# Thread: Show through C domain.

1. ## Show through C domain.

Show that $\displaystyle{ \int_0^\pi \sin^{2n}(\theta)d\theta=\pi\frac{(2n)!}{2^{2n}(n! )^2} }$.
From $\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ follows that $2i\sin(\theta)e^{i\theta}=e^{i2\theta}-1$.

Exponentiating by $2n$ gets, using $\theta$ and $-\theta$, respectively:
1. $\displaystyle{ 2^{2n}i^{2n}\sin^{2n}(\theta)e^{i2n\theta}=(e^{i2\ theta}-1)^{2n} }$
2. $\displaystyle{ 2^{2n}i^{2n}\sin^{2n}(\theta)e^{-i2n\theta}=(e^{-i2\theta}-1)^{2n} }$.

Subtracting 2. from 1.:
$\displaystyle{ 2^{2n}i^{2n}\sin^{2n}(\theta)\left[\underbrace{e^{in\theta}-e^{-in\theta}}_{2i\sin(2n\theta)}\right]=(e^{i2\theta}-1)^{2n}-(e^{-i2\theta}-1)^{2n} }$.

Integrating both sides from $0$ to $\pi$: $
2^{2n+1}i^{2n+1}\int_0^\pi \sin^{2n}(\theta)\sin(2n\theta)d\theta=\int_0^\pi\ left\{ (e^{i2\theta}-1)^{2n}-(e^{-i2\theta}-1)^{2n} \right\} \stackrel{?}{=}0$

What to do with $\sin(2n\theta)$ inside integral, and, is RHS really zero? Assuming everything is OK before that.

$?$: Two leading $1$'s cancel and all other terms in expansion would give $\displaystyle{ \int_0^\pi e^{i2k\theta}d\theta=0 \text{ ; } k=\pm 1, \pm 2, \hdots, \pm 2n }$

2. This result was known to Wallis. So, would you help me get it without Cauchy?

But I will surely read your (nicely written out) question, as I've just recently been introduced to Cauchy's two theorems.

3. Here is a way it can be done without using complex analysis

Define $I_n=\int_{0}^{\pi}\sin^{2n}(x)dx$

Note that $I_0=\pi$

Now if $n \ge 1$

$I_n=\int_{0}^{\pi}\sin^{2n-1}(x)\sin(x)dx$

Now use integration by parts with $u=\sin^{2n-1}(x) \implies du=(2n-1)\sin^{2n-2}(x)\cos(x)$ and
$dv=\sin(x)dx \implies v=-\cos(x)$

This gives

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

$I_n=(2n-1)\int_{0}^{\pi}\sin^{2n-2}(x)\cos^2(x)dx-(2n-1)\int_{0}^{\pi}\sin^{2n}(x)dx$

Now using the relationship above we get

$I_n=(2n-1)I_{n-1}-(2n-1)I_{n} \iff (2n)I_n=(2n-1)I_{n-1} \iff I_n=\frac{2n-1}{2n}I_{n-1}$

Now with

$I_0=\pi$
The rest can be proved via induction.

4. Originally Posted by TheEmptySet
This gives

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

$I_n=(2n-1)\int_{0}^{\pi}\sin^{2n-2}(x)\cos^2(x)dx-(2n-1)\int_{0}^{\pi}\sin^{2n}(x)dx$
I think you meant
$I_n=(2n-1)\int_{0}^{\pi}\sin^{2n-2}(x)dx-(2n-1)\int_{0}^{\pi}\sin^{2n}(x)dx$.

Originally Posted by TheEmptySet
Now using the relationship above we get

$I_n=\frac{2n-1}{2n}I_{n-1}$

Now with

$I_0=\pi$
The rest can be proved via induction.
I wrote it out ...
$I_1=\frac{1}{2}I_0=\frac{1}{2}\pi$

$I_2=\frac{3}{4}I_1=\frac{1\times 3}{2\times 4}\pi$

$I_3=\frac{5}{6}I_2=\frac{1\times 3\times 5}{2\times 4\times 6}\pi$

$I_4=\frac{7}{8}I_3=\frac{1\times 3\times 5\times 7}{2\times 4\times 6\times 8}\pi$ which is indeed $\neq \frac{(2\times 4)!}{2^{2\times 4}(4!)^2}\pi$, but I really don't know how would you "see" the general form of $\frac{(2n)!}{2^{2n}(n!)^2}\pi$.

5. Well, $I_4=\frac{7}{8}I_3=\frac{1\times 3\times 5\times 7}{2\times 4\times 6\times 8}\pi$ is actually $\neq \frac{(2\times 4)!}{2^{2\times 4}(4!)^2}\pi$, so the general formula $\frac{(2n)!}{2^{2n}(n!)^2}\pi$ isn't correct or I am (constantly) overlooking something.

I mean, the formula doesn't hold even for $I_1$! What am I doing wrong?

6. Originally Posted by courteous
Well, $I_4=\frac{7}{8}I_3=\frac{1\times 3\times 5\times 7}{2\times 4\times 6\times 8}\pi$ is actually $\neq \frac{(2\times 4)!}{2^{2\times 4}(4!)^2}\pi$, so the general formula $\frac{(2n)!}{2^{2n}(n!)^2}\pi$ isn't correct or I am (constantly) overlooking something.

I mean, the formula doesn't hold even for $I_1$! What am I doing wrong?

ummm..... I disagree

$I_0=\pi$

$I_1=\frac{1}{2}I_0=\frac{1}{2}\pi=\frac{1\cdot 2}{2\cdot 2}\pi=\frac{2!}{2^2\cdot 1!}\pi$

As I recommended here is the proof by induction.

Base case done

Assume

$I_k=\pi\frac{(2k)!}{2^{2k}(k!)^2}$

Show $I_{n+1}$

So we know from the relationship from before that

$I_{n+1}=\frac{2(n+1)-1}{2(n+1)}I_n=\frac{2n+1}{2(n+1)}I_n$

Now by the induction hypothesis we get

$\frac{2n+1}{2(n+1)}\cdot \pi \cdot \frac{(2n)!}{2^{2n}(n!)^2}$

Now multiply the numerator and denominator by $(2n+2)$ to get

$\displaystyle \pi \frac{(2n+1)(2n)!(2n+2)}{2(n+1)2^{2n}(n!)(n!)(2n+2 )}=\pi \frac{(2n+2)!}{2^{2n+1}(n+1)!(n!)2(n+1)}=\pi \frac{(2n+2)!}{2^{2n+2}(n+1)!(n+1)!}$

$\displaystyle \pi \frac{[2(n+1)]!}{2^{2(n+1)}[(n+1)!]^2}$

This completes the proof.