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I ran across an identity I had not saw before, and am wondering how it can be derived.

$\displaystyle \int_{0}^{\frac{\pi}{2}}\cos^{m}(x)\cos(nx)dx=\frac{\pi\Gamma(m+1)}{2^{m+1}\Gamma(\frac{m+n}{2}+1)\Gamma(\frac{m-n}{2}+1)}$.

For the case, $m=n$, then the result is $\frac{\pi}{2^{m+1}}$.

I can easily use parts, but I do not know how to connect it to Gamma. It would appear the

method must lie in generalizing somehow. It looks like the classic Beta/trig integral may

be in there somewhere.

I used parts and got:

$\displaystyle\int_{0}^{\frac{\pi}{2}}\cos^{m}(x)\cos(nx)dx=\frac{m}{n}\int_{0}^{\frac{\pi}{2}}\cos^{m-1}(x)\cos(n-1)xdx-\frac{m}{n}\int_{0}^{\frac{\pi}{2}}\cos^{m}(x)\cos(nx)dx$

$\displaystyle \left(1+\frac{m}{n}\right)I_{m,n}=\frac{m}{n}I_{m-1,n-1}$

and so on. Now, perhaps let $n=n-1, \;\ m=m-1$, then sub in and generalize?.

I still do not see how to tie it to Gamma unless it comes from the product of the m and n terms

Thanks to anyone who has a clever idea.

  • 0
    Determine whether the formula is true when $m=0$, whether it is true when $n=0$ and apply induction using your result $\displaystyle \left(1+\frac{m}{n}\right)I_{m,n}=\frac{m}{n}I_{m-1,n-1}$.2012-12-07
  • 0
    Thank you very much. I managed to work it out.2012-12-08

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