Solving Complex Integral: How to Approach?

[Bill Foster]

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

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?

 

[John Creighto]

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

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.

 

[Mystic998]

I believe this does require integration along a contour. I think it goes something like this:

\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

where \gamma is the contour from -R to R along the real axis and \sigma(t) = Re^{it}, 0 \leq t \leq \pi. 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.

 

[Gib Z]

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.

 

[Rainbow Child]

Break the integral to two ones one from (-\infty,0),(0,+\infty), transform the first to (0,\infty) and combine them to get
I=\int_0^\infty\frac{\ln(x^2+a^2)}{x^2+1}
Use a semi-circle to include the residue i in order to find I=2\,\pi\ln(a+1).

 

[Bill Foster]

Somebody point me to a site that explains contour integration since I am unfamiliar with it.

Thanks.