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












2












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Consider the following definite real integral:
$$I = int_{0}^infty dx frac{e^{-ix} - e^{ix}}{x}$$



Using the $text{Si}(x)$ function, I can solve it easily,
$$I = -2i int_{0}^infty dx frac{e^{-ix} - e^{ix}}{-2ix} = -2i int_{0}^infty dx frac{sin{x}}{x} = -2i lim_{x to infty} text{Si}(x) = -2i left(frac{pi}{2}right) = - i pi,$$
simply because I happen to know that $mathrm{Si}(x)$ 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:
$$I= int_{0}^infty dx frac{e^{-ix}}{x} - int_{0}^infty dx frac{ e^{ix}}{x} = color{red}{-int_{-infty}^0 dx frac{e^{ix}}{x}} - int_{0}^infty dx frac{ e^{ix}}{x}
= -int_{-infty}^infty dx frac{e^{ix}}{x} $$

Then I define $$I_epsilon := -int_{-infty}^infty dx frac{e^{ix}}{x-ivarepsilon}$$ 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
$$lim_{x to +iinfty} frac{e^{ix}}{x-ivarepsilon} = 0$$ and it contains the simple pole at $x_0 = i varepsilon$. Using the residue theorem with the contour enclosing $x_0$ $$I_varepsilon = -2 pi i , text{Res}_{x_0} left( frac{e^{ix}}{x-ivarepsilon}right) = -2 pi i left( frac{e^{ix}}{1} right)Biggrvert_{x=x_0=ivarepsilon}=-2 pi i , e^{-varepsilon}.$$
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?










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










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
















2












$begingroup$


Consider the following definite real integral:
$$I = int_{0}^infty dx frac{e^{-ix} - e^{ix}}{x}$$



Using the $text{Si}(x)$ function, I can solve it easily,
$$I = -2i int_{0}^infty dx frac{e^{-ix} - e^{ix}}{-2ix} = -2i int_{0}^infty dx frac{sin{x}}{x} = -2i lim_{x to infty} text{Si}(x) = -2i left(frac{pi}{2}right) = - i pi,$$
simply because I happen to know that $mathrm{Si}(x)$ 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:
$$I= int_{0}^infty dx frac{e^{-ix}}{x} - int_{0}^infty dx frac{ e^{ix}}{x} = color{red}{-int_{-infty}^0 dx frac{e^{ix}}{x}} - int_{0}^infty dx frac{ e^{ix}}{x}
= -int_{-infty}^infty dx frac{e^{ix}}{x} $$

Then I define $$I_epsilon := -int_{-infty}^infty dx frac{e^{ix}}{x-ivarepsilon}$$ 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
$$lim_{x to +iinfty} frac{e^{ix}}{x-ivarepsilon} = 0$$ and it contains the simple pole at $x_0 = i varepsilon$. Using the residue theorem with the contour enclosing $x_0$ $$I_varepsilon = -2 pi i , text{Res}_{x_0} left( frac{e^{ix}}{x-ivarepsilon}right) = -2 pi i left( frac{e^{ix}}{1} right)Biggrvert_{x=x_0=ivarepsilon}=-2 pi i , e^{-varepsilon}.$$
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?










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  • 1




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










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














2












2








2





$begingroup$


Consider the following definite real integral:
$$I = int_{0}^infty dx frac{e^{-ix} - e^{ix}}{x}$$



Using the $text{Si}(x)$ function, I can solve it easily,
$$I = -2i int_{0}^infty dx frac{e^{-ix} - e^{ix}}{-2ix} = -2i int_{0}^infty dx frac{sin{x}}{x} = -2i lim_{x to infty} text{Si}(x) = -2i left(frac{pi}{2}right) = - i pi,$$
simply because I happen to know that $mathrm{Si}(x)$ 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:
$$I= int_{0}^infty dx frac{e^{-ix}}{x} - int_{0}^infty dx frac{ e^{ix}}{x} = color{red}{-int_{-infty}^0 dx frac{e^{ix}}{x}} - int_{0}^infty dx frac{ e^{ix}}{x}
= -int_{-infty}^infty dx frac{e^{ix}}{x} $$

Then I define $$I_epsilon := -int_{-infty}^infty dx frac{e^{ix}}{x-ivarepsilon}$$ 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
$$lim_{x to +iinfty} frac{e^{ix}}{x-ivarepsilon} = 0$$ and it contains the simple pole at $x_0 = i varepsilon$. Using the residue theorem with the contour enclosing $x_0$ $$I_varepsilon = -2 pi i , text{Res}_{x_0} left( frac{e^{ix}}{x-ivarepsilon}right) = -2 pi i left( frac{e^{ix}}{1} right)Biggrvert_{x=x_0=ivarepsilon}=-2 pi i , e^{-varepsilon}.$$
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:
$$I = int_{0}^infty dx frac{e^{-ix} - e^{ix}}{x}$$



