Integrating $I(\alpha)=\int^{\infty}_{0} \frac{x^{\alpha}}{x^4+1}dx$ Here is the question:
Let $P(x)$ be a polynomial of degree $d>1$ with $P(x)>0$ for all $x>0$. For what values of $\alpha \in \mathbb{R}$ does the integral 
$I(\alpha)=\int^{\infty}_{0} \frac{x^{\alpha}}{P(x)}dx$
converge? Give a formula for $I(\alpha)$ in terms of residues. Compute $I({\alpha})$ for $P(x)=1+x^4$. 
I'm definitely a bit rusty...So far I have tried an integration technique from $[0, \infty)$, around a large circle, and back onto $[0, \infty)$. The last part should be shifted by an analytic continuation of $x^{\alpha}$. Now we need only consider our poles at $z=-1$ which are given by $\zeta_{8}$, $\zeta^{3}_{8}$, $\zeta^{5}_{8}$, and $\zeta^{7}_{8}$. Now can I just go ahead and sum the residues? Is anything strange happening here?
Thanks very much!
 A: The case $m=4$ of this answer yields
$$
\int_0^\infty\frac{x^\alpha}{x^4+1}\,\mathrm{d}x=\frac\pi4\csc\left(\pi\frac{\alpha+1}{4}\right)
$$
As long as $P(x)$ does not vanish on $[0,\infty)$, then
$$
\int_0^\infty\frac{x^\alpha}{P(x)}\,\mathrm{d}x
$$
converges when $-1\lt\alpha\lt\deg(P)-1$.
A: This would not probaly help a lot but could give you some ideas.  
The antiderivative of the integral in the title of your post is
$$\frac{x^{a+1} \, _2F_1\left(1,\frac{a+1}{4};\frac{a+5}{4};-x^4\right)}{a+1}$$  and the value of the integral is $$\frac{1}{4} \pi  \csc \left(\frac{1}{4} \pi  (a+1)\right)$$  provided $-1<\Re(a)<3$.
A: In this and all other similar situations, let $t=\dfrac1{x^4+1}$ . Then recognize the expression of the beta function in the new integral, and use the reflection formula for the $\Gamma$ function. All integrals of the form $\displaystyle\int_0^\infty\frac{x^a}{(x^b+1)^c}dx$ can be solved in this simple manner. Also, since the harmonic series is divergent, the integral only converges when $4-a>1$. On the other hand, analyzing its behavior near $x=0$, we conclude that $a>-1$. These observations are confirmed by the final result, which forbids $\dfrac{a+1}4\in\mathbb{Z}$, otherwise we have $I(a)\to\infty$ , since $\sin(k\pi)=0$ for integer values of k.
