Limit of $\int_0^\pi \sin^{2q}(x)\, dx$

Question

I believe (but am certainly not sure) that

$$ \lim_{q\rightarrow\infty} \int \limits_{0}^{\pi} \sin^{2q}{x}dx = 0 $$

simply from plotting each iteration of the function inside the integral as $n$ increases. I would like to prove it though.

My attempt:

I managed to derive the recursive formula $$ B_{2q} = \frac{2q-1}{2q} B_{2q-2} $$ through some long winded calculation involving the more general form of Wallis' integral, which I believe to be correct.

I then think I can safely say $B$ is monotone decreasing, and bounded below by $0$, so by the MCT a limit exists. It seems clear it would be $0$, but I'm having a hard time formulating the end of the proof because of the fractional $n$'s in the recursive relation when attempting to assume the limit exists.

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Furthermore, this whole approach feels really roundabout to me, as if I'm missing the point entirely. I was wondering if there are any hints towards a more direct approach to solving the problem than above, provided that I haven't made any critical mistakes already.

Answer 1

$\sin^{2q}(x)\to 0$ pointwise almost everywhere on $[0,\pi]$ and $|\sin^{2q}(x)|\leq 1$, hence $$ \lim_{q\to\infty}\int_0^{\pi}\sin^{2q}(x)\;dx=0 $$ by the dominated convergence theorem.

Answer 2

Hint (without the DCT): Break the integral into three pieces: The integral over $[\pi/2-\delta,\pi/2 + \delta]$ and the integral over the two complimentary intervals. On the first interval the integral is $\le 2 \delta.$ On the complementary intervals, $\sin^{2q}(x) \to 0$ uniformly.

Answer 3

Have a look here for the antiderivative $$I_n=\int \sin^n(x)\,dx$$ Apply the formula and use the bounds; this makes $$J_n=\int_0^\pi \sin^n(x)\,dx=\sqrt{\pi } \frac{\Gamma \left(\frac{n+1}{2}\right)}{\Gamma \left(\frac{n}{2}+1\right)}$$ Now, use Stirling expansion $$\log(\Gamma(x))=x (\log (x)-1)+\frac{1}{2} \left(-\log \left(x\right)+\log (2 \pi )\right)+O\left(\frac{1}{x}\right)$$ Apply and use Taylor again for large $n$; this will make $$\log(J_n)=\frac{1}{2} \left(\log \left(\frac{1}{n}\right)+\log (2 \pi )\right)+O\left(\frac{1}{n}\right)\implies J_n=\sqrt{\frac {2\pi} n}+O\left(\frac{1}{n}\right)$$ and $n$ does not need to be an integer.

For example, using $n=123.456$, the "exact" value would be $\approx 0.225141$ while the approximation formula gives $\approx 0.225597$.

Answer 4

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ The main contribution to the integral, as $\ds{q \to \infty}$, comes from $\ds{x\approx \pi/2}$. With Laplace Method: \begin{align} \int_{0}^{\pi}\sin^{2q}\pars{x}\,\dd x & \sim \int_{-\infty}^{\infty}\exp\pars{-q\bracks{x - {\pi \over 2}}^{2}}\,\dd x = {\root{\pi} \over q^{1/2}} + \,\mrm{O}\pars{q^{-3/2}}\to \require{enclose}\enclose{roundedbox}{\color{#f00}{0}}\ \mbox{as}\ q \to \infty. \end{align}