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So, I'm having a really difficult time trying to evaluate the following integral via contour integration (please, no other methods):

$$\int_0^\infty{\frac{\log{(x^2+1)}}{1+x^2}} dx$$

Obviously, we're going to have branch cuts at z= +/- i which extend to infinity, and I'm having trouble coming up with a nice contour to evaluate this. I already tried a double keyhole contour and split up the logarithm function, but this does not recover the real function I'm trying to evaluate. Could anyone please provide some insight?

Incognito
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  • See this: http://math.stackexchange.com/questions/358386/evaluating-int-0-infty-frac-lnx21x21dx/371327#371327 – Ron Gordon Mar 28 '14 at 01:53
  • Hmmm, I can follow that answer, but I'm having trouble showing that the R and epsilon integrals go to zero. Most of the inequality arguements which I use to show that don't work very well.... – Incognito Mar 28 '14 at 02:52
  • You just have to draw the contour and then parametrize each piece and evaluate. The epsilon integral, as you call it, does not go to zero, but rather cancels with another piece. You should readily see how The integral over the large arc vanishes as $R \to \infty$. – Ron Gordon Mar 28 '14 at 07:03
  • Yeah, I'm having some trouble showing that. Usually, for contour integrals, I use a nice inequality argument to show that the integral's magnitude < or = to 0, but it doesn't work out well at all for these integrals. Also, could you please elaborate on how you can assert that −πlog(R)+πlog(2+R)=0 as R approaches infinity? Normally, you can't conclude that infinity - infinity will lead to that result. – Incognito Mar 28 '14 at 15:02
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    Well, I would argue the following: $$\log{(2+R)} = \log{R} + \log{\left(1+\frac{2}{R}\right)} \sim \log{R} + \frac{2}{R}$$ as $R \to \infty$. – Ron Gordon Mar 28 '14 at 15:07
  • Ahhhh, that definitely makes sense and is a pretty clever explanation; thank you! How do you personally see how the integral over the large arc vanishes? That's what I'm really having trouble computing – Incognito Mar 28 '14 at 17:08

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