Is $\int_0^\infty \vert \sin x \vert^{x^2} \ dx$ convergent?

But I'm not able to get a good approximate of $v_k = \int_0^{\frac{\pi}{2}} \sin^{k^2 \pi^2} x \ dx\, $ ($k \to \infty$).

Hint. One may use the Euler beta function to get $$ \int_0^{\frac{\pi}{2}} \sin^{a} x \ dx=\frac{\sqrt{\pi}\: \Gamma\left(\frac{1+a}{2}\right)}{2 \: \Gamma\left(1+\frac{a}{2}\right)} $$ giving, as $a \to \infty$, $$ \int_0^{\frac{\pi}{2}} \sin^{a} x \ dx = \sqrt{\frac{\pi}2}\frac1{\sqrt{a}}+O\left(\frac1{a^{3/2}} \right). $$