Doubly Infinite Integral

$\sinh^{2n}x \geq 0$

Notice that when $|\sinh x| >1, \lim_{n\to\infty} \sinh^{2n}x = \infty$

Therefore $\lim_{n\to\infty} e^{x-\sinh^{2n}x}$ tends to $e^{-\infty} = 0$ in the range $|\sinh x| >1$ so we need not integrate in this region.

Also notice that when $-1 < \sinh x <1, \lim_{n\to\infty} \sinh^{2n}x = 0$

Therefore $\lim_{n\to\infty} e^{x-\sinh^{2n}x}$ tends to $e^{x}$ in the range $-1 < \sinh x <1$

Note $arsinh x$ is defined as $\ln(x + \sqrt{1+x^2})$

Therefore the range of values that we must integrate $e^x$ is $arsinh (-1)$ to $arsinh (1)$

$\lim_{n\to\infty} \int_{-\infty}^{\infty} e^{[x-\sinh^{2n}x]} dx = \int_{\ln(-1 + \sqrt{2})}^{\ln(1 + \sqrt{2})} e^x dx= \boxed{2}$