Complex Analysis Integration of x sinmx/ x4 + a4

Question

How to find this integral :- $\displaystyle \int_0^\infty \frac{x\sin(mx)}{x^4 + a^4} \, dx$

I know it is to be solved using Cauchy Residue Theorem but I am not able to get the answer. Thanks.

Answer 1

Assuming $a,m>0$ we have $$\int_{0}^{+\infty}\frac{x\sin(mx)}{x^4+a^4}\,dx\stackrel{\text{parity}}{=}\frac{1}{2}\int_{-\infty}^{+\infty}\frac{x\sin(mx)}{x^4+a^4}\,dx = \frac{1}{2}\text{Im}\int_{-\infty}^{+\infty}\frac{x e^{imx}}{x^4+a^4}\,dx $$ Now the function $f(x)=\frac{x e^{imx}}{x^4+a^4}$ fulfills the ML lemma and its poles in the upper half-plane are located at $\frac{a}{\sqrt{2}}(\pm 1+i)$. By computing the residues there it follows that $$ \int_{0}^{+\infty}\frac{x\sin(mx)}{x^4+a^4}\,dx = \color{red}{\frac{\pi}{2a^2} e^{-\frac{am}{\sqrt{2}}}\sin\left(\frac{am}{\sqrt{2}}\right)}. $$