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As you helped me so well last time, I might as well ask a final question! Today I'm trying to prove this:

$$ \int_0^\infty \frac{x^{p}}{ 1+x^{2}}dx = \frac{\pi}{2}\cos\left(p\frac{\pi}{2}\right) $$

For $-1 < p < 1$.

I have no idea how to handle the varying $p$. I've been able to prove the relationship in the case $p = 0$. So now I could try showing this for $-1 < p < 0$ and $0 < p < 1$ but both of those seem to be tricky.

Any tips on dealing with the non-constant $p$? There are poles at $x = i$ and $x = -i$, and if $p < 0$ also at $x = 0$.

I think the best approach would be a semicircle in the top half, maybe with a small inner radius as well in the case $p<0$? Or should I not be trying to prove the relationship for these separate parts but is there a general way to do it?

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    You have a branch point at $z=0$, so you need to use the "keyhole" contour. The condition $-1 is required for the convergence of the integral.2012-08-10
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    Take a look at this: http://math.stackexchange.com/questions/121976/how-to-integrate-int-0-infty-fracx1-3dx1x22012-08-10
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    Your statement is false because for $p=1$ the integral diverges but your LHS gives 0!2012-08-11
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    @Mercy, I think you meant above "your RHS gives 0"2012-08-11
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    @DonAntonio Yeah! thanks.2012-08-11

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