# Cauchy integral formula

• May 7th 2009, 06:53 PM
TheEmptySet
Cauchy integral formula
Okay this has been driving me crazy and I cant find my error.

Okay here it is.

$\displaystyle \int_{0}^{\pi}\sin^{2n}(\theta)d\theta ,n \in \mathbb{Z}^{+}$

So first Since sine is periodic and it is squared

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

So if I parameterize the unit circle in in the complex plane with

$\displaystyle z=e^{i\theta} \implies dz=ie^{i\theta}d\theta \iff \frac{dz}{iz}=d\theta$

Since $\displaystyle \sin(\theta)=\frac{e^{i\theta}-e^{i\theta}}{2i} \implies \sin^{2n}(\theta)=\left( \frac{e^{i\theta}-e^{i\theta}}{2i} \right)^{2n}$

Now if I sub z in I get

$\displaystyle \sin^{2n}(z)=\frac{(z-z^{-1})^{2n}}{(2i)^{2n}}$

$\displaystyle \int_{\Gamma}\frac{(z-z^{-1})^{2n}}{(2i)^{2n}}\frac{dz}{iz}=\frac{1}{i\cdot 2^{2n}}\int_{\Gamma}\left(z- \frac{1}{z}\right)^{2n}\frac{dz}{z}=$

$\displaystyle \frac{1}{i\cdot 2^{2n}}\int_{\Gamma}\frac{(z^2-1)^{2n}}{z^{2n}}\frac{dz}{z} = \frac{1}{i\cdot 2^{2n}}\int_{\Gamma}\frac{(z^2-1)^{2n}}{z^{2n+1}}dz$

Here I use the binomial theorem to expand the numerator

$\displaystyle (z^2-1)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}(z^2)^{2n-k}(-1)^{k}$

$\displaystyle \frac{1}{i\cdot 2^{2n}}\int_{\Gamma}\frac{1}{z^{2n+1}}\left(\sum_{ k=0}^{2n}\binom{2n}{k}(z^2)^{2n-k}(-1)^{k} \right)dz=$

$\displaystyle \frac{1}{i\cdot 2^{2n}}\int_{\Gamma}\left(\sum_{k=0}^{2n}\binom{2n }{k}(z)^{2n-2k-1}(-1)^{k} \right)dz=$

$\displaystyle \frac{1}{i\cdot 2^{2n}} \left(\sum_{k=0}^{2n}\binom{2n}{k}(-1)^{k} \right)\int_{\Gamma}(z)^{2n-2k-1} dz=$

Now by the cauchy integral theorem the only non zero integral is when k=n and it gives

$\displaystyle \int_{\Gamma}\frac{1}{z}dz =2\pi i$

$\displaystyle \frac{1}{i\cdot 2^{2n}} \left(\sum_{k=0}^{2n}\binom{2n}{k}(-1)^{k} \right)\int_{\Gamma}(z)^{2n-2k-1} dz=\frac{1}{i\cdot 2^{2n}}\binom{2n}{n}(-1)^{n}(2\pi i)=\frac{2\pi (2n)!}{(n!)^22^{2n}}(-1)^{n}$

So finally we have

$\displaystyle 2\int_{0}^{\pi}\sin^{2n}(\theta)d\theta=\frac{2\pi (2n)!}{(n!)^22^{2n}}(-1)^{n}$

$\displaystyle \int_{0}^{\pi}\sin^{2n}(\theta)d\theta=\frac{\pi (2n)!}{(n!)^22^{2n}}(-1)^{n}$

So I checked this formula numerically and it works except for the $\displaystyle (-1)^{n}$

I know that is can't be correct becuase the integrand is always non negative.

Thanks for looking
• May 7th 2009, 08:37 PM
TheEmptySet
Okay I see it (Giggle)

$\displaystyle \int_{\Gamma}\frac{(z-z^{-1})^{2n}}{(2i)^{2n}}\frac{dz}{iz}=\frac{1}{i\cdot 2^{2n}}\int_{\Gamma}\left(z- \frac{1}{z}\right)^{2n}\frac{dz}{z}=$

When I factored this out I made a mistake

$\displaystyle \frac{1}{(2i)^{2n}}=\frac{1}{2^{2n}(i)^{2n}}=\frac {1}{2^{2n}[(i)^2]^n}=\frac{1}{2^{2n}(-1)^n}$

I made that same mistake 3 times and never saw it(Crying)
• May 7th 2009, 08:47 PM
Banned for attempted hacking
I was going to say that (Cool).(not really)

What course is this ?
• May 8th 2009, 03:39 AM
mr fantastic
Quote:

Originally Posted by TheEmptySet
Okay this has been driving me crazy and I cant find my error.

Okay here it is.

