Question

A challenge for the new year: If possible, find a closed form or reduction formula for the integral,
\[\Large\int_0^{\pi/2}\sin^{\frac{2n+1}{2}}x\,dx\]

Answer 1

I'm trying something like this
\[\begin{align}
\int_0^{\pi/2}\sin^{\frac{2n+1}{2}}x\,dx &= \int_0^{\pi/2}(1-\cos ^2 x)^{\frac{2n-1}{4}}\cdot \sin x\,dx\\~\\
&=\int_0^{1}(1-u^2)^{\frac{2n-1}{4}}\,du\\~\\
&=\int_0^{1}\sum\limits_{k=0}^{\infty} \binom{(2n-1)/4}{k} (-u^2)^k\,du\\~\\
&=\sum\limits_{k=0}^{\infty} \binom{(2n-1)/4}{k} \frac{(-1)^k}{2k+1}\\~\\

\end{align}\]

whatever that equals hmm

Answer 2

Why did you make it an infinite sum?

Answer 3

thats because of the generalized binomial thm
http://en.wikipedia.org/wiki/Binomial_series

Answer 4

I'm still trying to see if the sum evaluates to something that looks more like "closed form"

Answer 5

i was confused all this time thinking 2n+1/2 is inside sin :/

Answer 6

i got a nice recurrence relation because of the symmetry after substituting away sin.. il post, one sec..

Answer 7

hmm 2n+1/2 is not integer , hmm i need to think of it

Answer 8

For recurrence relation :
\[\begin{align}
I=\int_0^{\pi/2}\sin^{\frac{2n+1}{2}}x\,dx &= \int_0^{\pi/2}(1-\cos ^2 x)^{\frac{2n-1}{4}}\cdot \sin x\,dx\\~\\
&=\int_0^{1}(1-u^2)^{\frac{2n-1}{4}}\,du\\~\\
&=\frac{1}{2}\int_0^{1}t^{\frac{-1}{2}}(1-t)^{\frac{2n-1}{4}}\,dt \\~\\
&=\frac{1}{2}\int_0^{1}t^{\frac{-1}{2}}(1-t)^{q}\,dt~~~;~~ \color{Red}{\text{ (2n-1)/4}=q}\\~\\

\end{align}\]

let \(I_q =\frac{1}{2}\int_0^{1}t^{\frac{-1}{2}}(1-t)^{q}\,dt \)
By parts we get
\[\begin{align}

I_q &=(1-t)^q \frac{t^{1/2}}{1/2}\Bigg|_0^1+\frac{1}{2}\int_0^{1}\frac{t^{\frac{1}{2}}}{1/2}\cdot q(1-t)^{q-1}\,dt \\~\\
&=0+q\int_0^{1}t^{\frac{1}{2}}\cdot (1-t)^{q-1}\,dt \\~\\
&=q\left[\int_0^{1}t^{\frac{-1}{2}}\cdot (1-t)^{q-1}\,dt -\int_0^{1}t^{\frac{-1}{2}}\cdot (1-t)^{q}\,dt \right]\\~\\
&=q\left[I_{q-1} -2I_q \right]\\~\\
\end{align}\]

Answer 9

haha i reached reduction formula only xD

Answer 10

lets see if we mess with the recurrence relation further
remember gamma function ?

Answer 11

yes , hmm how its involved here ?

Answer 12

wolfram says the integral can be represented using gamma function :
\[\sum\limits_{k=0}^{\infty} \binom{(2n-1)/4}{k} \frac{(-1)^k}{2k+1} ~~=~~ \dfrac{\sqrt{\pi} ~\Gamma ((2n+3)/4)}{2~\Gamma ((2n+5)/4)}\]

http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bk%3D0%7D%5E%7B%5Cinfty%7D+%5Cbinom%7B%282n-1%29%2F4%7D%7Bk%7D+%5Cfrac%7B%28-1%29%5Ek%7D%7B2k%2B1%7D

Answer 13

ok i understand how to convert this series to gama , but where did the series come from ?

Answer 14

look at my first reply

Answer 15

oh yes , i thught u start over >.<

Answer 16

yeah thats reduce it in nice formula , now to generate gama , here is in integral of it