Comparing the value of $\int _0 ^{\frac{\pi}{2}} x^n \sin^nx\,dx$ for different $n$

Question

Let $$I_n = \int _0 ^{\frac{\pi}{2}} x^n \sin^nx \, dx.$$

Could you tell me how to decide which one is true: $I_{2012}>I_{2013} $, or $I_{2013}>I_{2012}$?

Answer 1

Denote $f(x)=x\sin x$ and for every real number $p>0$, denote $$I_p:=\int_0^{\frac{\pi}{2}}\big(f(x)\big)^pdx.$$

Claim: $$I_n<I_{n+1},\quad\forall ~n\ge 2.\tag{1}$$

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Proof 1(using Jensen's inequality without the evaluation of $I_2$)

Note that $f> 0$ on $(0,\frac{\pi}{2}]$, and $$I_1=\int_0^{\frac{\pi}{2}}f(x)dx=-x\cos x\Big|_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}\cos xdx=1\tag{2}.$$ Then by Jensen's inequality, for every non-constant continuous function $g:[0,\frac{\pi}{2}]\to[0,\infty)$ and every strictly convex function $\varphi:[0,\infty)\to\mathbb{R}$, we have: $$\varphi\big(\int_0^{\frac{\pi}{2}}g(x)f(x)dx\big)<\int_0^{\frac{\pi}{2}}\varphi(g(x))f(x)dx.\tag{3}$$ In particular, for any $p> 0$, $g=f^p\ge 0$ is continuous and not constant on $[0,\frac{\pi}{2}]$, and $\varphi(x)=x^{\frac{p+1}{p}}$ is strictly convex on $[0,\infty)$. Substituting them into $(3)$, we have:

$$\left(I_{p+1}\right)^{\frac{p+1}{p}}< I_{p+2}\iff \left(I_{p+1}\right)^{p+1}<\left(I_{p+2}\right)^p.\tag{4}$$ Since $(4)$ holds for every $p>0$, letting $p\to 0$ in the right hand side of $(4)$ and using $(2)$, we obtain that $I_2\ge 1$. Staring with $n=2$, an induction on $n\in\mathbb N$ using $(4)$ shows that $$I_n\ge 1,\quad\forall~n\ge 2.\tag{5}$$ $(1)$ follows from $(4)$ and $(5)$ immediately. $\quad\square$

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Proof 2(using Cauchy-Schwarz inequality with the evaluation of $I_2$)

For every $n\ge 2$, since $f^{\frac{n-1}{2}}$ and $f^{\frac{n+1}{2}}$ are linearly independent on $[0,\frac{\pi}{2}]$, by Cauchy-Schwarz inequality, $$0<\left(I_n\right)^2=\big(\int_0^\frac{\pi}{2}f^{\frac{n-1}{2}}\cdot f^{\frac{n+1}{2}}dx\big)^2<\int_0^\frac{\pi}{2}f^{n-1}dx\cdot\int_0^\frac{\pi}{2}f^{n+1}dx=I_{n-1}\cdot I_{n+1}.$$ The inequality above implies that $$\frac{I_{n}}{I_{n-1}}<\frac{I_{n+1}}{I_n},\quad\forall~n\ge 2.\tag{6}$$ Due to $(6)$, to prove $(1)$, it suffices to show that $I_2\ge I_1$. A direct calculation shows that $$I_2=\int_0^{\frac{\pi}{2}}x^2\cdot\frac{1-\cos 2x}{2}dx=\frac{\pi^3}{48}+\frac{1}{2}\int_0^{\frac{\pi}{2}}x\sin 2x dx=\frac{\pi^3}{48}+\frac{\pi}{8}>1=I_1,$$ which completes to proof. $\quad\square$

Answer 2

The function $x \sin x$ is increasing in the interval $[0, \frac \pi 2]$, and there is a point $x_0 \approx 1.12$ which satisfies $x_0 \sin {x_0} = 1$.

Split the integral to a sum of two integrals: Integral from $0$ to $x_0$ plus integral from $x_0$ to $\frac \pi 2$.

The first one tends to $0$ as $n \to \infty$ (why?). The second one is monotone increasing in $n$ (transitioning from $n$ to $n+1$ we're multiplying by $x \sin x$ which is more than $1$).

Therefore, up to a small error term which is negligible for large $n$, the integral is monotone increasing in $n$.

Answer 3

Numerically, it would seem that $I_n > I_{n-1}$ for large enough $n$ and that moreover $$\lim_{n \to \infty} \frac{I_n}{I_{n-1}} \sim \frac{\pi}{2}$$ Why is this is so? Note that as $n \to \infty$ the $\sin^n x$ term is zero almost everywhere execpt for $x=\pi/2$ where it's equal to one. This means that the value of the integral is really just given by the $x^n$ at the end point, so we end up with the above limit.