$\displaystyle \int_{0}^{\pi}\sin^{2n}(\theta)d\theta ,n \in \mathbb{Z}^{+}$

So first Since sine is periodic and it is squared

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

So if I parameterize the unit circle in in the complex plane with

$\displaystyle z=e^{i\theta} \implies dz=ie^{i\theta}d\theta \iff \frac{dz}{iz}=d\theta$

Since $\displaystyle \sin(\theta)=\frac{e^{i\theta}-e^{i\theta}}{2i} \implies \sin^{2n}(\theta)=\left( \frac{e^{i\theta}-e^{i\theta}}{2i} \right)^{2n}$

Now if I sub z in I get

$\displaystyle \sin^{2n}(z)=\frac{(z-z^{-1})^{2n}}{(2i)^{2n}}$

$\displaystyle \int_{\Gamma}\frac{(z-z^{-1})^{2n}}{(2i)^{2n}}\frac{dz}{iz}=\frac{1}{i\cdot 2^{2n}}\int_{\Gamma}\left(z- \frac{1}{z}\right)^{2n}\frac{dz}{z}=$

$\displaystyle \frac{1}{i\cdot 2^{2n}}\int_{\Gamma}\frac{(z^2-1)^{2n}}{z^{2n}}\frac{dz}{z} = \frac{1}{i\cdot 2^{2n}}\int_{\Gamma}\frac{(z^2-1)^{2n}}{z^{2n+1}}dz$

Here I use the binomial theorem to expand the numerator

$\displaystyle (z^2-1)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}(z^2)^{2n-k}(-1)^{k}$

$\displaystyle \frac{1}{i\cdot 2^{2n}}\int_{\Gamma}\frac{1}{z^{2n+1}}\left(\sum_{ k=0}^{2n}\binom{2n}{k}(z^2)^{2n-k}(-1)^{k} \right)dz=$

$\displaystyle \frac{1}{i\cdot 2^{2n}}\int_{\Gamma}\left(\sum_{k=0}^{2n}\binom{2n }{k}(z)^{2n-2k-1}(-1)^{k} \right)dz=$

$\displaystyle \frac{1}{i\cdot 2^{2n}} \left(\sum_{k=0}^{2n}\binom{2n}{k}(-1)^{k} \right)\int_{\Gamma}(z)^{2n-2k-1} dz=$

Now by the cauchy integral theorem the only non zero integral is when k=n and it gives

$\displaystyle \int_{\Gamma}\frac{1}{z}dz =2\pi i$

$\displaystyle \frac{1}{i\cdot 2^{2n}} \left(\sum_{k=0}^{2n}\binom{2n}{k}(-1)^{k} \right)\int_{\Gamma}(z)^{2n-2k-1} dz=\frac{1}{i\cdot 2^{2n}}\binom{2n}{n}(-1)^{n}(2\pi i)=\frac{2\pi (2n)!}{(n!)^22^{2n}}(-1)^{n}$

So finally we have

$\displaystyle 2\int_{0}^{\pi}\sin^{2n}(\theta)d\theta=\frac{2\pi (2n)!}{(n!)^22^{2n}}(-1)^{n}$

$\displaystyle \int_{0}^{\pi}\sin^{2n}(\theta)d\theta=\frac{\pi (2n)!}{(n!)^22^{2n}}(-1)^{n}$

So I checked this formula numerically and it works except for the $\displaystyle (-1)^{n}$

I know that is can't be correct becuase the integrand is always non negative.

Thanks for looking

*Sigh* If only all question posters showed their working and what they tried .... (Clapping)
• May 8th 2009, 04:19 AM
running-gag
Quote:

Originally Posted by TheEmptySet
$\displaystyle \frac{1}{i\cdot 2^{2n}} \left(\sum_{k=0}^{2n}\binom{2n}{k}(-1)^{k} \right)\int_{\Gamma}(z)^{2n-2k-1} dz=$

Now by the cauchy integral theorem the only non zero integral is when k=n and it gives

$\displaystyle \int_{\Gamma}\frac{1}{z}dz =2\pi i$

Hi

I saw your mistake but you saw it before me (Wink)

Just a question anyway : I would say that the Cauchy integral theorem tells that the integrals are zero for $\displaystyle 0 \leq k \leq n-1$ ; the integrals for $\displaystyle n+1 \leq k \leq 2n$ are also zero but not for the same reason. Or could you explain how you can use the Cauchy integral theorem for $\displaystyle n+1 \leq k \leq 2n$ ?
• May 8th 2009, 03:02 PM
TheEmptySet
Quote:

Originally Posted by running-gag
Hi

I saw your mistake but you saw it before me (Wink)

Just a question anyway : I would say that the Cauchy integral theorem tells that the integrals are zero for $\displaystyle 0 \leq k \leq n-1$ ; the integrals for $\displaystyle n+1 \leq k \leq 2n$ are also zero but not for the same reason. Or could you explain how you can use the Cauchy integral theorem for $\displaystyle n+1 \leq k \leq 2n$ ?

You are absolutely correct they are zero, but not by the cauchy integral fromula.

They are zero becuase they are entire functions on $\displaystyle \mathbb{C}$ so the integral around any close path must be zero.

Thanks
• May 10th 2009, 09:42 AM
Moo
If you want more methods for calculating this integral, it is referred as "Wallis sine formula"