Integrate using complex

**Amer** · November 3rd 2009, 04:09 AM

I stuck with this question I need any hint in it thanks in advance

show that

$\int_{0}^{\pi} \sin ^{2n} \theta d\theta =\frac{(2n)! \pi}{2^{2n} (n!)^2}$

my work
let

$z=e^{i\theta}$

$d\theta = \frac{dz}{iz}$

and
$\sin \theta = \frac{1}{2i}\left(z-\frac{1}{z} \right)$

so

$\int_{\mid z \mid =1 } \left(\frac{1}{2i}\right)^{2n}\left(z-\frac{1}{z} \right)^{2n} \frac{dz}{zi}$

$\int_{\mid z \mid =1 }\left(\frac{1}{2i}\right)^{2n} \left(\frac{z^2-1}{z}\right)^{2n}\frac{dz}{zi}$

$\int_{\mid z \mid =1 } \frac{(z^2-1)^{2n}}{z^{2n+1}} dz$

the function is not analytic at z=0
let $g(z) =(z^2-1)^{2n}$ so

$\int_{\mid z \mid =1 } \frac{(z^2-1)^{2n}}{z^{2n+1}} dz= \frac{2\pi i g^{2n}(0)}{(2n)!}$

note $g^{2n}$ the 2n-th derivatives of g(z)

but I can't find the 2n-th derivative it is long I think there is a mistake in my work any help will be appreciated

**shawsend** · November 3rd 2009, 06:43 AM

I did this quick so check it well.

If you're going to use the Residue Theorem, go all the way round. Note:

$\int_0^{\pi} \sin^{2n}(t)dt=1/2 \int_0^{2\pi} \sin^{2n}(x)dx$

Now let $z=e^{it}$ and when I do that I get:

$-\frac{i}{2}\mathop\oint\limits_{|z|=1} \frac{(z^2-1)^{2n}}{z^{2n}} \frac{1}{2^{2n}}\frac{1}{i^{2n}} \frac{1}{z} dz$

For now, suppose we only wanted to solve:

$\mathop\oint\limits_{|z|=1}\frac{(z^2-1)^{2n}}{z^{2n+1}}dz$

That's still a messy derivative but we can simplify it if we make a change of variables. What happens to both the integrand and contour if I let $u=z^2-1$ ? If $z=e^{it}$ , then $u=-1+e^{2it}$ right? Isn't that a circular contour around the point $u=-1$ and going around twice as t goes from 0 to 2pi? Make all those substitutions and try and show:

$\mathop\oint\limits_{|z|=1} \frac{(z^2-1)^{2n}}{z^{2n+1}}dz=\frac{1}{2}\mathop\oint\limit s_{\tiny\begin{array}{c}u=-1+e^{2it} \\ 0\leq t \leq 2\pi\end{array}} \frac{u^{2n}}{(u+1)^n}du$

**Amer** · November 3rd 2009, 08:46 AM

Originally Posted by shawsend

I did this quick so check it well.

If you're going to use the Residue Theorem, go all the way round. Note:

$\int_0^{\pi} \sin^{2n}(t)dt=1/2 \int_0^{2\pi} \sin^{2n}(x)dx$

Now let $z=e^{it}$ and when I do that I get:

$-\frac{i}{2}\mathop\oint\limits_{|z|=1} \frac{(z^2-1)^{2n}}{z^{2n}} \frac{1}{2^{2n}}\frac{1}{i^{2n}} \frac{1}{z} dz$

For now, suppose we only wanted to solve:

$\mathop\oint\limits_{|z|=1}\frac{(z^2-1)^{2n}}{z^{2n+1}}dz$

That's still a messy derivative but we can simplify it if we make a change of variables. What happens to both the integrand and contour if I let $u=z^2-1$ ? If $z=e^{it}$ , then $u=-1+e^{2it}$ right? Isn't that a circular contour around the point $u=-1$ and going around twice as t goes from 0 to 2pi? Make all those substitutions and try and show:

$\mathop\oint\limits_{|z|=1} \frac{(z^2-1)^{2n}}{z^{2n+1}}dz=\frac{1}{2}\mathop\oint\limit s_{\tiny\begin{array}{c}u=-1+e^{2it} \\ 0\leq t \leq 2\pi\end{array}} \frac{u^{2n}}{(u+1)^n}du$

