# Proving that $\lim_{n \rightarrow\infty} \int_{0}^{\frac{\pi}{2}} \sin(t^n) dt =0$

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 ?

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.

• Nice. For those wondering how to deal with the $\int_0^1$, use the dominated convergence theorem. – Gabriel Romon Jan 29 '14 at 17:40
• @GabrielR. Or simply the bound $\sin u\leqslant u$ for $u=t^n$. – Did Jan 29 '14 at 17:46

Begin by doing this substitution $$\int_{0}^{\pi/2} \sin(t^n) dt = {1\over n}\int_0^{(\pi/2)^n} { \sin(t)\over t^{1 - 1/n}} dt.$$ The integrals we see oscillate and if you let the upper limit go to $\infty$, they are conditionally convergent. If they can be uniformly bounded, you have this, since we have the $1/n$ factor. I am not sure how to bound the integrals, but this seems like a viable route to solving the problem.

Nice problem. What I would try:

Pick l, m, r so that t^n = (2k)pi, (2k+1)pi, (2k+2)pi when t = l, m, r; k >= 1. Instead of integrating sin (t^n) from l to r, integrate sin (t^n) from l to m, then use a substitution so you integrate scalefactor (t) * sin (t^n + pi) from l to m instead of sin (t^n) from m to r. The integral from l to r is the integral of (1 - scalefactor (t)) * sin (t^n). Show that the scale factor is very close to 1, so the total integral is small. Sum for all integrals, show it is still small and converges to 0 as n -> infinity.

The integral from 0 to (2pi)^(1/n) also goes -> 0.