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Prove $$\lim_{n\to +\infty}\int_0^{\frac{\pi}{2}}\sin^n(x)\,dx=0$$ My attempt: $$I_n=\int_0^{\frac{\pi}{2}}\sin^n(x)\,dx = \left.- \int_0^{\frac{\pi}{2}} \sin^{n-1} x \,d(\cos x) = -\sin^{n-1} x\cos x \right|_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos x\,d(\sin^{n-1} x)$$ As $\left.-\sin^{n-1} x\cos x \vphantom{\dfrac11} \right|_0^{\frac{\pi}{2}} = 0$, hence $$\int_0^{\frac{\pi}{2}}\sin^n(x)\,dx = (n-1) \int_0^{\frac{\pi}{2}}\sin^{n-1}(x) \, dx - (n-1)\int_0^{\frac{\pi}{2}}\sin^n(x) \, dx \Rightarrow I_n = \frac{(n-1)I_{n-2}}{n}$$ hence if $n=2k$: $$I_{2k}=\frac{(2k-1)!!}{(2k)!!}\cdot\frac{\pi}{2}$$ if $n=2k+1$: $$I_{2k+1}=\frac{(2k)!!}{(2k+1)!!}$$ Got that by induction and because of $I_1=\int_0^{\frac{\pi}{2}}\sin x\,dx= \left.-\cos x \vphantom{\dfrac11} \right|_0^{\frac{\pi}{2}} = 1$. So I have to prove that if $k \to +\infty \Rightarrow I_{2k} \to 0 \text{ and } I_{2k+1} \to 0$. I don't know how to do that. Please help

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  • $\begingroup$ I'm not sure if this will help but maybe researching Wallis' formula $\endgroup$ – Teh Rod Mar 19 '17 at 20:04
  • $\begingroup$ The proof is fine but it lacks the reason why $I_{2k}$ or $I_{2k+1}$ is what you says, I mean, it will not hurt to add the value of $I_0$ and $I_1$, then just saying than using induction we reach the results. $\endgroup$ – Masacroso Mar 19 '17 at 20:09
  • $\begingroup$ I'm tempted to try this: Try to find a point $0<a_n<\pi/2$ such that $a_n \uparrow \pi/2$ as $n\to\infty$ and $\sin^n(a_n) \downarrow0$ as $n\to\infty.$ Then you have $$ 0 < \int_0^{\pi/2} \sin^n x\,dx \le \frac\pi 2 \sin^n(a_n) + \left( \frac \pi 2 - a_n\right) \to 0 \text{ as } n\to\infty. $$ $\endgroup$ – Michael Hardy Mar 19 '17 at 20:15
  • $\begingroup$ Take a look here for the inductive part, you had almost done. $\endgroup$ – Masacroso Mar 19 '17 at 20:27
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This is a direct consequence of the Dominated Convergence Theorem. Note $\forall x,n$, $\sin^n(x) \le 1$ which is integrable over $[0,\frac{\pi}{2}]$ and $\sin^n(x) \to 0 \hspace{1mm} \forall x \in (0,\frac{\pi}{2})$.

EDIT: I felt kinda bad just leaving my solution like this, for those who aren't familiar with DCT. The idea is very simple (although what I'm about to say slightly differs from DCT). Since $\sin^n(x) \to 0$ uniformly on $[0,1]$, for any $\epsilon > 0$, for $n$ large enough, $\sin^n(x) \le \epsilon$. Therefore, $|\int_0^{2\pi} \sin^n(x)| \le 2\pi \epsilon$ for $n$ large enough. Since this holds for all $\epsilon > 0$, this limit must be $0$.

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    $\begingroup$ But I can't use that theorem because I didn't have it on my course. I'm only first year undergraduate $\endgroup$ – ioleg19029700 Mar 19 '17 at 20:27
  • $\begingroup$ I made an edit to explain what's going on. I still don't think the answer you chose gets the main point across. $\endgroup$ – mathworker21 Apr 2 '17 at 19:23
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You can use the formule

$$(2k)!! =2^k k!, \quad (2k-1)!!=\frac{(2k)!}{2^k k!} $$

to obtain

$$I_{2k} =\frac{\pi}{2} \frac{(2k)!}{2^k k!} \frac{1}{2^k k!} = \frac{\pi}{2}\frac{(2k)(2k-1) \cdots (k+1)}{2^{2k} k!} \le \frac{\pi}{2}\frac{1}{2^k} \to 0. $$

$I_{2k+1}$ is similarly dealt with.

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A stronger result

Let $\displaystyle{W_n=\int_0^{\pi/2}\sin^n(x)\,dx}$.

