1. ## upper half plane

Compute the following integral.

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

I think that I should get $\frac{1}{4} \sqrt{1-\frac{1}{\sqrt 2}} \cdot \pi$. I see we know this is even. Then there is work to show that
$\int^{\infty}_{-\infty}$ is $2\pi i$ times the sum of the residues of $\frac{z^2}{z^8+1}$ in the upper half plane. Then I found that

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

Then I used

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

2. Originally Posted by xboxlive89128
Compute the following integral.

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

I think that I should get $\frac{1}{4} \sqrt{1-\frac{1}{\sqrt 2}} \cdot \pi$. I see we know this is even. Then there is work to show that
$\int^{\infty}_{-\infty}$ is $2\pi i$ times the sum of the residues of $\frac{z^2}{z^8+1}$ in the upper half plane. Then I found that

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

Then I used

$\text{Res}_{z=c_k}=\frac{c_k^2}{8c_k^7}=\frac{1}{8 c_k^5}=\frac{1}{8}c_k^{-5}$ where $k=0,1,2,3.$ Then when I add these residues up I did not get the answer above, I got $\frac{\sqrt{\sqrt{2}+2} \cdot \pi}{16} - \frac{\sqrt{2-\sqrt{2}} \cdot \pi }{16}i.$
That seems to be pretty much correct, but it's not obvious how to transform your answer to the given one. I would take the residues as $e^{ik\pi/8}$, where k = 1, 3, 5, 7. Their negative-fifth powers are $e^{ik\pi/8}$, where k = 1, 7, –3, –5. The sum of the first two of these is $2i\sin(\pi/8)$, and the sum of the other two is $-2i\sin(3\pi/8)$. So the sum of the four residues is $\tfrac i4(\sin(\pi/8) - \sin(3\pi/8)) = -\tfrac i2\cos(\pi/4)\sin(\pi/8)$ (addition formula for trig functions!). Now use $\cos(\pi/4) = 1/\sqrt2$ and $\sin(\pi/8) = \frac1{\sqrt2}\sqrt{1-\frac1{\sqrt2}}$ to see that $2\pi i$ times the sum of the residues is $\frac\pi2\sqrt{1-\frac1{\sqrt2}}$.

That is the integral from $-\infty$ to $\infty$, so you need to divide by 2 to get the integral from 0 to $\infty$.