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

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?

• what happens to the integrand when $x\rightarrow \pi$? The indefinite integral can be expressed in terms of dilogarithms after the substitution $t=e^{ix}$. – Start wearing purple Jul 19 '14 at 10:30

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.
• According to Wolfram Alpha, the integral converges to $117.281$. – Tunk-Fey Jul 19 '14 at 10:41
• Can you please explain the last step with the $\sim_0$ ? – rehband Jul 19 '14 at 11:03
• We use the Taylor series to get that $\sin t$ is asymptotically equivalent to $t$.@rehband – user63181 Jul 19 '14 at 11:11