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I would like to show $\lim\limits_{r\to\infty}\int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta=0$.

Now, of course, the integrand does not converge uniformly to $0$ on $\theta\in [0, \pi/2]$, since it has value $1$ at $\theta =0$ for all $r\in \mathbb{R}$.

If $F(r) = \int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta$, we can find the $j$th derivative $F^{(j)}(r) = (-1)^j\int_{0}^{\pi/2}\sin^{j}(\theta)e^{-r\sin\theta}\text d\theta$, but I don't see how this is helping.

The function is strictly decreasing on $[0,\pi/2]$, since $\partial_{\theta}(e^{-r\sin\theta})=-r\cos\theta e^{-r\sin \theta}$, which is strictly negative on $(0,\pi/2)$.

Any ideas?

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  • $\begingroup$ You may have a look to my answer:) $\endgroup$ – Guy Fsone Oct 31 '17 at 18:27
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It's only enough to show that

$$ \int\limits_{0}^{\pi/2}{e^{-r\sin\theta}\text d\theta}\le \int\limits_{0}^{\pi/2}{e^{-r\frac{2}{\pi}\theta}\text d\theta}=\frac{\pi}{2r}\left(1-e^{-r}\right) \to 0 \quad (r \to +\infty)$$

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  • $\begingroup$ Great answer! Very simple. $\endgroup$ – Eric Auld Jul 13 '13 at 20:06
  • $\begingroup$ I think the minus sign before $\frac{\pi}{2r}$ should be omitted, right? There's no way the integral would be negative for any $r>0$. $\endgroup$ – Eric Auld Jul 13 '13 at 20:29
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    $\begingroup$ Yes, a little mistake. Of course there must not be minus there. $\endgroup$ – cool Jul 13 '13 at 20:40
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On $[0,\pi/2]$ the sine is nonnegative and so $|e^{-r\sin \theta}| \leq 1$ for $r \geq 0$. It follows by dominated convergence that $$ \lim_{r \to \infty} \int_0^{{\pi/2}} e^{-r \sin \theta}\text d\theta = \int_0^{{\pi/2}}\lim_{r \to \infty} e^{-r \sin \theta}\text d\theta = \int_0^{\pi/2}0 \text d\theta = 0. $$

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    $\begingroup$ While correct, you must realize this doesn't help him. He doesn't know Lebesgue integration yet... $\endgroup$ – Potato Jul 13 '13 at 19:51
  • $\begingroup$ @Potato Actually, although your more elementary answer is appreciated (and is certainly what the exercise expected), I am familiar with Lebesgue integration. It's good to see things from two perspectives! $\endgroup$ – Eric Auld Jul 13 '13 at 19:52
  • $\begingroup$ @EricAuld Why are you using the uniform convergence criteria for interchanging limits and integrals then!? Use and abuse the dominated convergence theorem! $\endgroup$ – Potato Jul 13 '13 at 19:54
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    $\begingroup$ @EricAuld You can get around the continuous parameter problem by passing to subsequences. I forget the details, but Folland works them out in his book when he does a corollary of the DCT (differentiation under the integral sign). You might need to assume some things are continuous, yada yada, whatever. The point is you can do it. $\endgroup$ – Potato Jul 13 '13 at 20:00
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    $\begingroup$ @EricAuld To prove the continuous parameter case, consider that for every sequence $r_n \to \infty$ the $\mathbb{N}$ parameter dominated convergence theorem gives the result. Since it holds for all sequences $r_n \to \infty$, it holds as $r \to \infty$. $\endgroup$ – nullUser Jul 13 '13 at 20:00
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Split it into two pieces: one integral over $[0,\epsilon]$ and another over $[\epsilon, \pi/2]$. Since the integrand is bounded on the first piece, you can make it arbitrarily small by choosing $\epsilon$ small. On the other piece, it converges uniformly to zero. I think you can take it from there.

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Note that, the integral converges uniformly, since

$$ e^{-r\sin(\theta)} \leq e^{-\sin(\theta)}, \quad r\geq 1, $$

which justifies changing the limit with the integral and the answer is $0$.

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  • $\begingroup$ Hmmm...I thought I read that this was not a valid operation. For instance, define $F(x) = \int_0^\infty x^2 t e^{-xt}\, dt$ for $x\geq 1$. Then by substitution $s=xt$, we find the integral is equal to 1 for all values of $t$, so $\lim_{x\to\infty}F(x)=1$. Of course this disagrees with what we get if we interchange the limit, since this would be zero. And we can show that the integral converges uniformly on $x\geq 1$. (I've taken this example from p. 268-269 of Buck's Advanced Calculus.) $\endgroup$ – Eric Auld Jul 13 '13 at 20:04
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    $\begingroup$ @EricAuld I think this is just the DCT he's quoting. $\endgroup$ – Potato Jul 13 '13 at 20:05
  • $\begingroup$ Oh, OK. This makes sense, because the example I gave is not amenable to the Weierstrass test for convergence. I see now that anytime you can use the Weierstrass test to show uniform convergence, you have in fact satisfied the hypotheses of the DCT as well. $\endgroup$ – Eric Auld Jul 13 '13 at 20:10
  • $\begingroup$ @EricAuld: I expect you that you know M-test for uniform convergence which is more natural to tackle this problem since DCT is more advanced for you. $\endgroup$ – Mhenni Benghorbal Jul 13 '13 at 20:14
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    $\begingroup$ @MhenniBenghorbal: the limit function is not continuous: $$ \lim_{r\to\infty}e^{-r\sin(\theta)}=\left\{\begin{array}{} 0&\text{for }0\lt\theta\le\frac\pi2\\ 1&\text{for }\theta=0 \end{array}\right. $$ Therefore, the convergence is not uniform. $\endgroup$ – robjohn Sep 5 '13 at 10:41
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This can also be handled by Monotone Convergence.

The functions $f_r(\theta)=1-e^{-r\sin(\theta)}$ monotonically converge to $$ \lim_{r\to\infty}f_r(\theta)=f(\theta)=\left\{\begin{array}{} 1&\text{if }0\lt\theta\le\frac\pi2\\ 0&\text{if }\theta=0 \end{array}\right. $$ Thus, $$ \begin{align} \lim_{r\to\infty}\int_0^{\pi/2}e^{-r\sin(\theta)}\,\mathrm{d}\theta &=\lim_{r\to\infty}\int_0^{\pi/2}(1-f_r(\theta))\,\mathrm{d}\theta\\ &=\frac\pi2-\lim_{r\to\infty}\int_0^{\pi/2}f_r(\theta)\,\mathrm{d}\theta\\ &=\frac\pi2-\int_0^{\pi/2}f(\theta)\,\mathrm{d}\theta\\[4pt] &=\frac\pi2-\frac\pi2\\[9pt] &=0 \end{align} $$

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Very simple trick:

By studying the function $[0,\frac{\pi}{2}]\ni\theta \mapsto\frac{\sin\theta}{\theta}$ that

$$ \color{blue}{\sin\theta \geq \frac{2}{\pi}\theta ~~ \forall \theta\in [0,\frac{\pi}{2}] } $$ therefore we get that $$\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-R\sin\theta}d\theta\leq\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-\frac{2R}{\pi}\theta}d\theta =\lim_{R\to\infty}\frac{\pi}{2R}(1-e^{-R}) =0$$

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