test convergence of $\int_\pi^\infty{\frac{x \cos x}{x^2-1}dx}$

Question

$$\int_\pi^\infty{\frac{x \cos x}{x^2-1}dx}$$

So the only think I came up with was to take an absolute value of ${\frac{x \cos x}{x^2-1}}$ and by comparison test the integral does not converge.

But I see it's not very close to the solution, so what should I do?

Answer 1

Hint. Note that $$\int_\pi^\infty \frac{x \cos x}{x^2-1}dx= \frac{1}{2}\int_\pi^\infty \left(\frac{ \cos x}{x-1}+\frac{ \cos x}{x+1}\right)dx$$ Moreover, if $a<\pi$ then, by integrating by parts, $$\int^{\infty}_{\pi} \frac{\cos x}{x-a}\,dx=\left[\frac{\sin x}{x-a}\right]^{\infty}_{\pi}+\int^{\infty}_{\pi} \frac{\sin x}{(x-a)^2}\,dx=0+\int^{\infty}_{\pi} \frac{\sin x}{(x-a)^2}\,dx.$$ Can you take it from here?

Answer 2

Continuing from Robert Z's answer,

$$ \int_{\pi}^{+\infty}\frac{\sin x}{(x-a)^2}\,dx = -\int_{0}^{+\infty}\frac{\sin t}{(t+\pi-a)^2}\,dt \stackrel{\mathcal{L}}{=}-\int_{0}^{+\infty}\frac{s}{s^2+1}e^{(a-\pi)s}\,ds$$ and for every $a<\pi$ the Cauchy-Schwarz inequality gives a simple bound: $$\left| \int_{\pi}^{+\infty}\frac{\sin x}{(x-a)^2}\,dx\right|\leq \sqrt{\frac{1}{2(\pi-a)}\int_{0}^{+\infty}\frac{s^2\,ds}{(s^2+1)^2}}=\sqrt{\frac{\pi}{8(\pi-a)}}. $$