Computing $\int_{0}^{\infty} \frac{x}{x^{4}+1} dx$ using complex analysis. I want to compute $$\int_{0}^{\infty} \frac{x}{x^{4}+1} dx$$ using complex analysis. Now the first thing that strikes me is that $f(x)$ is not an even function. So this troubles me a bit since I would normally use 
$$\int_{0}^{\infty} f(x)dx = \frac{1}{2} \left[\lim_{R \rightarrow \infty} \int_{-R}^{R} f(z)dz + \int_{C_{R}}f(z)dz \right],
$$
where $C_{R}$ is the semi cirle connecting $R$ to $-R$ in the positive imaginary part. Now we see that we have to compute the singularities of $f(z)$, which we can do by computing the fourth root of $z$. We then find
$$
\begin{align*}
z^{4} &= e^{i (\pi + 2n\pi)} \\
z &= e^{i ( \frac{\pi}{4} + \frac{n\pi}{2} )}
\end{align*}.
$$
Since we are only interested in singularities above the real line, we find 
$z_{0} = e^{i \frac{\pi}{4}}$ and $z_{1} = e^{i \frac{3\pi}{4}}$. 
Then we let $p(z) = z$ and $q(z) = z^{4}+1$, which makes $q'(z) = 4z^{3}$. 
We then compute $p(z_{0}), q(z_{0})$ and $q'(z_{0})$ and finally $\frac{p(z)}{q'(z)}$ which equals the residue at $z_{0}$. 
However, when I do the above I find
$\text{Res}(z_{0}) = - \frac{i}{4}$ and $\text{Res}(z_{1}) = \frac{i}{4}$ but this would make the integral equal zero since $2\pi i (\frac{i}{4} - \frac{i}{4})=0$.
Can anybody point me to my mistake? Also, when would find the value for this integral, I would argue we can not simply take half of it, since the initial function is not even. How would we fix that?
 A: So i would take the following contour:
$[0,R],$ $C_R$ and $[iR,0]$ where $C_R$ connects $R$ and $iR.$ So, instead of a semicircle, you have a quarter of a circle. Notice that, by doing this, just one of your singularities is inside, mainly $e^{i\frac{\pi}{4}}$ hence your integral
$$\int _{[0,R]}+\int _{C_R}+\int _{[iR,0]}=2\pi i\frac{-i}{4}=\frac{\pi}{2}.$$ Check that  the integral vanishes in $C_R$ and make a change of variable, perhaps, $y=ix$ to convert the integral in $[iR,0]$ to an integral in $[0,R].$
A: It's somewhat easier with a "compromise" proof that substitutes before using complex analysis. The integral is$$\frac12\int_0^\infty\frac{du}{u^2+1}=\frac14\int_{\Bbb R}\frac{du}{u^2+1}=\frac{\pi i}{2}\lim_{u\to i}\frac{1}{u+i}=\frac{\pi}{4}.$$
A: Generally speaking, computing an integral $\int_a^b f(x)\;dx$ by residues will only work if $x=a$ and $x=b$ are special points for $f(x)$.  
If, for example, $b$ is not a special point, then you are essentially computing the indefinite integral.  If there is not nice formula for the indefinite integral, then trying to do it by residues will fail.
