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Get an approximation formula for the following integral: $ \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy $

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    @Srivastan: Thank you very mu$c$h.2011-12-27

2 Answers 2

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$\sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy $

First observe the powers of $cos$ and $sin$

$ cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) = cos^{2n-2k}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y) = (1-sin^{2})^{(n-k)}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y)$

$ \int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy = \int_0^{\frac{\pi}{2}} \left( (1-sin^{2})^{(n-k)}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y) \right) dy $

If we use the substitution $t = sin(y) \Rightarrow dt = cos(y) \hspace{4pt} dy$

$t = 0$ when $y=0$ and $t=1$ when $y=\frac{\pi}{2}$

$ \int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy = \int_0^{\frac{\pi}{2}} \left( (1-sin^2)^{(n-k)}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y) \right) dy $

$ = \int_0^1 t^{(2k-2)} (1-t^{2})^{(n-k)} dt $

The expression

$ \sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy = \sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1} \left( \int_0^1 t^{(2k-2)} (1-t^{2})^{(n-k)} dt \right) $

$ = \left( \frac{1}{35} \right)^{k-1}\int_0^1 \left( \sum_{k=1}^{n} t^{(2k-2)} (1-t^{2})^{(n-k)} \right) dt $

Notice that the sum is a geometric series with first term $\frac{1}{35}(1-t^{2})^{n}$ and the common ratio $\frac{t^{2}}{35(1-t^{2})^{2}}$

The expression therefore simplifies to

$ = \int_0^1 \left( \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1} t^{(2k-2)} (1-t^2)^{(n-k)} \right) dt $

$ \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy = \left( \frac{1}{35} \right)^{k-1} \sum_{k=1}^n \left( \int_0^1 t^{(2k-2)} (1-t^{2})^{(n-k)} dt \right) $

$ = \int_0^1 \left( \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1} t^{(2k-2)} (1-t^{2})^{(n-k)} \right) dt $

Notice that the sum is a geometric series with first term $(1-t^{2})^{(n-1)}$ and the common ratio $\frac{t^{2}}{35(1-t^{2})}$

The expression therefore simplifies to

$ = \int_0^1 \left( \sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1} \left( t^{(2k-2)} (1-t^{2})^{(n-k)} \right) \right) dt $

This expression turns out to be the summation of constant multiple of Beta function

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$ \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy $

We have that

$B(x,y) = 2 \int_0^{\pi/2} \sin^{2x-1} \theta \cos^{2y-1} \theta d\theta$

So we can write your expression as

$\sum\limits_{k = 1}^n {{{\left( {\frac{1}{{35}}} \right)}^{k - 1}}} \frac{1}{2}B\left( {k - \frac{1}{2},n - k + 1} \right)$

But we also know that

$B\left( {x,y} \right) = \frac{{\Gamma \left( x \right)\Gamma \left( y \right)}}{{\Gamma \left( {x + y} \right)}}$

so that

$\frac{1}{2}B\left( {k - \frac{1}{2},n - k + 1} \right) = \frac{1}{2}\frac{{\Gamma \left( {k - \frac{1}{2}} \right)\Gamma \left( {n - k + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}}$

And we have closed formulas for two of the three $\Gamma$ functions there, namely:

$\eqalign{ & \Gamma \left( {n + \frac{1}{2}} \right) = \frac{{\left( {2n - 1} \right)!!}}{{{2^n}}}\sqrt \pi \cr & \Gamma \left( {k - \frac{1}{2}} \right) = \Gamma \left( {k - 1 + \frac{1}{2}} \right) = \frac{{\left( {2k - 3} \right)!!}}{{{2^{k - 1}}}}\sqrt \pi \cr} $

For the last one, we simply put

$\Gamma \left( {n - k + 1} \right) = \left( {n - k} \right)!$

So we get

$\frac{1}{2}B\left( {k - \frac{1}{2},n - k + 1} \right) = \frac{{\left( {2k - 3} \right)!!\left( {n - k} \right)!}}{{{2^{k - n}}\left( {2n - 1} \right)!!}}$

And then

$\sum\limits_{k = 1}^n {{{\left( {\frac{1}{{35}}} \right)}^{k - 1}}} \frac{{\left( {2k - 3} \right)!!\left( {n - k} \right)!}}{{{2^{k - n}}\left( {2n - 1} \right)!!}}$

or

$\frac{{{2^n}}}{{\left( {2n - 1} \right)!!}}\sum\limits_{k = 1}^n {{{\left( {\frac{1}{{35}}} \right)}^{k - 1}}} \frac{{\left( {2k - 3} \right)!!\left( {n - k} \right)!}}{{{2^k}}}$