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The question is the one from the title i.e. : Compute $\displaystyle\lim_\limits{{n\to\infty}}\int_0^\frac{\pi}{2} \frac{\sin^nx} {\sqrt{1+x}}\, dx$

I do realise that I should probably use Monotone Convergence Theorem or Dominated Convergence Theorem. The problem is that:

  1. To use the first one I shoul prove first that $\frac{\sin^nx}{\sqrt{1+x}} \leq \frac{\sin^{n+1}x}{\sqrt{1+x}}$ for all $x\in(0,\frac{\pi}{2})$ which I think is not true. I would rather claim that the opposite is true but I am struggling with proving it, I mean : $\frac{\sin^nx}{\sqrt{1+x}} \geq \frac{\sin^{n+1}x}{\sqrt{1+x}}$ for all $x\in(0,\frac{\pi}{2})$.
  2. To use the second theorem I need $\frac{\sin^nx}{\sqrt{1+x}}$ to be bbd by some measurable function I can take $1$ so this bit is easy but also I need pointwise convergence to some function through all the $(0,\frac{\pi}{2})$ and here the funny part starts. Using some numerical (not sophisticated ;) ) methods ( by saying that I mean I just checked the behaviour for large n ;) ) It looks like $0$ (- zero) is a good candidate, but I couldn't prove it.

So the question is:

  • How to prove the second inequality in 1.)

Or

  • How to prove that $\lim_\limits{{n\to\infty}}\frac{\sin^nx}{\sqrt{1+x}}=0$ on $(0,\frac{\pi}{2})$
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    $\begingroup$ the integrand is not defined on $(0,\pi/2)$ $\endgroup$ – aziiri Jul 2 '17 at 17:10
  • $\begingroup$ One issue is that there is a singularity at $x = 1$, and the integrand is not defined for $x \gt 1$ unless you are working over $\mathbb{C}$. $\endgroup$ – Tob Ernack Jul 2 '17 at 17:10
  • $\begingroup$ Since $|\sin(x)| \le |x|$, certainly we will have pointwise convergence to $0$ in the range $[0, 1]$. Then dominated convergence would show that $f_n \to 0$ for $f_n(x) = \frac{\sin^n(x)}{\sqrt{1-x}}$ for any $I \subset [0, 1)$. $\endgroup$ – Chris Jul 2 '17 at 17:43
  • $\begingroup$ Sorry guys I just edited the question there was a mistake in the denominator it should be $\sqrt{1+x}$ not $\sqrt{1-x}$ now its correct $\endgroup$ – Kran Jul 2 '17 at 18:05
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The given limit is clearly zero.

$$\begin{eqnarray*} 0\leq \int_{0}^{\pi/2}\frac{\sin^n(x)}{\sqrt{1+x}}\,dx &\color{red}{\leq}& \int_{0}^{\pi/2}\sin^n(x)\,dx =\int_{0}^{\pi/2}\cos^n(x)\,dx \leq \int_{0}^{\pi/2}e^{-nx^2/2}\,dx\\ &\leq& \int_{0}^{+\infty} e^{-nx^2/2}\,dx = \color{red}{\sqrt{\frac{\pi}{2n}}}\end{eqnarray*}$$ As an alternative, the given integral is $\leq \int_{0}^{1}\frac{x^n}{\sqrt{1-x^2}}$. Over the interval $(0,1)$ the sequence of functions $f_n(x)=x^n$ is pointwise convergent to $0$ and bounded by $1$. Since $\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}=\frac{\pi}{2}$, the claim follows from the dominated convergence theorem, too.

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  • $\begingroup$ thanks Jack, How did you get the third inequality? The bound on $cos^n(x)$ $\endgroup$ – Kran Jul 2 '17 at 18:22
  • $\begingroup$ @Kacper: over the interval $I=\left(0,\frac{\pi}{2}\right)$ we have $\cos(x)\leq e^{-x^2/2}$, that is equivalent to $\log\cos(x)\leq -\frac{x^2}{2}$ and can be derived from $\tan(x)\geq x$ by integrating both sides. $\endgroup$ – Jack D'Aurizio Jul 2 '17 at 18:34
  • $\begingroup$ That is a useful inequality when dealing with powers of $\cos$. $\endgroup$ – Jack D'Aurizio Jul 2 '17 at 18:36
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    $\begingroup$ @JackD'Aurizio Very cool inequality, a little verbose to use on this, but still a nice inequality! $\endgroup$ – user335907 Jul 2 '17 at 18:38
  • $\begingroup$ @JackD'Aurizio Thanks $\endgroup$ – Kran Jul 2 '17 at 18:41
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As Jack D'Aurizio answered $$ 0\leq \int_{0}^{\pi/2}\frac{\sin^n(x)}{\sqrt{1+x}}\,dx\, {\leq} \int_{0}^{\pi/2}\sin^n(x)\,dx =\frac{\sqrt{\pi }} 2\frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma \left(\frac{n}{2}+1\right)}$$ Now, using Stirling approximation $$\frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma \left(\frac{n}{2}+1\right)}=\frac 1 {\sqrt{ n}}\left(\sqrt{2}-\frac{1}{2 \sqrt{2} n}+O\left(\frac{1}{n^2}\right)\right) $$

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The integrands converge pointwise to $0$ on $[0,\pi/2).$ By the dominated convergence theorem (the dominating function being $1/\sqrt {1+x}$), the limit is $0.$

For a more elementary solutionn, let $0<h<\pi/2.$ Then

$$ 0\le \int_0^{\pi/2} \frac{\sin^n x}{\sqrt {1+x}}\,dx \le \int_0^{\pi/2} \sin^n x\,dx$$ $$\tag 1 = \int_0^{\pi/2-h} \sin^n x\,dx + \int_{\pi/2-h}^{\pi/2} \sin^n x\,dx \le \sin^n(\pi/2 -h)\cdot (\pi/2) + 1\cdot h.$$

Because $0\le \sin(\pi/2 -h) <1,$ $\sin^n(\pi/2 -h) \to 0.$ It follows that the $\limsup$ of the left side of $(1)$ is $\le h.$ Because $h$ is arbitrarily small, this $\limsup$ is $0.$ Thus the desired limit is $0.$

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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}{\sin^{n}\pars{x} \over \root{1 + x}}\,\dd x & = \lim_{n \to \infty}\int_{0}^{\pi/2} {\cos^{n}\pars{x} \over \root{1 + \pars{\pi/2 - x}}}\,\dd x = {1 \over \root{1 + \pi/2}}\lim_{n \to \infty}\int_{0}^{\infty} \expo{-nx^{2}/2}\,\dd x \\[5mm] & = {1 \over \root{1 + \pi/2}}\lim_{n \to \infty}\root{\pi \over 2n} = \bbx{0} \end{align}

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