Equivalent of $ u_{n}=\int_0^{\pi/2} \cos\left(\frac{\pi}{2}\sin(x)\right)^n \mathrm dx $

Question

I would like to find an equivalent of the sequence $u_{n}$ where

$$ u_{n}=\int_0^{\pi/2} \cos\left(\frac{\pi}{2}\sin(x)\right)^n \mathrm dx $$

The substitution $x\rightarrow \frac{\pi}{2}\sin(x)$ gives:

$$ u_{n}=\frac{2}{\pi}\int_0^{\pi/2} \frac{\cos(x)^n}{\sqrt{1-(\frac{2x}{\pi})^2}} \mathrm dx$$

Numerical values seem to show that:

$$ \int_0^{\pi/2} \frac{\cos(x)^n}{\sqrt{1-(\frac{2x}{\pi})^2}} \mathrm dx \sim_{n\rightarrow \infty} \int_0^{\pi/2} \cos(x)^n \mathrm dx \sim_{n\rightarrow \infty} \sqrt{\frac{\pi}{2n}}$$

And $$ u_{n} \sim_{n\rightarrow \infty} \sqrt{\frac{2}{\pi n}}$$

So my question is:

How can I show that

$$ \int_0^{\pi/2} {\cos(x)^n}\left(\frac{1}{\sqrt{1-(\frac{2x}{\pi})^2}}-1 \right) \mathrm dx=_{n\rightarrow \infty} o(1/\sqrt{n}) $$ ?

Answer 1

For you to conclude $ \displaystyle u_n \sim \sqrt{\frac{2}{n\pi} }$ you will need to show $$I(n)=\int^{\pi/2}_0 \cos^n(x) \left( \frac{1}{\sqrt{1-4x^2/\pi^2} }-1 \right) dx = \mathcal{o}\left(\frac{1}{\sqrt{n}}\right)$$ i.e not just $\mathcal{o}(1).$ The main contribution to the integral is near $x=0$ and the integrand drops off exponentially away from there. Near $x=0$ we have (by Taylor expansion) $\cos^n (x) \approx(1-x^2/2)^n \approx \exp(-nx^2/2)$ and $\displaystyle \frac{1}{\sqrt{1-4x^2/\pi^2} }-1 \approx 2x^2/\pi^2$ so $$I(n) \approx \int^{\infty}_0 \frac{2x^2}{\pi^2} \exp(-nx^2/2) dx = \sqrt{\frac{2}{n^3\pi^3}}.$$

Thus one would reasonably conjecture that $$ u_n = \sqrt{\frac{2}{n\pi}} \left( 1+ \frac{2}{n\pi^2} + \mathcal{O}\left(\frac{1}{n^2} \right) \right).$$