Prove the Integral Inequality.

Question

Prove the Inequality: $$\frac{\pi}{2(n+1)}\leq\left(\int_{0}^{\frac{\pi}{2}}\sin^n(x)dx\right)^2\leq\frac{\pi}{2n},\text{ where }n\in\mathbb{N}.$$

My attempt: The only way I think one could get a $\pi$ term on both sides would be by using the following inequality: $$\frac{2x}{\pi}\leq\sin(x)\leq x,\text{ for }x\in[0,\pi/2]$$

And also the squaring of the integral suggests that there might be a subtle use of Cauchy-Schwarz-Bunyakovsky Inequality. I tried to find ways in which we could partition the function $\sin^nx$, but none of them worked. Any hints or suggestions regarding the problem would be extremly helpful.

Answer 1

HINT, using integration by parts (where $\text{n}\ge2$):

$$\mathcal{I}_\text{n}=\int_0^{\frac{\pi}{2}}\sin^\text{n}\left(x\right)\space\text{d}x=$$ $$\left[\sin^{\text{n}-1}\left(x\right)\left(-\cos\left(x\right)\right)\right]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}-\cos\left(x\right)\left(\text{n}-1\right)\sin^{\text{n}-2}\left(x\right)\cos\left(x\right)\space\text{d}x=$$ $$0+\left(\text{n}-1\right)\int_0^{\frac{\pi}{2}}\cos^2\left(x\right)\sin^{\text{n}-2}\left(x\right)\space\text{d}x=\left(\text{n}-1\right)\int_0^{\frac{\pi}{2}}\left(1-\sin^2\left(x\right)\right)\sin^{\text{n}-2}\left(x\right)\space\text{d}x=$$ $$\left(\text{n}-1\right)\left\{\int_0^{\frac{\pi}{2}}\sin^{\text{n}-2}\left(x\right)\space\text{d}x-\int_0^{\frac{\pi}{2}}\sin^\text{n}\left(x\right)\space\text{d}x\right\}=\left(\text{n}-1\right)\left(\mathcal{I}_{\text{n}-2}-\mathcal{I}_\text{n}\right)$$

So:

$$\color{red}{\mathcal{I}_\text{n}=\left(\text{n}-1\right)\left(\mathcal{I}_{\text{n}-2}-\mathcal{I}_\text{n}\right)\space\space\space\Longleftrightarrow\space\space\space\mathcal{I}_\text{n}=\frac{\text{n}-1}{\text{n}}\cdot\mathcal{I}_{\text{n}-2}}$$

And, use:

$$\text{As}\space\space\space\space\space0\le\sin(x)\le1\space\space\space\space\space\text{in}\space\space\space\space\space0\le x\le\pi$$

We have:

$$\sin^{2\text{n}+1}(x)\le\sin^{2\text{n}}(x)\le\sin^{2\text{n}-1}(x)$$

---

So, when we square it:

$$\mathcal{I}^2_\text{n}=\left(\int_0^{\frac{\pi}{2}}\sin^\text{n}\left(x\right)\space\text{d}x\right)^2=\left(\frac{\text{n}-1}{\text{n}}\right)^2\cdot\mathcal{I}^2_{\text{n}-2}$$

And, notice that:

$$\mathcal{I}_{\text{n}-2}=\int_0^{\frac{\pi}{2}}\sin^{\text{n}-2}\left(x\right)\space\text{d}x=\left(\text{n}-3\right)\int_0^{\frac{\pi}{2}}\left(1-\sin^2\left(x\right)\right)\sin^{\text{n}-4}\left(x\right)\space\text{d}x=$$ $$\left(\text{n}-3\right)\left\{\int_0^{\frac{\pi}{2}}\sin^{\text{n}-4}\left(x\right)\space\text{d}x-\int_0^{\frac{\pi}{2}}\sin^{\text{n}-2}\left(x\right)\space\text{d}x\right\}=\left(\text{n}-3\right)\left(\mathcal{I}_{\text{n}-4}-\mathcal{I}_{\text{n}-2}\right)$$

