I am trying to prove the convergence of the integral
$\displaystyle \int_1^\infty e^{-2^{2x} \sin(\pi x)^2} dx$.
The graph clearly oscillates between $0$ and $1$, but the width of the peaks gets smaller and smaller as $x$ tends to $\infty$. How can it be proven that the width gets sufficiently small for the integral to converge?
You can prove the integral converges with some fairly crude bounds. The function $\sin^2(\pi x)$ is $1$-periodic. Thus \begin{align*}I :=& \int^\infty_1 e^{-2^{2x}\sin^2(\pi x)}dx \\=& \sum_{n=1}^\infty \int_n^{n+1}e^{-2^{2x}\sin^2(\pi x)}dx \\=& \sum^\infty_{n=1} \int^1_0 e^{-2^{2(y+n)}\sin^2(\pi y)}dy\\ \le& \sum^\infty_{n=1}\int^1_0 e^{-2^{2n}\sin^2(\pi y)} dy \\ =& \,\,2 \sum^\infty_{n=1} \int^{1/2}_0 e^{-2^{2n} \sin^2(\pi y)} dy.\end{align*} Now for $y \in [0,1/2]$, we have $\pi y/2 \le \sin(\pi y).$ Thus continuing, \begin{align*}I \le& \,\,2 \sum^\infty_{n=1}\int^{1/2}_{0} e^{-2^{2n-2}\pi^2 y^2 } dy\\ \le&\,\,2\sum^\infty_{n=1} \int^\infty_{0}e^{-2^{2n-2}\pi^2 y^2 } dy \\ =&\,\, 2 \sum^\infty_{n=1} \frac{\sqrt \pi}{2\sqrt{2^{2n-2}\pi^2}} =\frac{1}{\sqrt \pi}\sum^\infty_{n=1} \frac{1}{2^{n-1}} <\infty. \end{align*} Here I used the Gaussian integral: $\int^\infty_0 e^{-ay^2}dy = \frac{1}{2} \sqrt{\frac{\pi}{a}}$.