Trigonometric Series : Trinity

That's real easy, Start with integral :

$\displaystyle {I}_{n} = \int _{ 0 }^{ \pi /2 }{ \sin ^{ n }{ x } dx }$

Integrating by parts we have :

$u=\sin^{n-1}{x} , v'=sin(x)$

$\displaystyle {I}_{n} = (n-1)\int _{ 0 }^{ \pi /2 }{ \sin ^{ n-2 }{ x } \cos ^{ 2 }{ x } dx }$

$\displaystyle \Rightarrow {I}_{n} = (n-1)\int _{ 0 }^{ \pi /2 }{ \sin ^{ n-2 }{ x } (1-\sin ^{ 2 }{ x } )dx }$

$\displaystyle \Rightarrow { I }_{ n }=\dfrac { (n-1) }{ n } { I }_{ n-2 }$

Put $n=3$ to get :

$\displaystyle { I }_{ 3 }=\frac { 2 }{ 3 } { I }_{ 1 } = \frac{2}{3}$

Put $n=5$ to get :

$\displaystyle { I }_{ 5 }=\frac { 4 }{ 5 } { I }_{ 3 }=\frac { 4 }{ 5 } .\frac { 2 }{ 3 } { I }_{ 1 }=\frac { 4 }{ 5 } .\frac { 2 }{ 3 }$

Hence we can write our summation as :

$\displaystyle f(x) = \sum _{ n=0 }^{ \infty }{ \sin ^{ 2n }{ x } { I }_{ 2n+1 } }$

$\displaystyle f(x) = \sum _{ n=0 }^{ \infty }{ \sin ^{ 2n }{ x } \int _{ 0 }^{ \pi /2 }{ { sin }^{ 2n+1 }y } dy }$

Interchanging integral and summation we have :

$\displaystyle \int _{ 0 }^{ \pi /2 }{ \sin { y } \sum _{ n=0 }^{ \infty }{ { (\sin { (x) } \sin { (y) } ) }^{ 2n } } } dy$

Applying formula for geometric progression we have :

$\displaystyle f(x) = \int _{ 0 }^{ \pi /2 }{ \frac { \sin { (y) } }{ 1-\sin ^{ 2 }{ x } \sin ^{ 2 }{ y } } } dy$

After a little manipulation it gets converted into.

$\displaystyle f(x) =\int _{ 0 }^{ \pi /2 }{ \frac { \sin { (y) } }{ \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } \cos ^{ 2 }{ y } } } dy$

Put $\cos(y) =t$ to get :

$\displaystyle f(x) = \int _{ 0 }^{ \pi /2 }{ \frac { dt }{ \cos ^{ 2 }{ x } +(\sin ^{ 2 }{ x }) { t }^{ 2 } } }$

Now this integral is trivial and finally $S$ comes out to be :

$\displaystyle f(x) =\frac { 2x }{ \sin { 2x } }$

Now rest work is trivial and left as an exercise to the reader.

Rewrite using Sigma notation:

$\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!}\sin^{2n}x$

$\displaystyle \sum_{n=0}^{\infty} \frac{n!2^n}{\frac{(2n+1)!}{n!2^n}}\sin^{2n}x$

$\displaystyle \sum_{n=0}^{\infty} \frac{n!n!}{(2n+1)!}(4\sin^{2}x)^n$

Recall the definition of the Beta Function :

$\displaystyle B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} dt = \frac{(x-1)!(y-1)!}{(x+y-1)!}$

Setting $x=y=n+1$ , We get:

$\displaystyle \frac{(n)!(n)!}{(2n+1)!} = \int_0^1 t^{n}(1-t)^{n} dt$

Replace this in our summation:

$\displaystyle f(x)= \sum_{n=0}^{\infty} \int_0^1 t^{n}(1-t)^{n} (4\sin^{2}x)^n dt$

$\displaystyle = \int_0^1 \Bigg(\sum_{n=0}^{\infty} (t(1-t)(4\sin^2x))^n\Bigg) dt$

Note that, on the interval $[0,\frac{\pi}{2} ]$ , the sum is a convergent geometric sum for $0<x<\frac{\pi}{6}$ .

$\displaystyle f(x) = \int_0^1 \frac{1}{1-t(1-t)(4\sin^2x)} dt$

Rearrange our integral, we get:

$\displaystyle f(x) =\frac{1}{\cos^2x} \int_0^1 \frac{dt}{ 1+ [(2t-1)\tan x]^2}$

Computing this integral yields :

$\displaystyle \boxed{f(x) = \frac{2x}{\sin 2x} }$

$\displaystyle \sin \frac{\pi}{12} = \sin \frac{\frac{\pi}{6}}{2} = \sqrt{\frac{1-\cos \frac{\pi}{6}}{2}}$

$\displaystyle = \frac{\sqrt{2-\sqrt{3}}}{2}$

Note: $\big( \frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{2} \big) ^2 = 2-\sqrt{3}$

$\displaystyle => \sin \frac{\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}$

$\displaystyle => f(\frac{\pi}{24}) = \frac{\frac{\pi}{12}}{\frac{\sqrt{6} -\sqrt{2}}{4} }$

$\displaystyle = \boxed{\frac{\pi(\sqrt{3}+1)}{6\sqrt{2}} }$