Evaluating a simple integral with the Cauchy residue theorem and a semicircular contour











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I am taking a course next week that requires some basic integral techniques from complex analysis and I've been trying to quickly teach it to myself. I was given this sample problem to test my understanding, and I'm obviously missing something.



Suppose we have the function:
$$ f(omega) = frac{e^{iomega t}}{omega - epsilon + igamma} $$
where $epsilon,gammainmathbb{R}$ and $gamma>0$. We want to integrate this function over the real line:
$$int_{-infty}^infty f(omega)domega$$
Here's how I would attempt it. To compute the integral, construct a countour $C$ consisting of a straight line from $-R$ to $R$ on the real line, and then close it a half-circle arc in the upper half of the complex plane. We want to consider $Rtoinfty$, in which case all poles in the upper half of the complex plane will be enclosed. The integrand approaches $0$ as $|omega|toinfty$, so in this limit, the semicircular arc has no contribution, and so the contour integral actually just approaches the above real integral we want to compute. The only pole of the function is at $epsilon-igamma$, which is in the lower half of the complex plane, so it's not enclosed in the contour, and so by the Cauchy-Goursat theorem we have overall:
$$int_{-infty}^infty f(omega) domega=oint_C f(z)dz=0$$
Here's the source of my confusion: suppose I instead closed the contour by a semicircular arc in the lower half of the complex plane, giving a new contour $C'$. I'd now enclose the pole of the function, so I'd instead obtain from the residue theorem that:
$$int_{-infty}^infty f(omega) domega=oint_{C'} f(z)dz=2pi itext{Res}(f,epsilon-igamma)$$
What have I misunderstood to get this discrepancy?










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  • What is $;t;$ ? Should it even be there?
    – DonAntonio
    Nov 17 at 16:57















up vote
2
down vote

favorite












I am taking a course next week that requires some basic integral techniques from complex analysis and I've been trying to quickly teach it to myself. I was given this sample problem to test my understanding, and I'm obviously missing something.



Suppose we have the function:
$$ f(omega) = frac{e^{iomega t}}{omega - epsilon + igamma} $$
where $epsilon,gammainmathbb{R}$ and $gamma>0$. We want to integrate this function over the real line:
$$int_{-infty}^infty f(omega)domega$$
Here's how I would attempt it. To compute the integral, construct a countour $C$ consisting of a straight line from $-R$ to $R$ on the real line, and then close it a half-circle arc in the upper half of the complex plane. We want to consider $Rtoinfty$, in which case all poles in the upper half of the complex plane will be enclosed. The integrand approaches $0$ as $|omega|toinfty$, so in this limit, the semicircular arc has no contribution, and so the contour integral actually just approaches the above real integral we want to compute. The only pole of the function is at $epsilon-igamma$, which is in the lower half of the complex plane, so it's not enclosed in the contour, and so by the Cauchy-Goursat theorem we have overall:
$$int_{-infty}^infty f(omega) domega=oint_C f(z)dz=0$$
Here's the source of my confusion: suppose I instead closed the contour by a semicircular arc in the lower half of the complex plane, giving a new contour $C'$. I'd now enclose the pole of the function, so I'd instead obtain from the residue theorem that:
$$int_{-infty}^infty f(omega) domega=oint_{C'} f(z)dz=2pi itext{Res}(f,epsilon-igamma)$$
What have I misunderstood to get this discrepancy?










share|cite|improve this question
























  • What is $;t;$ ? Should it even be there?
    – DonAntonio
    Nov 17 at 16:57













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I am taking a course next week that requires some basic integral techniques from complex analysis and I've been trying to quickly teach it to myself. I was given this sample problem to test my understanding, and I'm obviously missing something.



Suppose we have the function:
$$ f(omega) = frac{e^{iomega t}}{omega - epsilon + igamma} $$
where $epsilon,gammainmathbb{R}$ and $gamma>0$. We want to integrate this function over the real line:
$$int_{-infty}^infty f(omega)domega$$
Here's how I would attempt it. To compute the integral, construct a countour $C$ consisting of a straight line from $-R$ to $R$ on the real line, and then close it a half-circle arc in the upper half of the complex plane. We want to consider $Rtoinfty$, in which case all poles in the upper half of the complex plane will be enclosed. The integrand approaches $0$ as $|omega|toinfty$, so in this limit, the semicircular arc has no contribution, and so the contour integral actually just approaches the above real integral we want to compute. The only pole of the function is at $epsilon-igamma$, which is in the lower half of the complex plane, so it's not enclosed in the contour, and so by the Cauchy-Goursat theorem we have overall:
$$int_{-infty}^infty f(omega) domega=oint_C f(z)dz=0$$
Here's the source of my confusion: suppose I instead closed the contour by a semicircular arc in the lower half of the complex plane, giving a new contour $C'$. I'd now enclose the pole of the function, so I'd instead obtain from the residue theorem that:
$$int_{-infty}^infty f(omega) domega=oint_{C'} f(z)dz=2pi itext{Res}(f,epsilon-igamma)$$
What have I misunderstood to get this discrepancy?










share|cite|improve this question















I am taking a course next week that requires some basic integral techniques from complex analysis and I've been trying to quickly teach it to myself. I was given this sample problem to test my understanding, and I'm obviously missing something.



