Calculate limit of two integrale I have no idea how to calculate (I know that the answer is 1 but why ?).
Maybe with a theorem or with trigonometric properties ...

$$ \lim_{m\to\infty} \frac{\int_0^{\pi/2}(\sin x)^{2m}dx}{\int_0^{\pi/2}(\sin x)^{2m+1}dx} $$

 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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With Laplace's Method:

\begin{align}
\int_{0}^{\pi/2}\sin^{p}\pars{x}\,\dd x & =
\int_{0}^{\pi/2}\cos^{p}\pars{x}\,\dd x =
\int_{0}^{\pi/2}\exp\pars{p\ln\pars{\cos\pars{x}}}\,\dd x\quad
\pars{\begin{array}{l}
\mbox{The integral "main}
\\
\mbox{contribution"}
\\
\mbox{occurs at}\ x \gtrsim 0
\end{array}}
\\[5mm] & \stackrel{\mrm{as}\ p\ \to\ \infty}{\sim}\,\,\,
\int_{0}^{\infty}\exp\pars{-\,{p \over 2}\,x^{2}}\,\dd x = \root{\pi \over 2}\,
{1 \over p^{1/2}}
\end{align}

$\ds{\implies \lim_{m \to \infty}{\int_{0}^{\pi/2}\sin^{2m}\pars{x}\,\dd x \over \int_{0}^{\pi/2}\sin^{2m + 1}\pars{x}\,\dd x} = 
\lim_{m \to \infty}\root{1 + {1 \over 2m}} = \bbx{\large 1}}$

A: I'll use the result $2\int_0^{\pi/2}\sin^{2a-1}x\cos^{2b-1}xdx=\frac{\Gamma (a)\Gamma (b)}{\Gamma (a+b)}$ so $$\frac{\int_0^{\pi/2}\sin^{2m}xdx}{\int_0^{\pi/2}\sin^{2m+1}xdx}=\frac{\Gamma(m+\frac{1}{2})\Gamma(m+\frac{3}{2})}{\Gamma^2(m+1)}=\exp (\ln\Gamma (m+\tfrac{1}{2})-2\ln\Gamma (m+1)+\ln\Gamma (m+\tfrac{3}{2})).$$Since $x\mapsto x+1$ increases $\ln\Gamma (x)$ by $\ln x$, the right-hand side approximates $$\exp\frac{1}{2}\ln\frac{m+1}{m+1/2},$$where we have linearised the log-Gamma function. Since the exponential's argument is $O(\frac{1}{m})$, the limit is $1$.
