# Evaluating $\int_{-\infty}^{\infty} \frac{\log^{2}(1+ix^{2})}{1+ix^{2}} \ dx$ using contour integration

It was suggested to me to evaluate $\displaystyle \int_{-\infty}^{\infty} \frac{\log^{2}(1+ix^{2})}{1+ix^{2}} \ dx$ by integrating $\displaystyle f(z) = \frac{\log^{2}(1+z^{2})}{1+z^{2}}$around a semicircle with it's diameter along the line $z= e^{\frac{i \pi}{4}}t$.

Of course we need to take a detour around the branch point at $z=i$.

Choosing the principal branch of $\log (1+z^{2})$, there are cuts on the imaginary axis from $i$ to $i \infty$ and from $-i$ to $-i \infty$.

So letting the radius of the semicircle go to $\infty$ and the radius of the indentation around the branch point at $z=i$ go to $0$, I get

$$e^{ \frac{i \pi }{4}} \int_{-\infty}^{\infty} \frac{\log^{2}(1+it^{2})}{1+it^{2}} \ dt + \int^{e^{\frac{i \pi}{2}}}_{i \infty} \frac{(\log |1+z^{2}| + \pi i)^{2}}{1+z^{2}} dz + \int_{e^{\frac{i \pi}{2}}}^{i \infty} \frac{(\log |1+z^{2}| - \pi i)^{2}}{1+z^{2}} dz = 0$$

$$\displaystyle \implies e^{ \frac{i \pi }{4}} \int_{-\infty}^{\infty} \frac{\log^{2}(1+ix^{2})}{1+ix^{2}} \ dx = 4 \pi i \int_{e^{\frac{i \pi}{2}}}^{i \infty} \frac{\log|1+z^{2}|}{1+z^{2}} \ dz = -4 \pi \int_{1}^{\infty} \frac{\log(1-t^{2})}{1-t^{2}} \ dt$$

But $\displaystyle \int_{1}^{\infty} \frac{\log(1-t^{2})}{1-t^{2}} \ dt$ does not converge.

EDIT:

The problem is the indentation around the branch point. It's contribution is not zero nor finite.

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could you also take the semicircle the opposite direction too? but this way shod produce something meaningful still , how did you get the last equalities? –  Evan Aug 22 '13 at 14:52
What's wrong with integrating along the real axis here? Deform a semicircular contour to avoid the branch point at $z=e^{i \pi/4}$, but that's about as complicated as it needs to be. I think. –  Ron Gordon Aug 22 '13 at 14:55
Do you know what the answer is? –  Mhenni Benghorbal Aug 22 '13 at 14:58
I'm a bit confused: sometimes you have log, and sometimes log squared. –  user8268 Aug 22 '13 at 15:54

Easier: differentiate
$$2\int_0^{\infty}(1+i x^2)^s dx=-e^{3i\pi/4}\frac{\sqrt{\pi}}{2}\frac{\Gamma(-1/2-s)}{\Gamma(-s)}$$ twice wrt $s$ and set $s=-1$.

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