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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.

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Determine whether the formula is true when $m=0$, whether it is true when $n=0$ and apply induction using your result $\displaystyle \left(1+\frac{m}{n}\right)I_{m,n}=\frac{m}{n}I_{m-1,n-1}$. –  GEdgar Dec 7 '12 at 22:37
    
Thank you very much. I managed to work it out. –  Cody Dec 8 '12 at 10:02
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1 Answer 1

First, we note that:

$$z=e^{ix}$$

$$\cos x=\frac{e^{ix}+e^{-ix}}{2}=\frac{z+\frac{1}{z}}{2}$$

$$dx=\frac{dz}{iz}$$ Based on the above, we consider an integral:

$$I_1=\frac{1}{i2^m}\int_{0}^{i}(1+z^2)^mz^{n-m-1}dz$$ The original integral $I$ is the real part of $I_1$ because that $$z^n=\cos nx +i\sin nx$$

Next step is the change of variable: $z^2=-y$

$$I_1=\frac{i^{n-m-1}}{2^{m+1}}\int_{0}^{1}(1-y)^my^{\frac{n-m}{2}-1}dz=\frac{i^{n-m-1}}{2^{m+1}}B\left(m+1,\frac{n-m}{2}\right)$$

where $B(x,y)$ is the beta function.

From complex algebra we get:

$$i^{n-m-1}=\sin\left(\pi\frac{n-m}{2}\right)-i\cos\left(\pi\frac{n-m}{2}\right)$$

Remembering that we need a real part of $I_1$ we finally have:

$$I=\mathfrak{Re}(I_1)= \frac{\sin\left(\pi\frac{n-m}{2}\right)}{2^{m+1}}B\left(m+1,\frac{n-m}{2}\right)$$

To get the result in terms of gamma function as desired we use the common relations:

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

$$\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x}$$

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