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If $ 0 \le t \lt 1$ then, $$\lim\limits_{R \uparrow \infty}\int_{0}^{\pi}e^{-R\sin\theta}R^t\,\text d\theta = 0$$

I'm not sure how to solve this, I thought about starting to show that $ 0 \le \sin\theta \le \pi/2$, then $ 0 \le \sin\theta \le 2\theta/\pi$. But I'm still not sure how to continue from there.

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Using the fact that $\sin(\pi -\theta)=\sin \,\theta$ you can see that $\int_{\pi /2}^{\pi} e^{-R\sin\, \theta} R^{t}\text d\theta =\int_{0}^{\pi /2} e^{-R\sin\, \theta} R^{t}\text d\theta$. On $(0, \pi/2)$ we have $\sin\, \theta \geq \frac 2 {\pi} \theta$. Can you complete the argument now?

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    $\begingroup$ Like that: $\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-R\sin\theta}R^td\theta\leq\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-\frac{2R}{\pi}\theta}R^td\theta =\lim_{R\to\infty}\frac{\pi}{2R}(1-e^{-R})R^t =0$ $\endgroup$ – Ilan Aizelman WS May 18 at 12:20
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Kabo Murphy May 18 at 12:21
  • $\begingroup$ Wow, thank you. :) $\endgroup$ – Ilan Aizelman WS May 18 at 12:22
  • $\begingroup$ This also holds for integral from $0$ to $\pi$ right? $\endgroup$ – Ilan Aizelman WS May 18 at 12:24
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    $\begingroup$ As I mentioned in my answer the integral from $\pi/2$ to $\pi$ equals integral from $0$ to $\pi /2$. So integral from $0$ to $\pi$ is twice the integral form $0$ to $\pi /2$. So it is enough to show that integral from $0$ to $\pi /2$ tends to $0$. $\endgroup$ – Kabo Murphy May 18 at 12:31

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