Show that: $\lim\limits_{r\to\infty}\int\limits_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta=0$

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?

• You may have a look to my answer:) Oct 31, 2017 at 18:27

It's only enough to show that

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

• Great answer! Very simple. Jul 13, 2013 at 20:06
• I think the minus sign before $\frac{\pi}{2r}$ should be omitted, right? There's no way the integral would be negative for any $r>0$. Jul 13, 2013 at 20:29
• Yes, a little mistake. Of course there must not be minus there.
– cool
Jul 13, 2013 at 20:40

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 $$\lim_{r \to \infty} \int_0^{{\pi/2}} e^{-r \sin \theta}\text d\theta = \int_0^{{\pi/2}}\lim_{r \to \infty} e^{-r \sin \theta}\text d\theta = \int_0^{\pi/2}0 \text d\theta = 0.$$

• While correct, you must realize this doesn't help him. He doesn't know Lebesgue integration yet... Jul 13, 2013 at 19:51
• @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! Jul 13, 2013 at 19:52
• @EricAuld Why are you using the uniform convergence criteria for interchanging limits and integrals then!? Use and abuse the dominated convergence theorem! Jul 13, 2013 at 19:54
• @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. Jul 13, 2013 at 20:00
• @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 $r \to \infty$. Jul 13, 2013 at 20:00

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.

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 $0$.

• 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.) Jul 13, 2013 at 20:04
• @EricAuld I think this is just the DCT he's quoting. Jul 13, 2013 at 20:05
• 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. Jul 13, 2013 at 20:10
• @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. Jul 13, 2013 at 20:14
• @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.
– robjohn
Sep 5, 2013 at 10:41

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}

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 $$\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-R\sin\theta}d\theta\leq\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-\frac{2R}{\pi}\theta}d\theta =\lim_{R\to\infty}\frac{\pi}{2R}(1-e^{-R}) =0$$