Contour integration: $\int_0^{\infty} x^p /(x^4+1) dx$ where $-1 < p < 3$

Question

I want to calculate $\int_0^{\infty} x^p /(x^4+1) dx$ where $-1 < p < 3$. My first guess is to let $f(z) := \frac {z^p}{z^4+1}$ and integrate this over $\gamma_R$ where $$ \gamma_r = [-R,R] \cup \{Re^{i\theta} : \theta \in [0,\pi] \} $$ The simple poles of $f$ are in $e^{k \pi i /4}$ where $k\in \{1,3,5,7\}$ so two of the poles lie in $\gamma_R$ for $R > 1$.

Is this a good approach ?

Answer 1

Your approach will work just fine. However, you can reduce this to a problem where only one singularity needs to be considered.

In this answer, it is shown that $$ \int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x=\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right) $$ Setting $m=4$ yields $$ \int_0^\infty\frac{x^p}{1+x^4}\,\mathrm{d}x=\frac{\pi}{4}\csc\left(\pi\frac{p+1}{4}\right) $$