Intereting Posts

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A question related to Krull-Akizuki theorem

Show that $$\int_0^\pi \sin^{2n} \theta d\theta=\dfrac{\pi(2n)!}{(2^n n!)^2} $$

So far I have came up with:

$$\sin^{2n} \theta = \left(\dfrac {z-z^{-1}}{2i} \right)^{2n}$$ and I know I should be using:

- Upper bound for zeros of holomorphic function
- compute integral $\int_0^{2\pi} \frac{1}{z-\cos(\phi)} d\phi$
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$$(a+b)^n=\sum_{k=0}^n \dfrac{n!}{(n-k)!k!}a^kb^{n-k}$$

but I’m not sure how to get the conclusion. Any help will be greatly appreciated.

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This integral is easily done using residue theory. Rewrite the integral as

$$\begin{align} \frac12 \int_0^{2 \pi} d\theta \, \sin^{2 n}{\theta} &= \frac12 \frac{-i}{(2 i)^{2 n}} \oint_{|z|=1} \frac{dz}{z} (z-z^{-1})^{2 n}\\ &= \frac{(-1)^n}{2^{2 n+1}} (-i) \sum_{k=0}^{2 n} (-1)^k \binom{2 n}{k} \oint_{|z|=1} \frac{dz}{z} z^{2 n-2 k} \end{align}$$

Note that the contour integral in the RHS of the last equation is zero unless $k=n$, in which case it is equal to $i 2 \pi$. Therefore,

$$\int_0^{\pi} d\theta \, \sin^{2 n}{\theta} = \frac{(-1)^n}{2^{2 n+1}} (-i) (i 2 \pi) (-1)^n \binom{2 n}{n} = \frac{\pi}{2^{2 n}} \frac{(2 n)!}{(n!)^2}$$

as was to be shown.

This is my favorite example of integration by parts.

Let $I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^n(x) dx$.

$I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-1}(x) d(-\cos(x)) = -\sin^{n-1}(x) \cos(x) |_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} (n-1) \sin^{n-2}(x) \cos^2(x) dx$

The first expression on the right hand side is zero since $\sin(0) = 0$ and $\cos(\frac{\pi}{2}) = 0$.

Now rewrite $\cos^2(x) = 1 – \sin^2(x)$ to get

$$I_n = (n-1) \left(\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-2}(x) dx – \int_{0}^{\frac{\pi}{2}} \sin^{n}(x) dx \right) = (n-1) I_{n-2} – (n-1) I_n$$

Rearranging we get $n I_n = (n-1) I_{n-2}$, $I_n = \frac{n-1}{n}I_{n-2}$.

Using this recurrence we get

$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3} I_1$$

$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} I_0$$

$I_1$ and $I_0$ can be directly evaluated to be $1$ and $\frac{\pi}{2}$ respectively and hence,

$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3} = \dfrac{4^k (k!)^2}{(2k+1)!}$$

$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} \frac{\pi}{2} = \dfrac{(2k)!}{(k!)^2}\dfrac{\pi}{2^{2k+1}}$$

Your integral is twice the above integral, since the sine function is symmetric about $x=\pi/2$. Hence, the value of your integral is

$$J_{2k+1} = \dfrac{2^{2k+1} (k!)^2}{(2k+1)!}$$

$$J_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} \frac{\pi}{2} = \dfrac{(2k)!}{(k!)^2}\dfrac{\pi}{2^{2k}}$$

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