Hey guys, I was wondering if any of you could show me how I would go about using complex integration to show that:

$\displaystyle \int_{0}^{\infty}x^2/(1+x^4)dx = \pi /(2 \sqrt2)$

Thanks.

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- Sep 23rd 2009, 05:35 AMenjamComplex Integration
Hey guys, I was wondering if any of you could show me how I would go about using complex integration to show that:

$\displaystyle \int_{0}^{\infty}x^2/(1+x^4)dx = \pi /(2 \sqrt2)$

Thanks. - Sep 23rd 2009, 11:58 AMshawsend
Consider the contour integral over the upper half-disc as the radius extends to infinity:

$\displaystyle \mathop\oint\limits_{C} \frac{z^2}{z^4+1}dz$

By the Residue Theorem, that's equal to:

$\displaystyle \mathop\oint\limits_{C} \frac{z^2}{z^4+1}dz=2\pi i \left(\mathop \text{Res}_{z=e^{\pi i/4}} \frac{z^2}{z^4+1}+\mathop\text{Res}_{z=e^{3\pi i/4}} \frac{z^2}{z^4+1}\right)=\frac{\pi}{\sqrt{2}}$

But the integral is even so:

$\displaystyle \int_0^{\infty}\frac{x^2}{x^4+1}=\frac{1}{2}\frac{ \pi}{\sqrt{2}}$

But I've skipped steps that are important to fully understand the principle.