Use the Residue Theorem to evaluate the following integral: $$\int_{-∞}^{∞} \frac{x^4}{1+x^8} dx$$
I've found the zeros in the upper half plane to be 
$$e^{i \pi/8}, e^{i 3 \pi/8}, e^{i 5 \pi/8}, e^{i 7 \pi/8}$$ (right?)
But then the calculation got really ugly, so I think I'm doing something wrong...please help!!
 A: The good news is that this is as straightforward an application of the residue theorem as possible.  The bad news is that, yes, there are 4 simple poles for which you have to find residues, and they are the ones you listed.
I think things might get simpler is you use the formula for a simple pole:
$$\text{Res}_{z=z_k} \frac{f(z)}{g(z)} = \frac{f(z_k)}{g'(z_k)}$$
(You can prove this using the definition of a derivative.)
Here, $f(z)=z^4$ and $g'(z) = 8 z^7$.  The sum of the residues looks like
$$\frac{e^{i \pi/2}}{8 e^{i 7 \pi/8}} + \frac{e^{i 3 \pi/2}}{8 e^{i 5\pi/8}} + \frac{e^{i \pi/2}}{8 e^{i 3 \pi/8}} + \frac{e^{i 3 \pi/2}}{8 e^{i \pi/8}}$$
Note that I used periodic properties of the exponentials to simplify things.  Can you work with this?
The result I get is 
$$\int_{-\infty}^{\infty} dx \frac{x^4}{1+x^8} = \frac{\pi}{2} \sqrt{1-\frac{\sqrt{2}}{2}}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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The following integration $\ds{\underline{just}}$ need
$\ds{\underline{two}}$ poles. Namely, $\ds{p_{0} \equiv \expo{\pi\ic/8}\ \mbox{and}\ p_{1} \equiv \expo{3\pi\ic/8}}$.

\begin{align}
&\bbox[5px,#ffd]{%
\int_{-\infty}^{\infty}{x^{4} \over 1 + x^{8}}\,\dd x} =
2\,\Re\int_{0}^{\infty}{x^{4} \over 1 + x^{8}}\,\dd x
\\ &\
\mbox{"Close" the integration along a quarter circle}
\\ &\ \mbox{in the complex plane first quadrant.}
\\[5mm] = &\
2\Re\braces{2\pi\ic\sum_{j = 0}^{1}{p_{j}^4 \over 8p_{j}^{7}} - \int_{\infty}^{0}{\ic^{4}y^{4} \over
1 + \ic^{8}y^{8}}\,\ic\,\dd y}
\\[5mm] = &\
-4\pi\,\Im\sum_{j = 0}^{1}{p_{j}^{5} \over 8 \times \pars{-1}}
\\[5mm] = &\
{\pi \over 2}\,\Im\sum_{j = 0}^{1}p_{j}^{5} =
{\pi \over 2}\,\Im\pars{\expo{5\pi\ic/8} + \expo{15\pi\ic/8}}
\\[5mm] = &\
{\pi \over 2}\bracks{\sin\pars{5\pi \over 8} +
\sin\pars{15\pi \over 8}}
\\[5mm] = &\
\bbx{\root{1 - {\root{2} \over 2}}{\pi \over 2}} \approx 0.8501
\\ &\
\end{align}
