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