Complex Integral

  • #1
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Main Question or Discussion Point

[tex]\int_{-\infty}^{\infty}\frac{\ln{(a+ix)}}{x^2+1}dx[/tex]

I tried with integration by parts but go nowhere. I think it may require a branch cut and integrating along a contour.

How would you approach this?
 

Answers and Replies

  • #2
[tex]\int_{-\infty}^{\infty}\frac{\ln{(a+ix)}}{x^2+1}dx[/tex]

I tried with integration by parts but go nowhere. I think it may require a branch cut and integrating along a contour.

How would you approach this?
It's been a long time since I took a course on complex variables so my memory is hazy. Anyway, I recall it being helpful to find the Maclaurin Series.
 
  • #3
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I believe this does require integration along a contour. I think it goes something like this:

[tex]\int_{-\infty}^{\infty}\frac{\ln{(a+ix)}}{x^2+1}dx = \int_{\gamma + \sigma} \frac{\ln{(a+ix)}}{x^2+1}dx + \int_{-\sigma}\frac{\ln{(a+ix)}}{x^2+1}dx[/tex]

where [itex]\gamma[/itex] is the contour from -R to R along the real axis and [itex]\sigma(t) = Re^{it}[/itex], [itex]0 \leq t \leq \pi[/itex]. Then you evaluate the first integral with Cauchy's formula and take the limit as R goes to infinity. The second integral should go to zero, and there's your answer. Of course, it has been a long while since I've done this so I could be wrong.
 
  • #4
Gib Z
Homework Helper
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It's been a long time since I took a course on complex variables so my memory is hazy. Anyway, I recall it being helpful to find the Maclaurin Series.
The series for the natural log is only equal to the function itself if |z|< 1, which it is not throughout the domain of integration.
 
  • #5
Break the integral to two ones one from [itex](-\infty,0),(0,+\infty)[/itex], transform the first to [itex](0,\infty)[/itex] and combine them to get
[tex]I=\int_0^\infty\frac{\ln(x^2+a^2)}{x^2+1}[/tex]
Use a semi-circle to include the residue [itex]i[/itex] in order to find [itex]I=2\,\pi\ln(a+1)[/itex].
 
  • #6
338
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Somebody point me to a site that explains contour integration since I am unfamiliar with it.

Thanks.
 

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