Suppose we have the function:
$$ f(omega) = frac{e^{iomega t}}{omega - epsilon + igamma} $$
where $epsilon,gammainmathbb{R}$ and $gamma>0$. We want to integrate this function over the real line:
$$int_{-infty}^infty f(omega)domega$$
Here's how I would attempt it. To compute the integral, construct a countour $C$ consisting of a straight line from $-R$ to $R$ on the real line, and then close it a half-circle arc in the upper half of the complex plane. We want to consider $Rtoinfty$, in which case all poles in the upper half of the complex plane will be enclosed. The integrand approaches $0$ as $|omega|toinfty$, so in this limit, the semicircular arc has no contribution, and so the contour integral actually just approaches the above real integral we want to compute. The only pole of the function is at $epsilon-igamma$, which is in the lower half of the complex plane, so it's not enclosed in the contour, and so by the Cauchy-Goursat theorem we have overall:
$$int_{-infty}^infty f(omega) domega=oint_C f(z)dz=0$$
Here's the source of my confusion: suppose I instead closed the contour by a semicircular arc in the lower half of the complex plane, giving a new contour $C'$. I'd now enclose the pole of the function, so I'd instead obtain from the residue theorem that:
$$int_{-infty}^infty f(omega) domega=oint_{C'} f(z)dz=2pi itext{Res}(f,epsilon-igamma)$$
What have I misunderstood to get this discrepancy?







complex-analysis definite-integrals contour-integration






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edited Nov 17 at 15:32

























asked Nov 17 at 15:15









user502382

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  • What is $;t;$ ? Should it even be there?
    – DonAntonio
    Nov 17 at 16:57


















  • What is $;t;$ ? Should it even be there?
    – DonAntonio
    Nov 17 at 16:57
















What is $;t;$ ? Should it even be there?
– DonAntonio
Nov 17 at 16:57




What is $;t;$ ? Should it even be there?
– DonAntonio
Nov 17 at 16:57










2 Answers
2






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1
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accepted










You should know that, in the upper half plane, the integral vanishes over the circular arc as $R to infty$ when $t gt 0$. And analogously, the integral vanishes over the arc in the lower half plane when $t lt 0$. Thus, your function will be of the form $g(t) theta(-t)$, where $g$ is the result after taking the residue and $theta$ is the Heaviside step function.






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  • Ah, thank you! I had made an elementary error regarding the behaviour of the integrand.
    – user502382
    Nov 17 at 20:04


















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0
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Let $; C_R:=left{;z=Re^{itau};|;tauin [0,pi];right};$ be the upper canonical semicircle of radius $;RinBbb R^+;$ , and let $;-C_R;$ be corresponding lower canonical semicircle. Now



$$left|frac{e^{iwt}}{omega-(epsilon-igamma)}right|=frac{e^{-t Rsintau}}{|omega-epsilon+igamma|}gefrac{e^{-tRsintau}}{R+|epsilon+igamma|}$$



if $;t>0;$ and $;tauin [pi,2pi];$ then the above is unbounded on the lower circle, and the other way around if $;t<0;$ . Thus, you cannot deduce the complex integral, in the limit, is the improper real one.






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    2 Answers
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    2 Answers
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    up vote
    1
    down vote



    accepted










    You should know that, in the upper half plane, the integral vanishes over the circular arc as $R to infty$ when $t gt 0$. And analogously, the integral vanishes over the arc in the lower half plane when $t lt 0$. Thus, your function will be of the form $g(t) theta(-t)$, where $g$ is the result after taking the residue and $theta$ is the Heaviside step function.






    share|cite|improve this answer























    • Ah, thank you! I had made an elementary error regarding the behaviour of the integrand.
      – user502382
      Nov 17 at 20:04















    up vote
    1
    down vote



    accepted










    You should know that, in the upper half plane, the integral vanishes over the circular arc as $R to infty$ when $t gt 0$. And analogously, the integral vanishes over the arc in the lower half plane when $t lt 0$. Thus, your function will be of the form $g(t) theta(-t)$, where $g$ is the result after taking the residue and $theta$ is the Heaviside step function.






