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

Question

To study the question, I'm looking at the convergence of the series

$$u_k = \int_{k \pi}^{(k+1) \pi}\vert \sin x \vert^{x^2} \ dx,$$ using the inequalities

$$2 \int_0^{\frac{\pi}{2}} \sin^{(k+1)^2\pi^2} x \ dx \le u_k \le 2 \int_0^{\frac{\pi}{2}} \sin^{k^2 \pi^2} x \ dx.$$

But I'm not able to get a good approximate of

$$v_k = \int_0^{\frac{\pi}{2}} \sin^{k^2 \pi^2} x \ dx$$

Any good idea?

Answer 1

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 $a \to \infty$, $$ \int_0^{\frac{\pi}{2}} \sin^{a} x \ dx = \sqrt{\frac{\pi}2}\frac1{\sqrt{a}}+O\left(\frac1{a^{3/2}} \right). $$

Answer 2

I finally found another solution based on more basic technics.

$$\begin{align} v_k&= \int_0^{\frac{\pi}{2}} \sin^{k^2 \pi^2} x \ dx = \int_0^{\frac{\pi}{2}} \cos^{k^2 \pi^2} x \ dx\\ &\ge \int_0^{\frac{\sqrt{2}}{k \pi}} \cos^{k^2 \pi^2} x \ dx\\ &\ge \int_0^{\frac{\sqrt{2}}{k \pi}} (1-\frac{x^2}{2})^{k^2 \pi^2} \ dx\\ &\ge \frac{\sqrt{2}}{k\pi} \left(1-\frac{1}{k^2 \pi^2}\right)^{k^2\pi^2} \sim\frac{\sqrt{2}}{e k \pi} \end{align}$$

Based on the inequality

$$\cos x\ge 1-\frac{x^2}{2} \text{ for } x \ge 0$$

Hence the integral diverges as the harmonic series diverges.