Contour integration in complex analysis

Question

How to carry out the following integration?$$\int_{0}^{\infty}\frac{x^6}{(x^4+a^4)^2}dx$$I tried doing it using contour integration but it turned out to be extremely complicated. A help would be greatly appreciated!

Answer 1

$$\int_{0}^{+\infty}\frac{x^6}{(x^4+a^4)^2}\,dx \stackrel{x\mapsto az}{=}\frac{1}{|a|}\int_{0}^{+\infty}\frac{z^6 \,dz}{(z^4+1)^2}\stackrel{\text{IBP}}{=}\frac{3}{4|a|}\int_{0}^{+\infty}\frac{z^2}{1+z^4}\,dz$$ equals (by applying the substitution $z\mapsto 1/z$ on the subinterval $(1,+\infty)$ of the integration range) $$\begin{eqnarray*} \frac{3}{4|a|}\int_{0}^{1}\frac{1+z^2}{1+z^4}\,dz&=&\frac{3}{4|a|}\int_{0}^{1}\frac{1+z^2-z^4-z^6}{1-z^8}\,dz\\ &=& \frac{3}{4|a|}\sum_{n\geq 0}\left(\frac{1}{8n+1}+\frac{1}{8n+3}-\frac{1}{8n+5}-\frac{1}{8n+7}\right)\end{eqnarray*}$$ so the whole problem boils down to showing that the last (Dirichlet) series equals $\frac{\pi}{2\sqrt{2}}$, which can be done in a good number of ways, for instance by evaluating $-\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n}$ at the primitive $8$-th roots of unity.