# 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:) – Guy Fsone Oct 31 '17 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. – Eric Auld Jul 13 '13 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 $r>0$. – Eric Auld Jul 13 '13 at 20:29
• Yes, a little mistake. Of course there must not be minus there. – cool Jul 13 '13 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... – Potato Jul 13 '13 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! – Eric Auld Jul 13 '13 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! – Potato Jul 13 '13 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. – Potato Jul 13 '13 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 $r \to \infty$. – nullUser Jul 13 '13 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.) – Eric Auld Jul 13 '13 at 20:04
• @EricAuld I think this is just the DCT he's quoting. – Potato Jul 13 '13 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. – Eric Auld Jul 13 '13 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. – Mhenni Benghorbal Jul 13 '13 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 '13 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 $$\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-R\sin\theta}d\theta\leq\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-\frac{2R}{\pi}\theta}d\theta =\lim_{R\to\infty}\frac{\pi}{2R}(1-e^{-R}) =0$$