# Compute $\lim_{n\to\infty}\int_0^{\pi/2} \sin(x^n)dx$

How to compute that $$\lim_{n\to\infty}\int_0^{\pi/2} \sin(x^n)dx$$ $$\lim_{n\to\infty}n^{106}\int_0^{1/n^{2017}} f(x)dx$$ where $f$ is arbitrary integrable function on $[0,1]$

I have tried to swap the limit and integral sign, and get $\int_0^{\pi/2}\displaystyle\lim_{n\to\infty}\sin(x^n)dx$, but there seems no further result can get.. And also the uniform convergence seemingly cannot be justified.

• What is $f$ ? And why did swapping did not work ? – Atmos Feb 4 '18 at 14:46
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• Do you mean $n^{106}$ or $n^{2016}$? – Connor Harris Mar 12 '18 at 17:25

$$\frac{1}{n}\int_{0}^{(\pi/2)^n}\frac{\sin z}{z^{1-1/n}}\,dz=o\left(\frac{1}{n}\right)+\left(\frac{1}{n}-\frac{1}{n^2}\right)\underbrace{\int_{0}^{(\pi/2)^n}\frac{1-\cos z}{z^{2-1/n}}\,dz}_{O(1)}$$ holds by integration by parts and this gives that the first limit is zero. Try to apply this technique to the second limit, too.