    share|cite|improve this answer























    • Ah, thank you! I had made an elementary error regarding the behaviour of the integrand.
      – user502382
      Nov 17 at 20:04













    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    You should know that, in the upper half plane, the integral vanishes over the circular arc as $R to infty$ when $t gt 0$. And analogously, the integral vanishes over the arc in the lower half plane when $t lt 0$. Thus, your function will be of the form $g(t) theta(-t)$, where $g$ is the result after taking the residue and $theta$ is the Heaviside step function.






    share|cite|improve this answer














    You should know that, in the upper half plane, the integral vanishes over the circular arc as $R to infty$ when $t gt 0$. And analogously, the integral vanishes over the arc in the lower half plane when $t lt 0$. Thus, your function will be of the form $g(t) theta(-t)$, where $g$ is the result after taking the residue and $theta$ is the Heaviside step function.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 17 at 16:58

























    answered Nov 17 at 16:51









    Ron Gordon

    122k14152261




    122k14152261












    • Ah, thank you! I had made an elementary error regarding the behaviour of the integrand.
      – user502382
      Nov 17 at 20:04


















    • Ah, thank you! I had made an elementary error regarding the behaviour of the integrand.
      – user502382
      Nov 17 at 20:04
















    Ah, thank you! I had made an elementary error regarding the behaviour of the integrand.
    – user502382
    Nov 17 at 20:04




    Ah, thank you! I had made an elementary error regarding the behaviour of the integrand.
    – user502382
    Nov 17 at 20:04










    up vote
    0
    down vote













    Let $; C_R:=left{;z=Re^{itau};|;tauin [0,pi];right};$ be the upper canonical semicircle of radius $;RinBbb R^+;$ , and let $;-C_R;$ be corresponding lower canonical semicircle. Now



    $$left|frac{e^{iwt}}{omega-(epsilon-igamma)}right|=frac{e^{-t Rsintau}}{|omega-epsilon+igamma|}gefrac{e^{-tRsintau}}{R+|epsilon+igamma|}$$



    if $;t>0;$ and $;tauin [pi,2pi];$ then the above is unbounded on the lower circle, and the other way around if $;t<0;$ . Thus, you cannot deduce the complex integral, in the limit, is the improper real one.






    share|cite|improve this answer

























      up vote
      0
      down vote













      Let $; C_R:=left{;z=Re^{itau};|;tauin [0,pi];right};$ be the upper canonical semicircle of radius $;RinBbb R^+;$ , and let $;-C_R;$ be corresponding lower canonical semicircle. Now



      $$left|frac{e^{iwt}}{omega-(epsilon-igamma)}right|=frac{e^{-t Rsintau}}{|omega-epsilon+igamma|}gefrac{e^{-tRsintau}}{R+|epsilon+igamma|}$$



      if $;t>0;$ and $;tauin [pi,2pi];$ then the above is unbounded on the lower circle, and the other way around if $;t<0;$ . Thus, you cannot deduce the complex integral, in the limit, is the improper real one.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Let $; C_R:=left{;z=Re^{itau};|;tauin [0,pi];right};$ be the upper canonical semicircle of radius $;RinBbb R^+;$ , and let $;-C_R;$ be corresponding lower canonical semicircle. Now



        $$left|frac{e^{iwt}}{omega-(epsilon-igamma)}right|=frac{e^{-t Rsintau}}{|omega-epsilon+igamma|}gefrac{e^{-tRsintau}}{R+|epsilon+igamma|}$$



        if $;t>0;$ and $;tauin [pi,2pi];$ then the above is unbounded on the lower circle, and the other way around if $;t<0;$ . Thus, you cannot deduce the complex integral, in the limit, is the improper real one.






        share|cite|improve this answer












        Let $; C_R:=left{;z=Re^{itau};|;tauin [0,pi];right};$ be the upper canonical semicircle of radius $;RinBbb R^+;$ , and let $;-C_R;$ be corresponding lower canonical semicircle. Now



        $$left|frac{e^{iwt}}{omega-(epsilon-igamma)}right|=frac{e^{-t Rsintau}}{|omega-epsilon+igamma|}gefrac{e^{-tRsintau}}{R+|epsilon+igamma|}$$



        if $;t>0;$ and $;tauin [pi,2pi];$ then the above is unbounded on the lower circle, and the other way around if $;t<0;$ . Thus, you cannot deduce the complex integral, in the limit, is the improper real one.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 17 at 17:02









        DonAntonio

        176k1491224




        176k1491224






























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