Rate of decay of $I_n=\int^\infty_0 e^{-t} \big|\sin(nt +b_n)\big|^n\,dt$

Question

This is an old qualifier problem:

1. Show that the integral \begin{align*} I_n=\int^\infty_0 e^{-t}\big|\sin(nt+b_n)\big|^n\, dt \end{align*} converges to $0$ as $n\rightarrow\infty$, where $b_n$ is an arbitrary numerical sequence.
2. For which $p>0$ does $n^pI_n$ converge?

1. The change of variable $x=nt$ followed by the splitting $(0,\infty)=\bigcup^\infty_{k=0}(k\pi,(k+1)\pi]$ gives \begin{align*} I_n&=\frac{1}{n}\sum^\infty_{k=0}\int^{(k+1)\pi}_{k\pi} e^{-\frac{x}{n}}\big|\sin(x+b_n)\big|^n\,dx \\ &= \frac{1}{n}\sum^\infty_{k=0}e^{-\frac{k\pi}{n}}\int^{\pi}_0 e^{-\frac{x}{n}}\big|\sin(x+b_n)\big|^n\,dx \end{align*} Since $|\sin|$ has period $\pi$, $$ \frac1ne^{-\pi/n}\Big(\sum^\infty_{k=0} e^{-\frac{\pi(k+1)}{n}}\Big)\int^\pi_0|\sin x|^n\,dx \leq I_n\leq \frac1n\Big(\sum^\infty_{k=0} e^{-\frac{k\pi}{n}}\Big)\int^\pi_0 |\sin x|^n\, dx $$This provides answer to the first question.
2. The second part of the problem is where I am not making much progress. I think a good asymptotics on the right-hand side is all that is needed. A hint will be appreciated.

Answer 1

$$\int^\pi_0 \sin^n (x)\, dx=\sqrt{\pi }\,\frac{ \Gamma \left(\frac{n+1}{2}\right)}{\Gamma \left(\frac{n+2}{2}\right)}$$ So, we consider

$$A_n=\sqrt{\pi }\,\frac{ \Gamma \left(\frac{n+1}{2}\right)}{n\,\Gamma \left(\frac{n+2}{2}\right)}\,\,\frac{e^{\frac{\pi }{n}}}{e^{\frac{\pi }{n}}-1}$$

Take logarithms, use Stirling approximation and Taylor series, exponentiate to end with $$A_n=\sqrt{\frac{2}{n\pi }}\,\,\left(1+\frac{2 \pi -1}{4 n}+\frac{3-12 \pi +8 \pi ^2}{96 n^2}+O\left(\frac{1}{n^3}\right)\right)$$ which is extremely accurate.

The error of this approximation is $\sim \frac 1{10\, n^{7/2}}$

The absolute relative error is $1$% for $n=2$, $0.1$% for $n=5$, $0.01$% for $n=10$.

Answer 2

In $\int^\pi_0 |\sin x|^n\, dx $ you can drop the modulus signs and then there are formulae (different for even and odd $n$) for this integral in terms of factorials (or you can do this yourself with a reduction formula); then approximate the factorials with Stirling's formula. For $n$ even = $2r$, for instance, this gives, asymptotically, $\int^\pi_0 \sin^{2r} x\, dx \sim \sqrt{\dfrac {\pi }{r}}$. The sum $\sum^\infty_{k=0} e^{-\frac{k\pi}{n}}$ is geometric, and then using the first few terms of the Maclaurin series for $e^x$ will give you good bounds: $\frac{n}{\pi} <\sum^\infty_{k=0} e^{-\frac{k\pi}{n}} < 1+\frac{n}{\pi}$.