integration - Proving that $lim_{n rightarrowinfty} int_{0}^{frac{pi}{2}} sin(t^n) dt =0$

It is clear that $\lim_{n \rightarrow\infty} \int_{0}^{1} \sin(t^n) dt =0$. (which is not what is to be proved here)

I don't know how to proceed with the remaining part of the integral ie $\lim_{n \rightarrow\infty} \int_{1}^{\frac{\pi}{2}} \sin(t^n) dt$ (I tried substitution and partial integration).

Can anybody give me a hint ?

Answer

Substituting $u = t^n$, we obtain

$$\int_1^{\pi/2} \sin (t^n)\,dt = \frac{1}{n}\int_1^{(\pi/2)^n} \frac{\sin u}{u^{1-1/n}}\,du.\tag{1}$$

Integration by parts then brings us to (with $c_n = \left(\frac{\pi}{2}\right)^n$)

$$\begin{align}
\left\lvert\int_1^{c_n} \frac{\sin u}{u^{1-1/n}}\,du\right\rvert &= \left\lvert\frac{-\cos u}{u^{1-1/n}}\Biggl\lvert_1^{c_n} - \left(1-\frac{1}{n}\right)\int_1^{c_n}\frac{\cos u}{u^{2-1/n}}\,du\right\rvert\\
&= \left\lvert\cos 1 - \frac{\cos c_n}{c_n^{1-1/n}} - \left(1-\frac{1}{n}\right)\int_1^{c_n}\frac{\cos u}{u^{2-1/n}}\,du\right\rvert\\
&\leqslant 2 + \int_1^{\infty} \frac{du}{u^{2-1/n}}\\
&\leqslant 4,
\end{align}$$

for $n \geqslant 2$.

Use the factor $\frac{1}{n}$ from $(1)$, and the result for the $\int_0^1$ part to conclude.