Evaluate $\int_{0}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$ Evaluate :$$ \int_{0}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$$
using residue theorem. The integral is even and it can be written as :
$$ \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$$
The only pole with a positive imaginary part is $ \frac{i}{\sqrt2} $ and its degree is $4$. How can I evaluate this without differentiating 3 times when calculating the residue? 
I also tried calculating the infinite residue but that didn't help much.
 A: You can instead find the laurent series, and pick up the coefficient of the $z^{-1}$ term.
i.e. $(1+2z^2)^{-4} = (1+2(z-\frac{i}{\sqrt2})^2-2(\frac{-1}{2})+4\frac{iz}{\sqrt2})^{-4}= (1+2(z-\frac{i}{\sqrt2})^2+1+2\sqrt2iz)^{-4} = (2\sqrt{2}i(z-\frac{i}{\sqrt2})+2\sqrt2i\frac{i}{\sqrt2}+2+2(z-\frac{i}{\sqrt2})^2)^{-4}=(2\sqrt2i(z-\frac{i}{\sqrt2})+2(z-\frac{i}{\sqrt2})^2)^{-4}$
Put $y=z-\frac{i}{\sqrt2}$, so you need to find the laurent series of $(2\sqrt{2}iy+2y^2)^{-4}=(2\sqrt2iy)^{-4}(1+\frac{-i}{\sqrt2}y)^{-4}$.
Writing $\frac{z^4}{(1+2z^2)^{4}}=\frac{(y+\frac{i}{\sqrt2})^4}{(2\sqrt2iy)^{4}(1+\frac{-i}{\sqrt2}y)^{4}}=\frac{(y+\frac{i}{\sqrt2})^4}{64y^4(1+\frac{-i}{\sqrt2}y)^{4}}$
Therefore you are only interested in the $y^{-1}$ coefficient which corresponds to $y^3$ coefficient of $\frac{(y+\frac{i}{\sqrt2})^4}{(1+\frac{-i}{\sqrt2}y)^{4}}$ scaled by $\frac{1}{64}$.
Looking at $(y+\frac{i}{\sqrt2})^4(1+\frac{-i}{\sqrt2}y)^{-4}=(1/4 - i \sqrt2 y - 3 y^2 + 2 i \sqrt2 y^3 + y^4)(1+2\sqrt2iy-5y^2-5\sqrt2iy^3+O(y^4))=....-\frac{i}{2\sqrt2}y^3+....$
Scaling that by $\frac{1}{64}$, we find the residue to be $\frac{-i}{128\sqrt2}$
A: Note first that$$\frac{x}{1+2x^2}=\frac14\left(\frac{1}{x-i/\sqrt{2}}+\frac{1}{x+i/\sqrt{2}}\right).$$Hence$$\left(\frac{x}{1+2x^2}\right)^2=\frac{1}{16}\left(\frac{1}{(x-i/\sqrt{2})^2}+\frac{1}{(x+i/\sqrt{2})^2}+\frac{2}{x^2+1/2}\right)\\\frac{1}{16}\left(\frac{1}{(x-i/\sqrt{2})^2}+\frac{1}{(x+i/\sqrt{2})^2}+i\sqrt{2}\left(\frac{1}{x-i/\sqrt{2}}-\frac{1}{x+i/\sqrt{2}}\right)\right).$$Squaring this again, none of the diagonal terms matter, and many of the cross terms don't either. Let $f\sim g$ denote the condition that $f-g$ has zero residue at $\frac{i}{\sqrt{2}}$, so$$\left(\frac{x}{1+2x^2}\right)^4\sim\frac{i\sqrt{2}}{128}\frac{1}{(x+i/\sqrt{2})^2(x-i/\sqrt{2})}\\\implies\operatorname{Res}_{-i/\sqrt{2}}\left(\frac{x}{1+2x^2}\right)^4=\frac{-i\sqrt{2}}{256}.$$Now we just need to multiply by $\frac122\pi i$ to give $\frac{\pi\sqrt{2}}{256}$ as the value of the integral.
Just to double-check, let's solve the problem with $x=\frac{1}{\sqrt{2}}\tan t$ so the original integral is$$\frac{\sqrt{2}}{8}\int_0^{\pi/2}\sin^4t\cos^2tdt=\frac{\sqrt{2}}{16}\operatorname{B}\left(\frac52,\,\frac32\right)=\frac{\sqrt{2}}{16}\frac{\Gamma\left(\frac52\right)\Gamma\left(\frac32\right)}{\Gamma(4)}=\frac{\pi\sqrt{2}}{256}.$$