Using the $text{Si}(x)$ function, I can solve it easily,
$$I = -2i int_{0}^infty dx frac{e^{-ix} - e^{ix}}{-2ix} = -2i int_{0}^infty dx frac{sin{x}}{x} = -2i lim_{x to infty} text{Si}(x) = -2i left(frac{pi}{2}right) = - i pi,$$
simply because I happen to know that $mathrm{Si}(x)$ 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:
$$I= int_{0}^infty dx frac{e^{-ix}}{x} - int_{0}^infty dx frac{ e^{ix}}{x} = color{red}{-int_{-infty}^0 dx frac{e^{ix}}{x}} - int_{0}^infty dx frac{ e^{ix}}{x}
= -int_{-infty}^infty dx frac{e^{ix}}{x} $$

Then I define $$I_epsilon := -int_{-infty}^infty dx frac{e^{ix}}{x-ivarepsilon}$$ 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
$$lim_{x to +iinfty} frac{e^{ix}}{x-ivarepsilon} = 0$$ and it contains the simple pole at $x_0 = i varepsilon$. Using the residue theorem with the contour enclosing $x_0$ $$I_varepsilon = -2 pi i , text{Res}_{x_0} left( frac{e^{ix}}{x-ivarepsilon}right) = -2 pi i left( frac{e^{ix}}{1} right)Biggrvert_{x=x_0=ivarepsilon}=-2 pi i , e^{-varepsilon}.$$
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






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asked 5 hours ago









Ivan V.Ivan V.

811216




811216








  • 1




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










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














  • 1




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










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








1




1




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




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












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




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










3 Answers
3






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3












$begingroup$

You've replaced the converging integral $int_0^infty frac{mathrm{e}^{-mathrm{i} x} - mathrm{e}^{mathrm{i} x}}{x} ,mathrm{d}x$ with two divergent integrals, $int_0^infty frac{mathrm{e}^{-mathrm{i} x}}{x} ,mathrm{d}x$ and $int_0^infty frac{mathrm{e}^{mathrm{i} x}}{x} ,mathrm{d}x$. (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 $pm pi mathrm{i}$, not $pm 2 pi mathrm{i}$.






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  • $begingroup$
    Ah, of course! And thank you for the additional info, very useful.
    $endgroup$
    – Ivan V.
    4 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











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    0












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






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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 $int_0^infty frac{mathrm{e}^{-mathrm{i} x} - mathrm{e}^{mathrm{i} x}}{x} ,mathrm{d}x$ with two divergent integrals, $int_0^infty frac{mathrm{e}^{-mathrm{i} x}}{x} ,mathrm{d}x$ and $int_0^infty frac{mathrm{e}^{mathrm{i} x}}{x} ,mathrm{d}x$. (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 $pm pi mathrm{i}$, not $pm 2 pi mathrm{i}$.






      share|cite|improve this answer









      $endgroup$













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
















      3












      $begingroup$

      You've replaced the converging integral $int_0^infty frac{mathrm{e}^{-mathrm{i} x} - mathrm{e}^{mathrm{i} x}}{x} ,mathrm{d}x$ with two divergent integrals, $int_0^infty frac{mathrm{e}^{-mathrm{i} x}}{x} ,mathrm{d}x$ and $int_0^infty frac{mathrm{e}^{mathrm{i} x}}{x} ,mathrm{d}x$. (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 $pm pi mathrm{i}$, not $pm 2 pi mathrm{i}$.






      share|cite|improve this answer









      $endgroup$













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














      3












      3








      3





      $begingroup$

      You've replaced the converging integral $int_0^infty frac{mathrm{e}^{-mathrm{i} x} - mathrm{e}^{mathrm{i} x}}{x} ,mathrm{d}x$ with two divergent integrals, $int_0^infty frac{mathrm{e}^{-mathrm{i} x}}{x} ,mathrm{d}x$ and $int_0^infty frac{mathrm{e}^{mathrm{i} x}}{x} ,mathrm{d}x$. (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 $pm pi mathrm{i}$, not $pm 2 pi mathrm{i}$.






      share|cite|improve this answer









      $endgroup$



      You've replaced the converging integral $int_0^infty frac{mathrm{e}^{-mathrm{i} x} - mathrm{e}^{mathrm{i} x}}{x} ,mathrm{d}x$ with two divergent integrals, $int_0^infty frac{mathrm{e}^{-mathrm{i} x}}{x} ,mathrm{d}x$ and $int_0^infty frac{mathrm{e}^{mathrm{i} x}}{x} ,mathrm{d}x$. (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 $pm pi mathrm{i}$, not $pm 2 pi mathrm{i}$.







      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.
        4 hours ago


















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
















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




      $begingroup$
      Ah, of course! And thank you for the additional info, very useful.
      $endgroup$
      – Ivan V.
      4 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 4 hours ago









                  Kavi Rama MurthyKavi Rama Murthy

                  71k53170




                  71k53170






























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