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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.

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  • $\begingroup$ 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? $\endgroup$ Commented Oct 31, 2016 at 12:43
  • $\begingroup$ If you are interested it is possible to calculate the closed form of these integrals. $\endgroup$ Commented Oct 31, 2016 at 14:04
  • $\begingroup$ You mean using the Gamma Function, right? $\endgroup$
    – Student
    Commented Oct 31, 2016 at 14:41
  • $\begingroup$ Yes, using $$\int_{0}^{\pi/2}\sin^{n}(x)\cos^{m}(x)dx=\frac{B\left(\frac{n+1}{2},\frac{m+1}{2}\right)}{2}.$$ $\endgroup$ Commented Oct 31, 2016 at 16:10

2 Answers 2

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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 $\text{n}\in\mathbb{N}$: $$\mathcal{I}^2_{1+2\text{n}}=\prod_{\text{k}=1}^{\frac{1}{2}+\text{n}}\left\{\frac{2\left(1-\text{k}+\text{n}\right)}{3-2\text{k}+2\text{n}}\right\}^2$$
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  • $\begingroup$ Does this prove the double inequality the OP asks about? $\endgroup$
    – Did
    Commented Nov 1, 2016 at 21:10
  • $\begingroup$ @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 ) $\endgroup$ Commented Nov 1, 2016 at 21:40
  • $\begingroup$ 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... $\endgroup$
    – Did
    Commented Nov 1, 2016 at 21:42
  • $\begingroup$ @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. $\endgroup$
    – Student
    Commented Nov 4, 2016 at 14:50
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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

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