So:

$$\color{red}{\mathcal{I}_{\text{n}-2}=\left(\text{n}-3\right)\left(\mathcal{I}_{\text{n}-4}-\mathcal{I}_{\text{n}-2}\right)\space\space\space\Longleftrightarrow\space\space\space\mathcal{I}_{\text{n}-2}=\frac{\text{n}-3}{\text{n}-2}\cdot\mathcal{I}_{\text{n}-4}}$$

So, when we square it:

$$\mathcal{I}^2_\text{n}=\left(\int_0^{\frac{\pi}{2}}\sin^\text{n}\left(x\right)\space\text{d}x\right)^2=\left(\frac{\text{n}-1}{\text{n}}\right)^2\cdot\mathcal{I}^2_{\text{n}-2}=\left(\frac{\text{n}-1}{\text{n}}\right)^2\cdot\left(\frac{\text{n}-3}{\text{n}-2}\right)^2\cdot\mathcal{I}^2_{\text{n}-4}$$

And, notice that:

$$\mathcal{I}_{\text{n}-4}=\int_0^{\frac{\pi}{2}}\sin^{\text{n}-4}\left(x\right)\space\text{d}x=\left(\text{n}-5\right)\int_0^{\frac{\pi}{2}}\left(1-\sin^2\left(x\right)\right)\sin^{\text{n}-6}\left(x\right)\space\text{d}x=$$ $$\left(\text{n}-5\right)\left\{\int_0^{\frac{\pi}{2}}\sin^{\text{n}-6}\left(x\right)\space\text{d}x-\int_0^{\frac{\pi}{2}}\sin^{\text{n}-4}\left(x\right)\space\text{d}x\right\}=\left(\text{n}-5\right)\left(\mathcal{I}_{\text{n}-6}-\mathcal{I}_{\text{n}-4}\right)$$

So:

$$\color{red}{\mathcal{I}_{\text{n}-4}=\left(\text{n}-5\right)\left(\mathcal{I}_{\text{n}-6}-\mathcal{I}_{\text{n}-4}\right)\space\space\space\Longleftrightarrow\space\space\space\mathcal{I}_{\text{n}-4}=\frac{\text{n}-5}{\text{n}-4}\cdot\mathcal{I}_{\text{n}-6}}$$

So, when we square it:

$$\mathcal{I}^2_\text{n}=\left(\int_0^{\frac{\pi}{2}}\sin^\text{n}\left(x\right)\space\text{d}x\right)^2=\left(\frac{\text{n}-1}{\text{n}}\right)^2\cdot\mathcal{I}^2_{\text{n}-2}=$$ $$\left(\frac{\text{n}-1}{\text{n}}\right)^2\cdot\left(\frac{\text{n}-3}{\text{n}-2}\right)^2\cdot\mathcal{I}^2_{\text{n}-4}=\left(\frac{\text{n}-1}{\text{n}}\right)^2\cdot\left(\frac{\text{n}-3}{\text{n}-2}\right)^2\cdot\left(\frac{\text{n}-5}{\text{n}-4}\right)^2\cdot\mathcal{I}^2_{\text{n}-6}$$

Now, in general:

1. When $\text{n}\in\mathbb{N}$: $$\mathcal{I}_{2\text{n}}=\frac{\pi}{2}\prod_{\text{k}=1}^{\text{n}}\frac{1-2\text{k}+2\text{n}}{2\left(1-\text{k}+\text{n}\right)}$$
2. When $\text{n}\in\mathbb{N}$: $$\mathcal{I}_{1+2\text{n}}=\prod_{\text{k}=1}^{\frac{1}{2}+\text{n}}\frac{2\left(1-\text{k}+\text{n}\right)}{3-2\text{k}+2\text{n}}$$

So, when we square it:

1. When $\text{n}\in\mathbb{N}$: $$\mathcal{I}^2_{2\text{n}}=\frac{\pi^2}{4}\prod_{\text{k}=1}^{\text{n}}\left\{\frac{1-2\text{k}+2\text{n}}{2\left(1-\text{k}+\text{n}\right)}\right\}^2$$
2. When $\text{n}\in\mathbb{N}$: $$\mathcal{I}^2_{1+2\text{n}}=\prod_{\text{k}=1}^{\frac{1}{2}+\text{n}}\left\{\frac{2\left(1-\text{k}+\text{n}\right)}{3-2\text{k}+2\text{n}}\right\}^2$$