmathrme^-mathrmi xx ,mathrmdx$ and $int_0^infty fracmathrme^mathrmi xx ,mathrmdx$. (That something divergent has been introduced is evident in your need to sneak up on a singularity at $0$ that was not in the original integral.)



      Also, notice that your D-shaped contour does not go around your freshly minted singularity at $x = 0$. The singularity lands on your contour. See the Sokhotski–Plemelj theorem to find that the multiplier for the residue of the pole is $pm pi mathrmi$, not $pm 2 pi mathrmi$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 5 hours ago









      Eric TowersEric Towers

      33.3k22370




      33.3k22370











      • $begingroup$
        Ah, of course! And thank you for the additional info, very useful.
        $endgroup$
        – Ivan V.
        5 hours ago
















      • $begingroup$
        Ah, of course! And thank you for the additional info, very useful.
        $endgroup$
        – Ivan V.
        5 hours ago















      $begingroup$
      Ah, of course! And thank you for the additional info, very useful.
      $endgroup$
      – Ivan V.
      5 hours ago




      $begingroup$
      Ah, of course! And thank you for the additional info, very useful.
      $endgroup$
      – Ivan V.
      5 hours ago











      2












      $begingroup$

      You cannot shift the pole from the integration contour at will. Imagine that you shift it in the lower complex half-plane. Then instead of $-2pi i$ you would obtain for the integral the value $0$!



      The correct way to handle the pole is to take the half of its residue value, which is equivalent to bypassing the pole along a tiny semicircle around it (observe that the result does not depend on the choice between upper and lower semicircle).






      share|cite|improve this answer











      $endgroup$

















        2












        $begingroup$

        You cannot shift the pole from the integration contour at will. Imagine that you shift it in the lower complex half-plane. Then instead of $-2pi i$ you would obtain for the integral the value $0$!



        The correct way to handle the pole is to take the half of its residue value, which is equivalent to bypassing the pole along a tiny semicircle around it (observe that the result does not depend on the choice between upper and lower semicircle).






        share|cite|improve this answer











        $endgroup$















          2












          2








          2





          $begingroup$

          You cannot shift the pole from the integration contour at will. Imagine that you shift it in the lower complex half-plane. Then instead of $-2pi i$ you would obtain for the integral the value $0$!



          The correct way to handle the pole is to take the half of its residue value, which is equivalent to bypassing the pole along a tiny semicircle around it (observe that the result does not depend on the choice between upper and lower semicircle).






          share|cite|improve this answer











          $endgroup$



          You cannot shift the pole from the integration contour at will. Imagine that you shift it in the lower complex half-plane. Then instead of $-2pi i$ you would obtain for the integral the value $0$!



          The correct way to handle the pole is to take the half of its residue value, which is equivalent to bypassing the pole along a tiny semicircle around it (observe that the result does not depend on the choice between upper and lower semicircle).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 4 hours ago

























          answered 5 hours ago









          useruser

          6,09811031




          6,09811031





















              0












              $begingroup$

              There is a problem at the very first step. You cannot split the integral because both integrals are divergent.






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                There is a problem at the very first step. You cannot split the integral because both integrals are divergent.






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  There is a problem at the very first step. You cannot split the integral because both integrals are divergent.






                  share|cite|improve this answer









                  $endgroup$



                  There is a problem at the very first step. You cannot split the integral because both integrals are divergent.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 5 hours ago









                  Kavi Rama MurthyKavi Rama Murthy

                  71k53170




                  71k53170



























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