I have a contour integral $\int_{0}^{\infty} \frac{x^3}{x^5-a^5}dx $ where $ a>0$ Here I tried to find residue, however I failed since the roots are $z=a,-(-1)^{1/5},(-1)^{2/5},-(-1)^{3/5},(-1)^{4/5}$ It's not possible to calculate $\lim_{x->x_0} (x-x_0)\frac{x^3}{(x-a)(x^4+x^3a+x^2a^2+xa^3+a^4)}$
So is there any suggested method to calculate this integral with contour integral method?
On
Solutions of $x^5=a^5$ are
$a,-\sqrt[5]{-1} a,(-1)^{2/5} a,-(-1)^{3/5} a,(-1)^{4/5} a$
Residue at $x=a$ is, for instance $$\underset{x\to a}{\text{lim}}\frac{x^3 (x-a)}{x^5-a^5}=\underset{x\to a}{\text{lim}}\frac{x^3 (x-a)}{(x-a)(x^4+ax^3+a^2x^2+a^3x+a^4)}=\frac{a^3}{5a^4}=\frac{1}{5 a}$$ At $x=(-1)^{2/5} a$ it is $$\underset{x\to (-1)^{2/5} a}{\text{lim}}\frac{x^3 \left(x-(-1)^{2/5} a\right)}{x^5-a^5}=$$ $$=\underset{x\to (-1)^{2/5} a}{\text{lim}}\frac{x^3 \left(x-(-1)^{2/5} a\right)}{\left(x-(-1)^{2/5} a\right)(-(-1)^{3/5} a^4-\sqrt[5]{-1} a^3 x+(-1)^{4/5} a^2 x^2+(-1)^{2/5} a x^3+x^4)}=$$ $$=-\frac{(-1)^{3/5}}{5 a}$$
Hope this helps
The roots are $ae^{2\pi i/5}$, $ae^{4\pi i/5}$, $ae^{6\pi i/5}$, and $ae^{8\pi i/5}$. If $r$ is any of them, then it is a simple root of the denominator, and therefore\begin{align}\operatorname{res}_{z=r}\left(\frac{z^3}{z^5-a^5}\right)&=\frac{r^3}{5r^4}\\&=\frac1{5r}.\end{align}