Integration by parts leads to the formula :

$$\forall n\ge2,\,nW_n=(n-1)W_{n-2}\tag{1}$$

which implies :

$$\forall n\ge1,\,nW_nW_{n-1}=\frac\pi 2$$

On the other hand, it's easy to see that the sequence $(W_n)_{n\ge0}$ is decreasing, and so :

$$W_{n+1}\le W_n\le W_{n-1}$$

After division by $W_{n-1}$ and combining with (1), we get :

$$\lim_{n\to\infty}\frac{W_n}{W_{n-1}}=1$$

Finally :

$$nW_n^2=\left(nW_nW_{n-1}\right)\left(\frac{W_n}{W_{n-1}}\right)\underset{n\to\infty}{\longrightarrow}\frac\pi2$$

We have proved the (very classical) result :

$$\int_0^{\pi/2}\sin^n(x)\,dx\sim\sqrt{\frac\pi{2n}}$$

In particular, we have $\displaystyle{\lim_{n\to\infty}W_n=0}$

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    $\begingroup$ I think your argument can be simplified a bit: since $W_n$ is decreasing and non-negative, it has a nonnegative limit $W$. But then, from $n W_n W_{n-1} = \frac{\pi}{2}$ you get $$ W^2 = \lim_{n \to\infty} W_nW_{n-1} = \lim_{n \to \infty} \frac{\pi}{2n} = 0, $$ which is the desired result. $\endgroup$ – Stefano Mar 19 '17 at 21:17
  • $\begingroup$ @Stefano: Nice observation ! $\endgroup$ – Adren Mar 19 '17 at 21:57
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Different approach: For any $a\in [0,\pi/2),$ we have

$$\int_0^{\pi/2} \sin^n(x)\, dx = \int_0^{a} \sin^n(x)\, dx + \int_a^{\pi/2} \sin^n(x)\, dx \le a\cdot\sin^n(a) + 1\cdot (\pi/2-a).$$

Now the first term on the right $\to 0.$ Taking the $\limsup$ then shows

$$\tag 1 \limsup_{n\to \infty} \int_0^{\pi/2} \sin^n(x)\, dx \le 0 +(\pi/2-a).$$

Since $a$ is arbitrarily close to $\pi/2,$ we see the $\limsup$ is $0,$ which is the desired result.

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$$\color{red}{\int_{0}^{\pi/2}\sin(x)^n\,dx} = \int_{0}^{\pi/2}\cos(x)^n\,dx \color{red}{\leq} \int_{0}^{\pi/2}e^{-nx^2/2}\,dx\leq \int_{0}^{+\infty}e^{-nx^2/2}\,dx=\color{red}{\sqrt{\frac{\pi}{2n}}}.$$

The crucial inequality $\cos(x)\leq e^{-x^2/2}$ over $(0,\pi/2)$ is equivalent to $\log\cos(x)\leq -\frac{x^2}{2}$, that is a consequence of $\tan(x)\geq x$ (just integrate both sides).

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  • $\begingroup$ Is it difficult to prove $cos^n x \le e^{\frac{-nx^2}{2}}$? $\endgroup$ – ioleg19029700 Mar 19 '17 at 21:06
  • $\begingroup$ @ioleg19029700: I was expecting such question - no, it is not difficult. Now explained. $\endgroup$ – Jack D'Aurizio Mar 19 '17 at 21:06
  • $\begingroup$ And if you do not know that $C=\int_{0}^{+\infty}e^{-x^2/2}\,dx=\sqrt{\frac{\pi}{2}}$ you may simply denote it as $C$ and notice that the RHS equals $\frac{C}{\sqrt{n}}$ by the substitution $x=\frac{z}{\sqrt{n}}$. $\endgroup$ – Jack D'Aurizio Mar 19 '17 at 21:14
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\underline{\mrm{Laplace\ Method}}$:

\begin{align} \lim_{n \to \infty}\int_{0}^{\pi/2}\sin^{n}\pars{x}\,\dd x & = \lim_{n \to \infty}\int_{0}^{\pi/2}\exp\pars{n\ln\pars{\cos\pars{x}}}\,\dd x = \lim_{n \to \infty}\int_{0}^{\infty}\expo{-nx^{2}/2} \pars{1 - {n \over 12}\,x^{4}}\,\dd x \\[5mm] & = \root{\pi \over 2}\lim_{n \to \infty} \pars{{1 \over n^{1/2}} - {1 \over 4n^{3/2}}} = \bbx{\ds{\large 0}} \end{align}

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