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Math Help - Integrate using complex

  1. #1
    MHF Contributor Amer's Avatar
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    Integrate using complex

    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
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  2. #2
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    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
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by shawsend View Post
    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
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  4. #4
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    Quote Originally Posted by shawsend View Post

    \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
    Last edited by shawsend; November 3rd 2009 at 12:32 PM. Reason: Corrected residue results
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  5. #5
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by shawsend View Post
    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.
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  6. #6
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    Quote Originally Posted by Amer View Post
    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!
    Last edited by Enrique2; November 4th 2009 at 06:16 AM. Reason: Correcting the last expression
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