Convergence of an improper integral.

Question

$$\int_{-1}^1\sin\left(\frac{1+x}{1-x}\right)\frac{1}{(1-x^2)^2}\,dx$$ I've tried Abel and Dirichlet tests but it doesn't seem to be that easy. Any thoughts?

Answer 1

Let $t=x+1$. Then, as $t\to 0^+$, $$\sin\left(\frac{1+x}{1-x}\right)\frac{1}{(1-x^2)^2} =\sin\left(\frac{t}{2-t}\right)\frac{1}{t^2(2-t)^2}\sim \frac{t/2}{4t^2}=\frac{1}{8t}$$ and may we conclude that the integral over $(-1,a]$ is divergent when $-1<a<1$ .

Moreover, if we restrict the integral over $[a,1)$ with $-1<a<1$, then by letting $t=\frac{1+x}{1-x}$, we get $$\int_{a}^1\sin\left(\frac{1+x}{1-x}\right)\frac{1}{(1-x^2)^2}\,dx= \frac{1}{8}\int_{b}^{+\infty}\frac{\sin\left(t\right)(t+1)^2}{t^2}\,dt=\\ \frac{1}{8}\int_{b}^{+\infty}\left(\sin(t)+\frac{2\sin(t)}{t}+\frac{\sin(t)}{t^2}\right)\,dt$$ where $b=\frac{a+1}{a-1}>0$, which is not convergent because $\frac{\sin(t)}{t}$ and $\frac{\sin(t)}{t^2}$ are integrable over $[b,+\infty)$, whereas $\int_b^r\sin(t)\,dt$ is not convergent as $r\to +\infty$.