# Prove the Integral Inequality.

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.

• To get $\pi$ in the middle, how about doing the $\left(\int_{0}^{\pi/2}\sin^n(x)dx\right)^2$ as a double integral, one in $x$ and one in $y$, then change to polar coordinates? – Lee David Chung Lin Oct 31 '16 at 12:43
• If you are interested it is possible to calculate the closed form of these integrals. – Marco Cantarini Oct 31 '16 at 14:04
• You mean using the Gamma Function, right? – nls Oct 31 '16 at 14:41
• Yes, using $$\int_{0}^{\pi/2}\sin^{n}(x)\cos^{m}(x)dx=\frac{B\left(\frac{n+1}{2},\frac{m+1}{2}\right)}{2}.$$ – Marco Cantarini Oct 31 '16 at 16:10

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 $\text{n}\in\mathbb{N}$: $$\mathcal{I}^2_{1+2\text{n}}=\prod_{\text{k}=1}^{\frac{1}{2}+\text{n}}\left\{\frac{2\left(1-\text{k}+\text{n}\right)}{3-2\text{k}+2\text{n}}\right\}^2$$
• Does this prove the double inequality the OP asks about? – Did Nov 1 '16 at 21:10
• @Did It gives him clues. And a downvote is for wrong answers everything I state in this answer is right so.... but I know your problem (downvoting my answer is part of your hobby ) – Jan Eerland Nov 1 '16 at 21:40
• Clues for what? Please be specific. "a downvote is for wrong answers" No. Not at all the idea. Did you try to read how the site functions? After 18 months, it could be about time to do it... – Did Nov 1 '16 at 21:42
• @JanEerland thank you for writing the answer. After the last results I am still finding it difficult to establish the inequality. Please help. Once again thank you for your time and consideration. – nls Nov 4 '16 at 14:50

Using Wallis sine formula (https://mathworld.wolfram.com/WallisSineFormula.html): $$\int_0^{\pi/2} \sin^n x \mathrm{d} x = \frac{\Gamma(\tfrac{1}{2} + \tfrac{n}{2})}{2\Gamma(1 + \tfrac{n}{2})}\sqrt{\pi}.$$

Thus, it suffices to prove that $$\sqrt{\frac{\pi}{2(n+1)}} \le \frac{\Gamma(\tfrac{1}{2} + \tfrac{n}{2})}{2\Gamma(1 + \tfrac{n}{2})}\sqrt{\pi} \le \sqrt{\frac{\pi}{2n}}$$ or $$\sqrt{\frac{n}{2}}\le \frac{\Gamma(1 + \tfrac{n}{2})}{\Gamma(\tfrac{1}{2} + \tfrac{n}{2})} \le \sqrt{\frac{n+1}{2}}.\tag{1}$$ Recall Kershaw's improvement of Gautschi's inequality [1]: for $$x > 0$$ and $$s \in (0, 1)$$, $$\left(x + \frac{s}{2}\right)^{1-s} < \frac{\Gamma(x+1)}{\Gamma(x+s)} < \left(x - \frac{1}{2} + \sqrt{s + \frac{1}{4}}\right)^{1-s}.$$ By letting $$x = \frac{n}{2}$$ and $$s = \frac{1}{2}$$ in Kershaw's improvement of Gautschi's inequality, we have $$\sqrt{\frac{n}{2}+\frac{1}{4}} < \frac{\Gamma(1 + \tfrac{n}{2})}{\Gamma(\tfrac{1}{2} + \tfrac{n}{2})} < \sqrt{\frac{n}{2}+\frac{\sqrt{3}-1}{2}}.$$ Clearly, the inequality in (1) is true.

[1] D. Kershaw, "Some extensions of W. Gautschi’s inequalities for the gamma function", Math. Comp. 41 (1983) 607–611. https://en.wikipedia.org/wiki/Gautschi%27s_inequality