How do I find $$\int_{0}^{2\pi} (\cos x)^{2n}\,\mathrm dx$$ using contour integration ? Should I turn it into an integral around the unit circle? Or should I integrate some other function with real part $(\cos x)^{2n}$. Not sure where to start, please help.
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1I don't think there's a nice function with real part $(\cos(x))^{2n}$, as $\Re(e^{i2n x})\ne (\Re( e^{ix}))^{2n}$, you'll have to struggle with binomial expansions. – Meow Jun 09 '13 at 18:36
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5Yes, convert it into an integral over unit circle. You will indeed need to use the binomial expansion, but don't be afraid because of the previous comment, only one term of the expansion will contribute to the integral. – Start wearing purple Jun 09 '13 at 18:44
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Oh, so I expand (1/2(e^iz + e^-iz))^2n , I get lots of exponentials but these are holomorphic on C so their integrals become zero, so only the constant term contribute to the integral, right? – John Michael Jun 09 '13 at 21:23
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Almost the same problem. – Mhenni Benghorbal Jun 11 '13 at 04:38
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After the change of variables $z=e^{ix}$, you will have something like \begin{align} I=2^{-2n}\oint_C\left(z+\frac1z\right)^{2n}\frac{dz}{iz}&=2^{-2n}\sum_{k=0}^{2n}\oint_C{2n \choose k} \cdot z^k\cdot z^{-(2n-k)}\,\frac{dz}{iz}=\\ &=\frac{2^{-2n}}{i}\sum_{k=0}^{2n}{2n \choose k}\oint_Cz^{2(k-n)-1}dz=\\ &=2\pi\cdot 2^{-2n}\sum_{k=0}^{2n}{2n \choose k}\mathrm{res}_{z=0} z^{2(k-n)-1}, \end{align} where $C$ denotes the unit circle oriented counterclockwise and $\displaystyle{n \choose k}=\frac{n!}{k!(n-k)!}$ is the binomial coefficient. Now the residue is non-zero only for $2k-2n-1=-1$, i.e. for $k=n$, in which case it is equal to $1$. Therefore out of the sum of $2n+1$ terms only one term (with $k=n$) will remain and the answer will be $$I=2\pi\cdot 2^{-2n}\cdot {2n \choose n}=\frac{(2n)!\,\pi}{2^{2n-1}(n!)^2}.$$
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