Compute: as $\displaystyle{\lim_{n\to \infty} \int_0^\pi \sin^n x dx}$

Question

$$\lim_{n\to \infty} \int_0^\pi \sin^n x dx$$

I have been trying to compute this integral for months and have gotten nowhere. I seem to be going in circles. Any help at all would be seriously appreciated.

Answer 1

This integral will be zero- the result follows from the dominated convergence theorem:

$|(\sin(x))^n|<1$ and $1$ is integrable on the finite measure space $[0,\pi]$ and $\sin(x))^n \rightarrow 0$ a.e.

So, by dominated convergence theorem, the integral is $0$.

Answer 2

As derived here, we have

$$\int_0^{\pi} d\theta \, \sin^n{\theta} = \begin{cases}\displaystyle \frac{\pi}{2^{2 k}} \binom{2 k}{k} & n=2 k \\ \displaystyle \frac{2^{2 k}}{\displaystyle k \binom{2 k}{k}} & n = 2 k-1 \end{cases} $$

Now use the Stirling approximation, i.e.,

$$\binom{2 k}{k} \sim \frac{(2 k)^{2 k} e^{-2 k} \sqrt{2 \pi 2 k}}{k^{2 k} e^{-2 k} (2 \pi k)} = \frac{2^{2 k}}{\sqrt{\pi k}}$$

Thus,

$$\int_0^{\pi} d\theta \, \sin^n{\theta} \sim \begin{cases}\displaystyle \sqrt{\frac{\pi}{k}} & n=2 k \\ \displaystyle\sqrt{\frac{\pi}{k}} & n = 2 k-1 \end{cases} $$

i.e.,

$$\int_0^{\pi} d\theta \, \sin^n{\theta} \sim \sqrt{\frac{2 \pi}{n}} \quad (n \to \infty)$$

which vanishes in the limit of $n \to \infty$.

ADDENDUM

This agrees with analysis using Laplace's method, i.e., express the integral as

$$\int_0^{\pi} d\theta \, e^{n \log{\sin{\theta}}}$$

Expand about the critical point, and we have, to third order,

$$\int_0^{\pi} d\theta \, e^{-(n/2) (\theta-\pi/2)^2}$$

We may then extend the integration interval to the real line with exponentially small error; the above result follows.

Answer 3

You don't need to compute the integral: $$\int_0^\pi\sin^n x=2\int_0^{\pi/2}\sin^n x=2\int_0^{\pi/2-\varepsilon}\sin^n x+2\int_{\pi/2-\varepsilon}^{\pi/2}\sin^n x\le\\2\left(\frac\pi2-\varepsilon\right)\sin^n\left(\frac\pi2-\varepsilon\right)+2\varepsilon\overset{n\to\infty}{\longrightarrow}2\varepsilon$$ for arbitrarily small $\varepsilon>0$.

Answer 4

Just think logically. $\sin x \le 1 $ from $0$ to $\pi$, and when you take something below $1$ and above $0 $to an extremely high value, it approaches $0$ because each time you multiply it by itself it gets smaller. There is only $1$ point where $\sin x=1$, at $x = \pi / 2$. These points have no width, so as n goes to infinity $\int^\pi_0 (\sin x)^n$ goes to $0$.

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\lim_{n \to \infty} \int_{0}^{\pi}\sin^{n}\pars{x}\,\dd x} = \lim_{n \to \infty} \int_{-\pi/2}^{\pi/2}\cos^{n}\pars{ x}\,\dd x = 2\lim_{n \to \infty}\int_{0}^{\pi/2}\cos^{n}\pars{ x}\,\dd x \end{align} The main contribution to the integration comes from values of $\ds{x \gtrsim 0}$. Then, \begin{align} &\bbox[5px,#ffd]{\lim_{n \to \infty} \int_{0}^{\pi}\sin^{n}\pars{x}\,\dd x} = 2\lim_{n \to \infty} \int_{0}^{\pi/2}\expo{n\ln\pars{\cos\pars{x}}}\,\,\,\dd x \\[5mm] = &\ 2\lim_{n \to \infty} \int_{0}^{\infty}\expo{-nx^{2}\,/2}\,\,\dd x\qquad \pars{~Laplace\,'s\ Method~} \\[5mm] = &\ 2\lim_{n \to \infty}\pars{\root{2 \over n}{\root{\pi} \over 2}} = \bbx{0} \\ & \end{align}