Computing integral unbounded on both sides!

I get $\sqrt2$ as follows:

Since the domain of arcsin is [-1,1], you know that $1/\sqrt{2} \le t \le 1 + 1/\sqrt{2}$

Also, by definition of improper integrals,

$\int_{-\infty}^\infty e^{tx} f(x) dx = \lim_{a \to -\infty} \int_a^0 e^{tx} f(x) dx + \lim_{b \to \infty} \int_0^b e^{tx} f(x) dx = \sin^{-1} \left( t - \frac{1}{\sqrt{2}} \right)$

$\int_{-\infty}^\infty xf(x) dx = \lim_{c \to -\infty} \int_c^0 f(x) dx + \lim_{d \to \infty} \int_0^d xf(x) dx$

With some more bounding, you can get that $\int_{-\infty}^\infty xf(x) dx = \dfrac{\pi \sqrt{2}}{4}$