Improper integral $\int_{0}^{\pi} \frac{x}{\sin x} dx$

Question

Find out whether or not the following integral exists $$\int_{0}^{\pi} \frac{x}{\sin x} dx.$$

I'm pretty sure this integral doesn't exist but I can't seem to find a good way to prove this. It certainly seems way too hard to find the indefinite integral. Can someone please share a hint?

Answer 1

There's two problems to treat: on $0$ and on $\pi$.

• On $0$ the function is extended since $\lim\limits_{x\to0}\frac{x}{\sin x}$ exists so the integral $$\int_0^{\frac12}\frac{x}{\sin x}dx$$ exists
• On $\pi$ and by the change of variable $t=\pi-x$ we find $$\int_{\frac12}^\pi\frac{x}{\sin x}dx=\int_0^{\pi-\frac12}\frac{\pi-t}{\sin t}dt$$ and this integral isn't convergent since $$\frac{\pi-t}{\sin t}\sim_0\underbrace{\frac{\pi}{t}}_{\text{this term gives a divergent integral}}-1$$ Conclusion: The given integral is divergent.