Finding the value of $\displaystyle \lim\limits_{n\rightarrow \infty}\sqrt{n}\int^{\frac{\pi}{4}}_{0}\cos^{2n-2}(x)\mathrm dx$
What I tried
Let $\displaystyle I_{k} =\int^{\frac{\pi}{4}}_{0}\cos^{k}(x)\mathrm dx=\int^{\frac{\pi}{4}}_{0}\cos^{k-1}(x)\cdot \cos (x)\mathrm dx$
$$ I_{k}=\left.\cos^{k-1}(x)\sin x\right|^{\frac{\pi}{4}}_{0}+(k-1)\int^{\frac{\pi}{4}}_{0}\cos^{k-2}(x)\left(1-\cos^2 x\right)\mathrm dx$$
$$ k I_{k}=\bigg(\frac{1}{2}\bigg)^{\frac{k}{2}}+(k-1)I_{k-2}$$
How do I solve it? Help me please.
Best Answer
Put \begin{equation*} I_{n}=\sqrt{n}\int_{0}^{\pi/4}\cos^{2n-2}(x)\,\mathrm{d}x. \end{equation*} Via the substitutions $ y=\sin x $ and $ y=\frac{z}{\sqrt{n-1}} $ we get \begin{gather*} I_{n}=\sqrt{n}\int_{0}^{\pi/4}(1-\sin^2(x))^{n-1}\,\mathrm{d}x = \sqrt{n}\int_{0}^{1/\sqrt{2}}(1-y^2)^{n-1}\cdot\dfrac{1}{\sqrt{1-y^2}}\,\mathrm{d}y =\\[2ex] \dfrac{\sqrt{n}}{\sqrt{n-1}}\int_{0}^{\sqrt{n-1}\left/\sqrt{2}\right.}\left(1-\dfrac{z^2}{n-1}\right)^{n-1}\cdot\dfrac{1}{\sqrt{1-\dfrac{z^2}{n-1}}}\,\mathrm{d}z = \dfrac{\sqrt{n}}{\sqrt{n-1}}\int_{0}^{\infty}f_{n(z)}\,\mathrm{d}z \end{gather*} where \begin{equation*} f_{n}(z)=\begin{cases} \left(1-\dfrac{z^2}{n-1}\right)^{n-1}\cdot\dfrac{1}{\sqrt{1-\dfrac{z^2}{n-1}}}&\mbox{ if } 0<z<\sqrt{n-1}\left/\sqrt{2}\right.\\ 0&\mbox{ if } z>\sqrt{n-1}\left/\sqrt{2}\right. \end{cases} \end{equation*}
Then $ 0 \le f_{n}(z)<e^{-z^2}\cdot \dfrac{1}{\sqrt{1-1/2}} $ and $\displaystyle \lim_{n\to \infty}f_{n}(z) = e^{-z^2}.$
Consequently, according to Lebesgue's dominated convergence theorem \begin{equation*} \lim_{n\to \infty}I_{n} = \int_{0}^{\infty}e^{-z^2}\,\mathrm{d}z =\dfrac{\sqrt{\pi}}{2}. \end{equation*}
Remark. This is an alternative answer where we use the beta function and the gamma function. From
https://en.wikipedia.org/wiki/Beta_function
we get \begin{equation*} \sqrt{n}\int_{0}^{\pi/2}\cos^{2n-2}(x)\,\mathrm{d}x = \dfrac{\sqrt{n}\,\Gamma(n-\frac{1}{2})}{\Gamma(n)}\cdot\dfrac{\sqrt{\pi}}{2}\to \dfrac{\sqrt{\pi}}{2}, \mbox{ as } n\to \infty \end{equation*} where we find the limit here https://en.wikipedia.org/wiki/Gamma_function
Since \begin{equation*} 0 \le \sqrt{n}\int_{\pi/4}^{\pi/2}\cos^{2n-2}(x)\,\mathrm{d}x \le \sqrt{n}\,2^{1-n}\cdot\dfrac{\pi}{4} \to 0, \mbox{ as } n\to \infty \end{equation*} we are ready.