Thanks very much

**shawsend** · November 3rd 2009, 09:10 AM

Originally Posted by shawsend

$\mathop\oint\limits_{|z|=1} \frac{(z^2-1)^{2n}}{z^{2n+1}}dz=\frac{1}{2}\mathop\oint\limit s_{\tiny\begin{array}{c}u=-1+e^{2it} \\ 0\leq t \leq 2\pi\end{array}} \frac{u^{2n}}{(u+1)^n}du$

After checking it, I think this should be:

$\mathop\oint\limits_{|z|=1} \frac{(z^2-1)^{2n}}{z^{2n+1}}dz=\frac{1}{2}\mathop\oint\limit s_{\tiny\begin{array}{c}u=-1+e^{2it} \\ 0\leq t \leq 2\pi\end{array}} \frac{u^{2n}}{(u+1)^{n+1}}du$

Here's a numeric check of this relationship with n=6 (because this sort of thing is new for me too). Remember, we're going around twice in the second integral so I just multiplied it by two and went around once:

Code:

In[65]:= NIntegrate[(u^(2*n)/(u + 1)^(n + 1))*I*
       Exp[I*t] /. {n -> 6, 
       u -> -1 + Exp[I*t]}, {t, 0, 2*Pi}]
N[2*Pi*I*((-1)^n(2*n)!/n!^2) /. n -> 6]
NIntegrate[((z^2 - 1)^(2*n)/z^(2*n + 1))*
       I*Exp[I*t] /. {n -> 6, 
       z -> Exp[I*t]}, {t, 0, 2*Pi}]

Out[65]= -9.379164112033322*^-13 + 
   5805.66322740536*I

Out[66]= 0. + 5805.663223833938*I

Out[67]= -1.3515644212702682*^-6 + 
   5805.663226625173*I

**Amer** · November 3rd 2009, 09:52 AM

Originally Posted by shawsend

After checking it, I think this should be:

$\mathop\oint\limits_{|z|=1} \frac{(z^2-1)^{2n}}{z^{2n+1}}dz=\frac{1}{2}\mathop\oint\limit s_{\tiny\begin{array}{c}u=-1+e^{2it} \\ 0\leq t \leq 2\pi\end{array}} \frac{u^{2n}}{(u+1)^{n+1}}du$

Here's a numeric check of this relationship with n=6 (because this sort of thing is new for me too). Remember, we're going around twice in the second integral so I just multiplied it by two and went around once:

ya I figured that but what I want is the idea just and you gave me it, thanks.

**Enrique2** · November 3rd 2009, 04:46 PM

Originally Posted by Amer

I stuck with this question I need any hint in it thanks in advance

show that

$\int_{0}^{\pi} \sin ^{2n} \theta d\theta =\frac{(2n)! \pi}{2^{2n} (n!)^2}$

my work
let

$z=e^{i\theta}$

$d\theta = \frac{dz}{iz}$

and
$\sin \theta = \frac{1}{2i}\left(z-\frac{1}{z} \right)$

so

$\int_{\mid z \mid =1 } \left(\frac{1}{2i}\right)^{2n}\left(z-\frac{1}{z} \right)^{2n} \frac{dz}{zi}$

$\int_{\mid z \mid =1 }\left(\frac{1}{2i}\right)^{2n} \left(\frac{z^2-1}{z}\right)^{2n}\frac{dz}{zi}$

$\int_{\mid z \mid =1 } \frac{(z^2-1)^{2n}}{z^{2n+1}} dz$

the function is not analytic at z=0
let $g(z) =(z^2-1)^{2n}$ so

$\int_{\mid z \mid =1 } \frac{(z^2-1)^{2n}}{z^{2n+1}} dz= \frac{2\pi i g^{2n}(0)}{(2n)!}$

note $g^{2n}$ the 2n-th derivatives of g(z)

but I can't find the 2n-th derivative it is long I think there is a mistake in my work any help will be appreciated

You are really near from the solution!!

First, one small mistake in the calculations, take in account $\left(\frac{1}{2i}\right)^{2n}\frac{1}{i}=\frac{(-1)^n}{i2^{2n}}$ . This factor is missed in the integral.

Second, as Shawsend obseved, the integral you want is one half of the integral around the circle.

Third, observe that your $g(z)=b_0+b_1z+\cdots+b_{4n}z^{4n}$ , i.e. is a polynomial of degree $4n$ . Thus, $g^{(2n)} (0)=(2n)!b_{2n}$ . By the Newton binomial formula applied to $(z^2+(-1))^{2n}$ we have

$b_{2n}=\left(\begin{array}{cc}2n \\ n\end{array}\right)(-1)^n=\frac{(2n)!}{n!^2}(-1)^n$ .

All together lead you to the desired formula!

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