real analysis - Find $lim_{r to infty} int_{0}^{pi} e^{-rcos^2(theta)} dtheta$

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

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?

Answer

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:

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