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$\displaystyle \lim_{r \to \infty} \int_{0}^{\pi} e^{-r\cos^2(\theta)} d\theta$

Here is what I did:

Since $e^{-r\cos^2(\theta)} $ is continuous on $[0,\pi]$ for any fixed $r$, we can use MVT: There is $c \in (0,\pi)$ such that

$\displaystyle \int_{0}^{\pi} e^{-r\cos^2(\theta)} d\theta = e^{-r\cos^2(c)}\pi$

Also $e^x>0 \forall x \in \mathbb{R}$, so:

$\displaystyle 0 \leq \int_{0}^{\pi} e^{-r\cos^2(\theta)} d\theta = e^{-r\cos^2(c)}\pi$

Using squeeze theorem, $\displaystyle \lim_{r \to \infty} 0 = 0 = \lim_{r \to \infty} e^{-r\cos^2(\theta)}$, then we have $\displaystyle \lim_{r \to \infty} \int_{0}^{\pi} e^{-r\cos^2(\theta)} d\theta=0$

Is this correct? Thanks.

Also: I cannot pass the limit inside the integral, right? But why not, exactly?

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    $\begingroup$ Don't forget that $c$ depends on $r$. This means computing the limit as $r \to \infty$ of $e^{-r\cos^2(c)}\pi$ is a bit harder than it looks. Also, if $c = \pi/2$, then the limit is $1$ not $0$. $\endgroup$ – JimmyK4542 Sep 13 '18 at 3:30
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    $\begingroup$ Do you know Lebesgue's dominated convergence theorem? If not, split the integral in three with a tiny neighborhodd arounf $\pi/2$. $\endgroup$ – amsmath Sep 13 '18 at 3:37
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For any $r\ge 0$ and for any $\theta$ the following inequality holds $e^{-r\cos^2\theta}\le 1$. Choose a sequence of positive real numbers $(r_n)$ that runs to the infinity. Then $$ e^{-r_n\cos^2\theta}\longrightarrow \begin{cases} 0 \text{ if } \theta\neq\frac{\pi}{2}\\ 1 \text{ if } \theta = \frac{\pi}{2} \end{cases}$$ pointwise as $n$ goes to infinity (but not uniformly). Moreover such a limit does not depend on the sequence we have chosen before. Now apply the dominated Lebesgue’s theorem to any sequence to pass the limit under the integral (indeed, the pointwise convergence is sufficient, you don’t need the uniform convergence). Hence:

$$\lim_{r_n\to\infty}\int_0^\pi e^{-r_n\cos^2\theta}=\int_0^\pi \lim_{r_n\to\infty} e^{-r_n\cos^2\theta}=0$$

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  • $\begingroup$ The limit do not depending on the sequence we have chosen means the sequence converges uniformly, and that is why I can pass the limit inside? $\endgroup$ – creepyrodent Sep 13 '18 at 4:12
  • $\begingroup$ Also, can you explain this theorem in undergrad level? I did not take measure theory yet. $\endgroup$ – creepyrodent Sep 13 '18 at 4:33
  • $\begingroup$ No, it only means that the limit function is always the same. The dominated Lebesgue’s theorem states that: if you have a sequence of functions $f_n$ such that the modulus of every function is bounded by the same function $g$, the the limit passes under the sign of integral. $\endgroup$ – InsideOut Sep 13 '18 at 4:46
  • $\begingroup$ See also en.m.wikipedia.org/wiki/Dominated_convergence_theorem $\endgroup$ – InsideOut Sep 13 '18 at 4:46
  • $\begingroup$ "Notice that $e^{-r\cos^2\theta}\le e^{-r}$ for any $\theta\in [0,\pi]$." False. $\endgroup$ – zhw. Sep 13 '18 at 5:09
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We may notice that

$$ 0<I(r)=\int_{0}^{\pi}e^{-r\cos^2\theta}\,d\theta = \int_{-\pi/2}^{\pi/2}e^{-r\sin^2\theta}\,d\theta = 2\int_{0}^{\pi/2}e^{-r\sin^2\theta}\,d\theta$$ but for any $\theta\in\left[0,\pi/2\right]$ we have $\sin^2\theta\geq\frac{\theta^2}{3} $, hence $$ I(r) \leq 2\int_{0}^{\pi/2} e^{-r\theta^2/3}\,d\theta < 2\int_{0}^{+\infty} e^{-r\theta^2/3}\,d\theta = \sqrt{\frac{3\pi}{r}}$$ and the wanted limit is clearly zero by squeezing.

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