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 $
approximation formula for the integral
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0@Srivastan: Thank you very mu$c$h. – 2011-12-27
2 Answers
$\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
$ \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
$\frac{{{2^n}}}{{\left( {2n - 1} \right)!!}}\sum\limits_{k = 1}^n {{{\left( {\frac{1}{{35}}} \right)}^{k - 1}}} \frac{{\left( {2k - 3} \right)!!\left( {n - k} \right)!}}{{{2^k}}}$