Real integrals with two poles in the complex plane












2














I'm studying the Cauchy Integral Theorem / Formula, but realised I have a misunderstanding.



Consider an integral over the function $f: mathbb{R} to mathbb{C}$
$$
I = int^infty_{-infty} f(x) , dx = int^infty_{-infty} frac{e^{ix}}{x^2 + 1} , dx quad.
$$



This can be considered in the complex plane, where $f(z)$ is holomorphic except at its two poles at $z = pm i$. Choosing to consider the positive case, we therefore factorise as



$$
I = int_C frac{frac{e^{iz}}{z+i}}{z-i} , dz
= 2pi i frac{e^{-1}}{2i} =pi e^{-1}
quad ,
$$



where the contour $C = C_R + C_+$ is taken anticlockwise, $C_R$ is the real axis between $pm R$ and $C_+$ is the positive semicircle in the complex plane with $left|zright| = R$. It's clear from inspection that $C_+$ does not contribute to the path integral if we let $R to infty$, so we can equate the integral in the complex plane to the real integral $I$.



Now my question: I could equally well have chosen the negative half of the complex plane ($C'=C_R + C_-$) to evaluate my integral, negating the answer since the anticlockwise integral would otherwise calculate $int_{infty}^{-infty}$. This would give



$$
I = - int_{C'} frac{frac{e^{iz}}{z-i}}{z+i} , dz
= - 2pi i frac{e^{+1}}{-2i}
=pi e^{+1}
quad .
$$



But clearly, that's different to my previous calculation. I thought these two methods should give the same answer, so what went wrong?






complex-analysis contour-integration cauchy-integral-formula







share|cite|improve this question



























    2














    I'm studying the Cauchy Integral Theorem / Formula, but realised I have a misunderstanding.



    Consider an integral over the function $f: mathbb{R} to mathbb{C}$
    $$
    I = int^infty_{-infty} f(x) , dx = int^infty_{-infty} frac{e^{ix}}{x^2 + 1} , dx quad.
    $$



    This can be considered in the complex plane, where $f(z)$ is holomorphic except at its two poles at $z = pm i$. Choosing to consider the positive case, we therefore factorise as



    $$
    I = int_C frac{frac{e^{iz}}{z+i}}{z-i} , dz
    = 2pi i frac{e^{-1}}{2i} =pi e^{-1}
    quad ,
    $$



    where the contour $C = C_R + C_+$ is taken anticlockwise, $C_R$ is the real axis between $pm R$ and $C_+$ is the positive semicircle in the complex plane with $left|zright| = R$. It's clear from inspection that $C_+$ does not contribute to the path integral if we let $R to infty$, so we can equate the integral in the complex plane to the real integral $I$.



    Now my question: I could equally well have chosen the negative half of the complex plane ($C'=C_R + C_-$) to evaluate my integral, negating the answer since the anticlockwise integral would otherwise calculate $int_{infty}^{-infty}$. This would give



    $$
    I = - int_{C'} frac{frac{e^{iz}}{z-i}}{z+i} , dz
    = - 2pi i frac{e^{+1}}{-2i}
    =pi e^{+1}
    quad .
    $$



    But clearly, that's different to my previous calculation. I thought these two methods should give the same answer, so what went wrong?






    complex-analysis contour-integration cauchy-integral-formula







    share|cite|improve this question

























      2












      2








      2







      I'm studying the Cauchy Integral Theorem / Formula, but realised I have a misunderstanding.



      Consider an integral over the function $f: mathbb{R} to mathbb{C}$
      $$
      I = int^infty_{-infty} f(x) , dx = int^infty_{-infty} frac{e^{ix}}{x^2 + 1} , dx quad.
      $$



      This can be considered in the complex plane, where $f(z)$ is holomorphic except at its two poles at $z = pm i$. Choosing to consider the positive case, we therefore factorise as



      $$
      I = int_C frac{frac{e^{iz}}{z+i}}{z-i} , dz
      = 2pi i frac{e^{-1}}{2i} =pi e^{-1}
      quad ,
      $$



      where the contour $C = C_R + C_+$ is taken anticlockwise, $C_R$ is the real axis between $pm R$ and $C_+$ is the positive semicircle in the complex plane with $left|zright| = R$. It's clear from inspection that $C_+$ does not contribute to the path integral if we let $R to infty$, so we can equate the integral in the complex plane to the real integral $I$.



