4

$$I_{n} = \int_{0}^{\frac{\pi}{2}} \sin^{2n-1}x + \sin^{2n-3}x + ... + \sin x dx, n\in \mathbb{N}$$

I'm using this integral to form part of answer to someone's question, but I'm struggling to find a formula for this integral (if one exists).

Edit:

I believe the formula for $J_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{2n+1}xdx$ is:

$J_{n} = \frac{2^{2n+1}n!(n+1)!}{(2n+2)!}$

Nosrati
  • 29,995
mrnovice
  • 5,773

3 Answers3

2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\pi/2}\sum_{k = 1}^{n}\sin^{2k - 1}\pars{x}\,\dd x = \int_{0}^{\pi/2}\sin\pars{x}\,{\sin^{2n}\pars{x} - 1 \over \sin^{2}\pars{x} - 1} \,\dd x\,\,\, \stackrel{\cos\pars{x}\ \mapsto\ x}{=}\,\,\, \int_{0}^{1}{1 - \pars{1 - x^{2}}^{n} \over x^{2}}\,\dd x \\[5mm] = &\ -1 + \int_{0}^{1}{1 \over x}\bracks{-n\pars{1 - x^{2}}^{n - 1}}\pars{-2x}\,\dd x = -1 + 2n\int_{0}^{1}\pars{1 - x^{2}}^{n - 1}\,\dd x \\[5mm] \stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, &\ -1 + n\int_{0}^{1}x^{-1/2}\,\pars{1 - x}^{n - 1}\,\dd x = -1 + n\,{\Gamma\pars{1/2}\Gamma\pars{n} \over \Gamma\pars{1/2 + n}} = -1 + {\pars{-1/2}!\,n! \over \pars{n - 1/2}!} \\[5mm] = &\ {1 \over \ds{{n - 1/2 \choose n}}} - 1= \bbx{\ds{{2^{2n} \over \ds{{2n \choose n}}} - 1}} \end{align}

A proof of the last identity can be seen in one of my previous answers.

Community
  • 1
Felix Marin
  • 89,464
  • $\int_{0}^{\pi/2}\sin\pars{x},{\sin^{2n}\pars{x} - 1 \over \sin^{2}\pars{x} - 1} ,\dd x,,, \stackrel{\cos\pars{x}\ \mapsto\ x}{=},,, \int_{0}^{1}{1 - \pars{1 - x^{2}}^{n} \over x^{2}},\dd x \ $ Could you explain how you arrived at this? – mrnovice Feb 26 '17 at 15:07
  • @mrnovice It's equivalent to $t = \cos\left(x\right)$. Note that $\sin^{2}\left(x\right) = 1 - \cos^{2}\left(x\right) = 1 - t^{2}$ and $\sin^{2n}\left(x\right) = \left[1 - \cos^{2}\left(x\right)\right]^{n} = \left(1 - t^{2}\right)^{n}$. Moreover, $\mathrm{d}t = -\sin\left(x\right),\mathrm{d}x$. – Felix Marin Feb 27 '17 at 16:57
  • Ah okay thanks, also what is $(-\frac{1}{2})!$? – mrnovice Feb 27 '17 at 17:03
  • @mrnovice In general, the factorial is defined as $z! = \Gamma\left(z +1\right)$. So, $\left(-,{1 \over 2}\right)! = \Gamma\left(1 \over 2\right) = ,\sqrt{,\pi,},$. – Felix Marin Mar 02 '17 at 16:39
  • Ah... I didnt noticed that $\sqrt\pi=\Gamma(1/2)$. Sorry but it is not correct to write $(-1/2)!$ because factorial is not defined for negative numbers, but the binomial is correct because the generalized binomial is defined from falling factorials, what are well defined for any $x\in\Bbb C$. – Masacroso Mar 07 '17 at 23:59
  • @Masacroso $z!$ is defined $\forall z \in \mathbf{C}\setminus\left{-1,-2,-3,\ldots\right}$. I already talk about it in a previous comment. See this link. Thanks for your remark. – Felix Marin Mar 08 '17 at 00:28
  • @FelixMarin this is not the general consensus about the definition of the factorial. Just one example. The reason is that exists infinite possible extensions of the factorial function to other values. – Masacroso Mar 08 '17 at 00:35
1

