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Math Help - Cauchy integral formula

  1. #1
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    Cauchy integral formula

    Okay this has been driving me crazy and I cant find my error.

    Okay here it is.

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

    So first Since sine is periodic and it is squared

    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

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

    Since \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

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

    \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}=

    \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

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

    \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=

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

    \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

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

    \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

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

    \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 (-1)^{n}

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

    Thanks for looking
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  2. #2
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    Okay I see it

    \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

    \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
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  3. #3
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    I was going to say that .(not really)

    Here have a cookie *cookie*

    What course is this ?
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    Okay this has been driving me crazy and I cant find my error.

    Okay here it is.

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

    So first Since sine is periodic and it is squared

    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

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

    Since \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

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

    \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}=

    \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

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

    \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=

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

    \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

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

    \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

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

    \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 (-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 ....
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    \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

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

    I saw your mistake but you saw it before me

    Just a question anyway : I would say that the Cauchy integral theorem tells that the integrals are zero for 0 \leq k \leq n-1 ; the integrals for 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 n+1 \leq k \leq 2n ?
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  6. #6
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    Quote Originally Posted by running-gag View Post
    Hi

    I saw your mistake but you saw it before me

    Just a question anyway : I would say that the Cauchy integral theorem tells that the integrals are zero for 0 \leq k \leq n-1 ; the integrals for 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 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 \mathbb{C} so the integral around any close path must be zero.

    Thanks
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  7. #7
    Moo
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    If you want more methods for calculating this integral, it is referred as "Wallis sine formula"
    Though I can't find interesting links about it on google -_-
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