Improper integral: $\int_0^{+\infty} \frac{\sin x \log|x- \pi|}{x(x-1)}dx$

Question

Can I get help determining the nature of this improper integral?

(¿Podrían ayudarme a determinar el carácter de está integral impropia?)

$$\int_0^{+\infty} \frac{\sin x \log|x- \pi|}{x(x-1)}$$

Answer 1

Your integral is not an improper one since the integrand function $ f $ is not locally integrable at $(0,+\infty)$. You can speak only about $$\int_0^1f$$ or $$\int_1^{\pi}f$$ or $$\int_{\pi}^{+\infty}f$$

Answer 2

Show that $$\int_1^{3/2}\frac{\sin x\log|x-\pi|}{x(x-1)}\,dx $$ diverges which would imply your integral diverges as well. Notice that the integrand is positive in $[1,3/2]$ and $$\frac{\sin x\log|x-\pi|}{x(x-1)}\sim \frac{\sin(1)\log(\pi-1)}{x-1}\;\;(x\to 1^+) $$ So you need to show that $$\sin(1)\log(\pi-1)\int_1^{3/2}\frac{1}{x-1}\,dx $$ is divergent.