Proving $\int_0^{\pi/4}\sin^n(x)\,dx>\frac{1}{2^{n/2}(n+2)}$

Question

I want to prove:$$\int_0^{\pi/4}\sin^n(x)\,dx>\frac{1}{2^{n/2}(n+2)}$$

This came up when I was working on this question that only asked for elementary calculus solution. Some trivial lower bounds such as: $$ \sin(x)\geq\frac{2\sqrt{2}}{\pi}x$$ or even a slightly stronger one: $$\sin(x)\geq \dfrac{3}{\pi}x\cdot 1_{[0,\pi/6]}+\left(\frac{6(\sqrt{2}-1)}{\pi}x+\dfrac{3-2\sqrt{2}}{2}\right)\cdot 1_{[\pi/6, \pi/4]}$$ were not tight enough to prove the assertion.

Ideally, one could find the asymptotic expansion of: $$n2^{n/2}\int_0^{\pi/4}\sin^n(x)\,dx$$ which can be seen to be converging to $1.$

Answer 1

I would (optionally) substitute $\sin x=t$ and integrate by parts twice: \begin{align}\int_0^{1/\sqrt{2}}\frac{t^n\,dt}{(1-t^2)^{1/2}}&=\frac{1}{n+1}\left(\left.\frac{t^{n+1}}{(1-t^2)^{1/2}}\right|_0^{1/\sqrt{2}}-\int_0^{1/\sqrt{2}}\frac{t^{n+2}\,dt}{(1-t^2)^{3/2}}\right)\\&=\frac{1}{n+1}\left(2^{-n/2}-\frac{1}{n+3}\left.\frac{t^{n+3}}{(1-t^2)^{3/2}}\right|_0^{1/\sqrt{2}}+\ldots\right)\\&>\frac{2^{-n/2}}{n+1}\left(1-\frac{1}{n+3}\right)=2^{-n/2}\frac{n+2}{(n+2)^2-1}>\begin{bmatrix}\text{what}\\ \text{we}\\ \text{need }\end{bmatrix}\end{align}

Answer 2

$$I(n)=\int_0^{\pi/4}\sin^n(x)dx=\int_0^{\pi/4}\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^ndx$$ $$I(n)=2^{-n}(-i)^n\int_0^{\pi/4}\left(e^{ix}-e^{-ix}\right)^ndx=2^{-n}(-i)^n\int_0^{\pi/4}\sum_{r=0}^n\begin{pmatrix}n\\r\end{pmatrix}(e^{ix})^{n-r}(e^{-ix})^rdx$$ $$I(n)=2^{-n}(-i)^n\int_0^{\pi/4}\sum_{r=0}^n\begin{pmatrix}n\\r\end{pmatrix}e^{ix(n-2r)}dx$$ You could try and get something from this

Answer 3

Remark: @metamorphy gave a very nice solution by IBP. Can we use bounds for the integrand instead? I want to give some alternative solutions.

With the substitution $\sin x = t$, we have $$\int_0^{\pi/4} \sin^n x \mathrm{d} x = \int_0^{1/\sqrt{2}} \frac{t^n}{\sqrt{1 - t^2}} \mathrm{d} t.$$

Since $\frac{1}{\sqrt{1 - t^2}} \ge 2t$ for all $t$ in $[0, 1/\sqrt{2}]$ (with equality if and only if $t = 1/\sqrt{2}$), we have $$\int_0^{1/\sqrt{2}} \frac{t^n}{\sqrt{1 - t^2}} \mathrm{d} t > \int_0^{1/\sqrt{2}} 2t^{n+1} \mathrm{d} t = \frac{1}{2^{n/2}(n + 2)}.$$

We are done.

Note: If we use better lower bounds for the integrand, we get better lower bounds for the integral.

About the lower bound $2t$: The Taylor expansion of $\frac{1}{\sqrt{1 - t^2}}$ around $t = \tfrac{1}{\sqrt{2}}$ is $\sqrt{2} + 2(t - \tfrac{1}{\sqrt{2}}) + \cdots$. It is easy to prove that $\sqrt{2} + 2(t - \tfrac{1}{\sqrt{2}}) = 2t$ is a lower bound for $\frac{1}{\sqrt{1 - t^2}}$.