Originally Posted by

**xboxlive89128** Compute the following integral.

$\displaystyle \int^{\infty}_{0} \frac{x^2}{x^8+1}dx. $

I think that I should get $\displaystyle \frac{1}{4} \sqrt{1-\frac{1}{\sqrt 2}} \cdot \pi$. I see we know this is even. Then there is work to show that

$\displaystyle \int^{\infty}_{-\infty}$ is$\displaystyle 2\pi i$ times the sum of the residues of $\displaystyle \frac{z^2}{z^8+1}$ in the upper half plane. Then I found that

$\displaystyle c_k = \text{exp}[i(\frac{\pi}{8}+\frac{2k\pi}{8}]$ where $\displaystyle k=0,1, \ldots, 7.$ Then the zeros in the upper half plane occur at $\displaystyle k=0,1,2,3$.

Then I used

$\displaystyle \text{Res}_{z=c_k}=\frac{c_k^2}{8c_k^7}=\frac{1}{8 c_k^5}=\frac{1}{8}c_k^{-5}$ where $\displaystyle k=0,1,2,3.$ Then when I add these residues up I did not get the answer above, I got $\displaystyle \frac{\sqrt{\sqrt{2}+2} \cdot \pi}{16} - \frac{\sqrt{2-\sqrt{2}} \cdot \pi }{16}i.$