Complex analysis

Dec 2009
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I have trouble showing \(\displaystyle \int_0^{\infty}\frac{(\log(x))^2}{1+x^2}dx =\frac{\pi^3}{8}\)

We have first order singularities \(\displaystyle P=\pm i\). And I found \(\displaystyle 2\pi i\sum_{i}Res(f,i) =2\pi^3\)

And I want to integrate over the following contour:
http://upload.wikimedia.org/wikipedia/commons/0/0d/Keyhole_contour.png

However I need some help, how to find the desired integral from the key-hole contour, using some properties of log(z). Some guiding hints would be greatly appreciated, as allways.
 

Bruno J.

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I think that to evaluate this integral you will have to integrate another function along the keyhole contour. Here's why I think this :

The integral along the big circle will go to 0 as the radius increases to infinity. Same for the integral along the small circle. That's no problem.

Just under the positive real axis, the function \(\displaystyle \log z\) is \(\displaystyle 2\pi i\) more than it was just above the positive real axis; i.e. \(\displaystyle \log e^{2\pi i}x = 2\pi i + \log x\). So, in the limit, the integral along the contour is equal to \(\displaystyle \int_0^\infty \frac{(\log x)^2}{1+x^2}dx + \int_\infty^0 \frac{(2\pi i+\log x)^2}{1+x^2}dx\), where here \(\displaystyle \log x\) is the usual (real-valued) logarithm. But the \(\displaystyle (\log x)^2\) terms cancel here! So you actually end up evaluating \(\displaystyle \int_0^\infty \frac{\log x}{1+x^2}dx\).

You might first have to evaluate \(\displaystyle \int_0^\infty \frac{\log x}{1+x^2}dx\), and then integrate \(\displaystyle \int_0^\infty \frac{(\log x)^3}{1+x^2}dx\) over the keyhole contour using the residue theorem so the \(\displaystyle (\log x)^3\) terms cancel, and then use the known integral \(\displaystyle \int_0^\infty \frac{\log x}{1+x^2}dx\) to solve for \(\displaystyle \int_0^\infty \frac{(\log x)^2}{1+x^2}dx\).
 
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May 2010
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Bruno J is right. However, there's an easier way. Take your branch cut to be along the negative imaginary axis and your contour to be semicircular in the upper half plane. That way you don't get your integral cancelling out.
 

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Bruno J.

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Good observation! I didn't think of using a different contour; I just figured this one was the suggested one...

Perhaps the the process I described is useful for other integrals though.
 
May 2010
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Bruno J, your process is definitely useful; that's how \(\displaystyle \int_0^\infty\frac{\log x}{(x+a)^2}dx\) is computed, for example.
 

Bruno J.

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Bruno J, your process is definitely useful; that's how \(\displaystyle \int_0^\infty\frac{\log x}{(x+a)^2}dx\) is computed.
Yeah, silly me! I even said it myself. (Rofl)
 
Dec 2009
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Thank you both for your kind help.

However, the other contour suggested doesn't contain both singularities. I suspect this may not be necessary.

So I guess this contour integral becomes: \(\displaystyle 2 \pi i *Res(f,i) = \frac{-\pi ^3}{4}\)?

Somehow I was stuck on the idea I needed to contain both sinularities.
But why do we need to leave out a hole around z=0 when it's not even a singularity?

Anyway, I think I'll stick to Bruno's Method, since I'm not sure how to manipulate the line-integrals of the semi-circular contour. I did however try that method before, but evaluating \(\displaystyle \log(x)^3\) along the keyhole contour was somewhat bothersome.

The other suggested contour gives me: \(\displaystyle -\frac{\pi^3}{4}= \int_{-\infty}^0 \frac{(\log(x)+i\pi)^2}{1+x^2}+\int_0^{\infty}\frac{\log(x)}{1+x^2} \)

But from here I'm not sure how I may manipulate these integrals.
 
May 2010
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Los Angeles, California
Your residue calculation is correct. You should study the notion of a branch cut: \(\displaystyle z=0\) is a problem for \(\displaystyle \log z\).

The method I suggest is the recommneded one; Bruno's will work but not with some sweat as you're finding out.

Continue with mine, you're on the right track.

\(\displaystyle \int_{-\infty}^0\frac{(\log z)^2}{1+z^2}\,dz = \int_0^\infty\frac{(\log x +i\pi)^2}{1+x^2}\, dx\)

Expand the numerator and use (or prove) the fact that \(\displaystyle \int_0^\infty\frac{\log x}{1+x^2}\, dx = 0.\)
 
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Dec 2009
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I made a typo:

\(\displaystyle -\frac{\pi^3}{4}= \int_{-\infty}^0 \frac{(\log(x)+i\pi)^2}{1+x^2}+\int_0^{\infty}\frac{\log(x)^2}{1+x^2} \)

\(\displaystyle \int_0^{\infty}\frac{\log(x)^2}{1+x^2}=\frac{\pi^3}{8}\) is what we're trying to prove here. Or am I misunderstanding you?
 
May 2010
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I think you have the limits wrong in your first integral: it should be from 0 to \(\displaystyle \infty\) like I have. I used the fact that \(\displaystyle \log (-x)=\log x +i\pi\) for \(\displaystyle x>0\). You will get a term 2 times the original integral coming up.
 
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