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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?

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    $\begingroup$ what happens to the integrand when $x\rightarrow \pi$? The indefinite integral can be expressed in terms of dilogarithms after the substitution $t=e^{ix}$. $\endgroup$ – Start wearing purple Jul 19 '14 at 10:30
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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.
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  • $\begingroup$ According to Wolfram Alpha, the integral converges to $117.281$. $\endgroup$ – Tunk-Fey Jul 19 '14 at 10:41
  • $\begingroup$ @Tunk-Fey this only shows that Wolfram Alpha sucks. $\endgroup$ – Start wearing purple Jul 19 '14 at 10:45
  • $\begingroup$ @Tunk-Fey it's interesting that Wolfram-Alpha says that, Mathematica says that it diverges $\endgroup$ – DanZimm Jul 19 '14 at 10:50
  • $\begingroup$ Can you please explain the last step with the $\sim_0$ ? $\endgroup$ – rehband Jul 19 '14 at 11:03
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    $\begingroup$ We use the Taylor series to get that $\sin t$ is asymptotically equivalent to $t$.@rehband $\endgroup$ – user63181 Jul 19 '14 at 11:11

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