      Now my question: I could equally well have chosen the negative half of the complex plane ($C'=C_R + C_-$) to evaluate my integral, negating the answer since the anticlockwise integral would otherwise calculate $int_{infty}^{-infty}$. This would give



      $$
      I = - int_{C'} frac{frac{e^{iz}}{z-i}}{z+i} , dz
      = - 2pi i frac{e^{+1}}{-2i}
      =pi e^{+1}
      quad .
      $$



      But clearly, that's different to my previous calculation. I thought these two methods should give the same answer, so what went wrong?






      complex-analysis contour-integration cauchy-integral-formula







      share|cite|improve this question













      I'm studying the Cauchy Integral Theorem / Formula, but realised I have a misunderstanding.



      Consider an integral over the function $f: mathbb{R} to mathbb{C}$
      $$
      I = int^infty_{-infty} f(x) , dx = int^infty_{-infty} frac{e^{ix}}{x^2 + 1} , dx quad.
      $$



      This can be considered in the complex plane, where $f(z)$ is holomorphic except at its two poles at $z = pm i$. Choosing to consider the positive case, we therefore factorise as



      $$
      I = int_C frac{frac{e^{iz}}{z+i}}{z-i} , dz
      = 2pi i frac{e^{-1}}{2i} =pi e^{-1}
      quad ,
      $$



      where the contour $C = C_R + C_+$ is taken anticlockwise, $C_R$ is the real axis between $pm R$ and $C_+$ is the positive semicircle in the complex plane with $left|zright| = R$. It's clear from inspection that $C_+$ does not contribute to the path integral if we let $R to infty$, so we can equate the integral in the complex plane to the real integral $I$.



      Now my question: I could equally well have chosen the negative half of the complex plane ($C'=C_R + C_-$) to evaluate my integral, negating the answer since the anticlockwise integral would otherwise calculate $int_{infty}^{-infty}$. This would give



      $$
      I = - int_{C'} frac{frac{e^{iz}}{z-i}}{z+i} , dz
      = - 2pi i frac{e^{+1}}{-2i}
      =pi e^{+1}
      quad .
      $$



      But clearly, that's different to my previous calculation. I thought these two methods should give the same answer, so what went wrong?






      complex-analysis contour-integration cauchy-integral-formula




      complex-analysis contour-integration cauchy-integral-formula






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      asked 2 hours ago









      CharlieB

      1183




      1183






















          2 Answers
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          No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.






          share|cite|improve this answer





















          • Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
            – CharlieB
            1 hour ago



















          1














          If you choose the contour in the upper half-plane, the integral over the
          semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
          on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
          for the absolute value of the integral, when $R>1$.



          But on the lower half-plane, the integral over the semicircle does not tend to
          zero as $Rtoinfty$.






          share|cite|improve this answer





















          • In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
            – CharlieB
            1 hour ago











          Your Answer





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          2 Answers
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          2 Answers
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          active

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          3














          No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.






          share|cite|improve this answer





















          • Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
            – CharlieB
            1 hour ago
















          3














          No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.






          share|cite|improve this answer





















          • Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
            – CharlieB
            1 hour ago














          3












          3








          3






          No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.






          share|cite|improve this answer












          No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          José Carlos Santos

          149k22117219




          149k22117219












          • Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
            – CharlieB
            1 hour ago


















          • Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
            – CharlieB
            1 hour ago
















          Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
          – CharlieB
          1 hour ago




          Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
          – CharlieB
          1 hour ago











          1














          If you choose the contour in the upper half-plane, the integral over the
          semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
          on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
          for the absolute value of the integral, when $R>1$.



          But on the lower half-plane, the integral over the semicircle does not tend to
          zero as $Rtoinfty$.






          share|cite|improve this answer





















          • In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
            – CharlieB
            1 hour ago
















          1














          If you choose the contour in the upper half-plane, the integral over the
          semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
          on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
          for the absolute value of the integral, when $R>1$.



          But on the lower half-plane, the integral over the semicircle does not tend to
          zero as $Rtoinfty$.






          share|cite|improve this answer





















          • In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
            – CharlieB
            1 hour ago














          1












          1








          1






          If you choose the contour in the upper half-plane, the integral over the
          semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
          on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
          for the absolute value of the integral, when $R>1$.



          But on the lower half-plane, the integral over the semicircle does not tend to
          zero as $Rtoinfty$.






          share|cite|improve this answer












          If you choose the contour in the upper half-plane, the integral over the
          semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
          on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
          for the absolute value of the integral, when $R>1$.



          But on the lower half-plane, the integral over the semicircle does not tend to
          zero as $Rtoinfty$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Lord Shark the Unknown

          101k958131




          101k958131












          • In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
            – CharlieB
            1 hour ago


















          • In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
            – CharlieB
            1 hour ago
















          In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
          – CharlieB
          1 hour ago




          In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
          – CharlieB
          1 hour ago


















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