The terms of your integral can be evaluated one by one \begin{eqnarray*} \int_{0}^{\pi/2} \sin^{2n+1}x dx =\frac{(2n)!!}{(2n+1)!!} \end{eqnarray*} So your integral can be written as a sum \begin{eqnarray*} I_n = \sum_{i=0}^{n-1}\frac{(2i)!!}{(2i+1)!!} \end{eqnarray*}

Donald Splutterwit
  • 36,613
  • 2
  • 26
  • 73
0

HINT: by integration by parts we have that

$$I_n:=\int(\sin x)^n\mathrm dx=-\cos x(\sin x)^{n-1}+(n-1)\int (\cos x)^2(\sin x)^{n-2}=\\=-\cos x(\sin x)^{n-1}+(n-1)\int (1-\sin x)^2(\sin x)^{n-2}=\\=-\cos x(\sin x)^{n-1}+(n-1)(I_{n-2}-I_n)$$

hence

$$I_n=\frac1n((n-1)I_{n-2}-\cos x(\sin x)^{n-1})\tag{1}$$

where clearly $I_1=-\cos x$ and $I_0=x$. Evaluating in $A:=[0,\pi/2]$ we have that $I_1|_A=1$, then because (1) we have that $I_3|_A=2/3$, $I_5|_A=\frac23\cdot\frac45$, and in general $$I_{2k+1}|_A=\frac{(2k)!!}{(2k+1)!!}=2^k\frac{k!}{(2k+1)!!}=\frac{2^kk!(2k)!!}{(2k+1)!}=\\=\frac{4^k(k!)^2}{(2k+1)!}=\frac{4^kk!}{(2k+1)^\underline{k+1}}=4^kk!k^\underline{-k-1}$$

Hence

$$\int_0^{\pi/2}\sum_{k=0}^n (\sin x)^{2k+1}\mathrm dx=\sum_{k=0}^{n}I_{2k+1}\big|_A$$

what dont seems to have a closed form of elementary functions. From Wolfram Mathemathica I get a "closed" solution in terms of the gamma function

$$\begin{align}\int_0^{\pi/2}\sum_{k=0}^n (\sin x)^{2k+1}\mathrm dx&=\frac{2 \sqrt{\pi } n \Gamma (n+2)+3 \sqrt{\pi } \Gamma (n+2)-2 \Gamma \left(\frac{1}{2} (2 n+5)\right)}{2 \Gamma \left(\frac{1}{2} (2 n+5)\right)}\\&=\frac{\sqrt\pi(n+1)!}{\Gamma(n+3/2)}-1\end{align}$$

Masacroso
  • 30,417
  • Your formula for $I_{2k+1}$ seems to be incorrect, unless I'm making a mistake. – mrnovice Feb 25 '17 at 23:50
  • @mrnovice I think my formulas are correct. I checked it with mathematica from both approaches: the integral and the summation, and both coincide. – Masacroso Feb 26 '17 at 00:11
  • By your own formula $I_{3} = 4^{1}1!1^{-2} = 4$ But $ I_{3} = \frac{2}{3}$ – mrnovice Feb 26 '17 at 00:12
  • @mrnovice the elements $a^\underline k$ are falling factorials, not powers, then $I_3|_A=4/6=2/3$, because $1^\underline{-2}=\frac12\cdot\frac13=\frac16$ – Masacroso Feb 26 '17 at 00:15
  • Oh ok, there's some more factorial notation I have now learned lol, sorry for the confusion. – mrnovice Feb 26 '17 at 00:29
  • @mrnovice it was my fault, sorry... its too late now and I forget that not everyone know this notation. This notation was introduced many years ago by Donald Knuth and it is common nowadays. – Masacroso Feb 26 '17 at 00:30