Evaluate $\int_0^\infty \frac{x^2}{x^4+16}\,dx$

Question

It's a very simple problem. I put it into wolfram alpha and it gave me a hugely complicated answer. This makes me think that I am supposed to be doing it differently.

I believe I am supposed to use the residue theorem for this answer but have no idea how. $$\int_{0}^{\infty}\frac{x^2}{x^4+16}dx = 2\pi i \sum_{n=0}^{\infty} Res\big(\frac{x^2}{x^4+16}\big)$$ I have no idea what Res() is or how to use this in any way.

Any and all help is appreciated, thank you.

Sorry, I typed the equation in wrong, I just corrected it from $\frac{x^4}{x^2+16}$ to $\frac{x^2}{x^4+16}$

Answer 1

With substitution $x^2=4\tan u$ then \begin{align} \int_0^\infty\dfrac{x^2}{x^4+16}\ dx &= \dfrac14\int_0^{\pi/2}\sqrt{\tan u} \ du \\ &= \dfrac14\int_0^{\pi/2}\sin^{2(\frac34)-1}u \cos^{2(\frac14)-1}u\ du \\ &= \dfrac18\beta(\dfrac34,\dfrac14) \\ &= \dfrac18\Gamma(\dfrac34)\Gamma(\dfrac14) \\ &= \color{blue}{\dfrac{\pi}{4\sqrt{2}}} \end{align} here $\beta(x,y)$ is Beta function and $$\Gamma(1+x)\Gamma(1-x)=\dfrac{\pi x}{\sin\pi x}$$

Answer 2

If you call the integral $I$, then $$2I=\int_{-\infty}^\infty\frac{z^2\,dz}{z^4+16}.$$ This can be done by the "semicircle" method in contour integration, and equals $2\pi i$ times the sum of the residues of the integrand of the poles in the upper half plane. These poles are at $2\zeta$ and $2\zeta^3$ where $\zeta=\exp(\pi i/4)$. The residue at $2\zeta$ is $$\lim_{z\to2\zeta}\frac{(z-2\zeta)z^2}{z^4+16}=\frac{(2\zeta)^2} {4(2\zeta)^3}=\frac{1}{8\zeta}.$$ Now do the very similar calculation for the other residue and put the pieces together.

Answer 3

I will provide another complex method.
Let $f(z)=\frac{z^2\ln (-z)}{z^4+16}$. Taking $C$ as the keyhole contour (small circle around the origin, upward of positive real axis, big circle, downward of positive real axis). $$\int_C f(z)dz=2\pi i\sum_{z:\text{roots of $z^4+16$}}\operatorname{Res}(f(z))=2\pi i(-\frac{\pi}{4\sqrt2}),$$ $$I=\frac{2\pi i}{-2\pi i}(-\frac{\pi}{4\sqrt2})=\frac{\pi}{4\sqrt2}$$