Prove convergence of $\int_{-\infty}^{\infty} \frac{x\sin(x)}{x^{4}-\pi^{4}}dx$

Question

Given $\int_{-\infty}^{\infty} \frac{x\sin(x)}{x^{4}-\pi^{4}}dx$, prove that the integral converges. I thought about splitting the integral and using the direct comparison test, but I'm not sure what function to use.

Answer 1

Since $\left\lvert\dfrac{x\sin x}{x^4-\pi^4}\right\rvert\leqslant\dfrac x{x^4-\pi^2}<\dfrac1{x^3}$ when $x\geqslant 4$, the integral $\displaystyle\int_4^\infty\frac{\mathrm dx}{x^3}$ converges, then the integral $\displaystyle\int_4^\infty\frac{x\sin x}{x^4-\pi^4}\,\mathrm dx$ converges. By almost the same argument, the integral $\displaystyle\int_{-\infty}^{-4}\frac{x\sin x}{x^4-\pi^4}\,\mathrm dx$ converges too.

Answer 2

Hint:

It's simpler with some asymptotic analysis:

• Near $\infty$, we have $$\frac{x\sin x}{x^4-\pi^4}=O\Bigl(\frac1{x^3}\Bigr),$$ and the latter has a convergent integral on $\mathbf R$, so the given integral is absolutely convergent.
• Near $\pi$ $$\frac{x\sin x}{x^4-\pi^4}=-\frac{x\sin(x-\pi)}{(x-\pi)(x+\pi)(x^2+\pi^2)}\sim_\pi-\frac{x}{(x+\pi)(x^2+\pi^2)},$$ and the latter function is defined in a small neighbourhood of $\pi$ (it is equivalent to $-\frac1{4\pi^2}$.

Similarly, near $-\pi$, we have the equivalence $$\frac{x\sin x}{x^4-\pi^4}\sim_{-\pi}-\frac{x}{(x-\pi)(x^2+\pi^2)}\sim -\frac1{4\pi^2}.$$

Answer 3

$x \not =π$;

0) $f(x):=\dfrac{x\sin x}{x^4-π^4}$;

$f(-x)=f(x)$, even;

Consider $2\displaystyle{\int_{0}^{\infty}}f(x)dx$;

$f(x):=\dfrac{-x\sin (x-π)}{(x-π)(x+π)(x^2+π^2)}$;

Define:

$f(π):=\dfrac{-π}{(2π)(2π^2)}$.

$f$ is continuos on $[0,\infty)$.

Split:

1) $\displaystyle{\int_{0}^{2π}}f(x)dx +\int_{2π}^{\infty}f(x)dx;$

For $x \ge 2π$:

$|\dfrac{-x\sin (x-π)}{(x-π)(x+π)(x^2+π^2)}|\le$

$\dfrac{x \cdot 1}{πx(x^2)}=\dfrac{1}{πx^2}$;

$\displaystyle{\int_{2π}^{\infty}}\dfrac{1}{πx^2}dx$ converges.

Used: $\lim_{y \rightarrow 0} \dfrac{\sin y}{y}=1$, and

$|\sin y| \le 1.$