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I'm trying to compute $\lim_{n\rightarrow\infty}\int_{0}^{\frac{\pi}{2}}\frac{n\cos x \sin^n x}{1+x}dx$.

In the given interval $(0,\frac{\pi}{2})$, $\frac{n\cos x \sin^n x}{1+x}$ is non negative, and increasing, I used the monotone convergence theorem to interchange the limit and the integral to get $$\lim_{n\rightarrow\infty}\int_{0}^{\frac{\pi}{2}}\frac{n\cos x \sin^n x}{1+x}dx=\int_{0}^{\frac{\pi}{2}}\lim_{n\rightarrow\infty}\frac{n\cos x \sin^n x}{1+x}dx$$

But when I'm trying to compute the limit, I'm stuck. I even tried to do through the "$u$ substitution" as well but the denominator isn't helping. Any input is appreciated.

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  • $\begingroup$ The integrands are not increasing. Also, the pointwise limit of the integrands is $0.$ That's because $na^n\to 0$ if $0\le a <1.$ $\endgroup$ – zhw. May 7 '18 at 18:37
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Here's a possible approach. We have $$ \int_0^{\pi/2} \frac{\cos x \sin^n x}{1+x } dx = \int_0^{\pi/2} \frac{\sin^n x}{1+x } d \sin x = (\text{set } y = \sin x) = \int_0^1 \frac{y^n}{ 1 + \arcsin y} dy. $$

Now, the last integral does not seem a very pleasant one to compute explicitly, but we can compute the desired limit. Just observe that the main contribution comes from the neighborhood of 1.

Namely, fix any $0<\alpha <1$, then $$ n \int_0^\alpha \frac{y^n}{ 1 + \arcsin y} dy \leq n \int_0^\alpha y^n dy =\frac{n}{n+1} \alpha^n \to 0 , \text{ as } n \to \infty, $$ as $\alpha < 1$. Hence, the contribution comes from $\int_\alpha^1$, where we have $$ \frac{y^n}{1 + \pi/2} \leq \frac{y^n}{1 + \arcsin y} \leq \frac{y^n}{1 + \arcsin \alpha}. $$ Now integrating the last ineqalities over $[\alpha, 1]$ we obtain $$ \frac{n}{n+1} \frac{1 - \alpha^{n+1}}{1+\pi/2} \leq n \int_{\alpha}^1 \frac{y^n}{1 + \arcsin y} dy\leq \frac{n}{n+1} \frac{1 - \alpha^{n+1}}{1 + \arcsin \alpha}. $$ Finally, taking $n\to \infty$ in the last line, we get $$ \frac{1}{1+\pi/2} \leq \lim n \int_{\alpha}^1 \frac{y^n}{1 + \arcsin y} dy\leq \frac{1 }{1 + \arcsin \alpha}. $$ But we also proved that the limit of the integral over $[0, \alpha]$ is vanishing, hence the limit in the last inequality is the actual limit in question. But as $\alpha<1$ is arbitrary, we get that the limit has to be equal to $$ \frac{1}{1+\pi/2}. $$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\int_{0}^{\pi/2} {n\cos\pars{x}\sin^{n}\pars{x} \over 1 + x}\,\dd x & = \lim_{n \to \infty}\bracks{{n \over n + 1}\int_{x\ =\ 0}^{x\ =\ \pi/2} {\dd\sin^{n + 1}\pars{x} \over 1 + x}} \\[5mm] & = \lim_{n \to \infty}\bracks{{n \over n + 1}\,\color{red}{1 \over 1 + \pi/2} + {n \over n + 1}\int_{0}^{\pi/2} {\cos^{n + 1}\pars{x} \over \pars{1 + \pi/2 - x}^{2}}\,\dd x} \\[5mm] & = \bbx{\large{1 \over 1 + \pi/2}} \approx 0.3890 \end{align}


$$ \overbrace{\int_{0}^{\pi/2} {\cos^{n + 1}\pars{x} \over \pars{1 + \pi/2 - x}^{2}}\,\dd x \sim {1 \over \pars{1 + \pi/2}^{\, 2}}\ \underbrace{\int_{0}^{\infty}{\expo{-\pars{n + 1}x^{2}/2}}\dd x} _{\ds{\root{\pi \over 2}{1 \over \pars{n + 1}^{1/2}}}} \color{red}{\to 0} \quad\mbox{as}\quad n \to \infty}^{\ds{\large\mbox{Laplace Method}}} $$

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