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Does $I=\displaystyle \int_0^{+\infty} e^{-t\sin t}\;dt$ converge?

I haven't got the correction so, I would like to know if it's correct.

$\forall t\in \big[(2k+1)\pi,\; (2k+2)\pi\big],\quad t\sin t\ \le 0\implies e^{-t\sin t}\ge 1$

Then $\displaystyle \sum_{k=0}^n\int_{(2k+1)\pi}^{(2k+2)\pi}e^{-t\sin t}\;dt\ge \sum_{k=0}^n\int_{(2k+1)\pi}^{(2k+2)\pi}1\;dt=(n+1)\pi$

since $(n+1)\pi\underset{n\to+\infty}{\longrightarrow}+\infty\implies I$ diverges

Stu
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    Since you tagged the question with [tag:improper-integrals], the context is probably improper Riemann integrals, not Lebesgue integrals? – Daniel Fischer Aug 31 '17 at 20:38
  • Yes I haven't studied yet, the Lebesgue integrals. – Stu Aug 31 '17 at 20:41
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    For completeness you should point out that the integrand is nonnegative, so $\int_0^{\infty} e^{-t \sin t}\ dt$ is at least as large as your sum. Aside from that, your argument is fine. –  Aug 31 '17 at 20:46
  • I wanted to write that first, but as it's obvious that for any $t\in\mathbb{R}, \quad e^t>0$. I haven't done it. – Stu Aug 31 '17 at 20:50
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    You should have, since for $f(t) = \frac{\sin t}{t}$ the observation that $$\sum_{k = 0}^n \int_{2k\pi}^{(2k+1)\pi} f(t),dt \geqslant \sum_{k = 0}^n \frac{1}{(2k+1)\pi}\int_0^\pi \sin t,dt = \frac{2}{\pi} \sum_{k = 0}^n \frac{1}{2k+1} > \frac{1}{\pi}\log n \to +\infty$$ doesn't imply the divergence of the integral. Your argument shows that for $\varepsilon < \pi$, there is no $K$ such that for $K \leqslant x < y$ the bound $$\Biggl\lvert\int_x^y f(t),dt\Biggr\rvert <\varepsilon$$ holds, hence the integral doesn't converge, but you don't explicitly say that either. Partial marks, I'm afraid. – Daniel Fischer Aug 31 '17 at 20:56
  • @Daniel Fischer Do you mean, my proof is not sufficient??? – Stu Aug 31 '17 at 21:06
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    That depends on what the purpose of the proof is. If it is to show fellow mathematicians that the integral doesn't converge, it would be sufficient for that purpose since the readers can fill in the gap on their own. But in an exercise, the purpose is to give a complete argument, and then it matters to state the obvious little things that you use in your argument, like $e^x > 0$ for all $x\in\mathbb{R}$. – Daniel Fischer Aug 31 '17 at 21:27

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As you said

$$I_k=\int_{(2k+1)\pi}^{(2k+2)\pi}e^{-t\sin (t)}dt\ge 1$$

$$\implies \lim_{k\to+\infty}I_k\ne 0$$

thus, by Cauchy criterion, the integrale diverges.

hamam_Abdallah
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