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Math Help - Definite integral - tricky substitutions

  1. #1
    Member Pranas's Avatar
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    Definite integral - tricky substitutions

    Hi.

    So I am asked to analyze

    \displaystyle \[\int\limits_0^{2\pi } {\frac{{dx}}{{1 + \varepsilon  \cdot \cos x}}} \]

    However I believe substitution \displaystyle \[u = \tan \frac{x}{2}\] might be very wrong in this situation because of discontinuity.

    Also I guess this may diverge when \displaystyle \[\left| \varepsilon  \right| \geqslant 1\]

    Unfortunately at this very moment I am not aware of a nice proper solution, maybe you have an idea?
    Last edited by Pranas; March 23rd 2011 at 11:45 AM.
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  2. #2
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    Quote Originally Posted by Pranas View Post
    Hi.

    So I am asked to analyze

    \displaystyle \[\int\limits_0^{2\pi } {\frac{{dx}}{{1 + \varepsilon  \cdot \cos x}}} \]

    However I believe substitution \displaystyle \[u = \tan \frac{x}{2}\] might be very wrong in this situation because of discontinuity.

    Also I guess this may diverge when \displaystyle \[\left| \varepsilon  \right| \geqslant 1\]

    Unfortunately at this very moment I am not aware of a nice proper solution, maybe you have an idea?
    This integral is perfect to use a little complex analysis and the residue theorem.

    First note that the substitution \displaystyle z=e^{ix} \implies dz=ie^{ix}dx \iff \frac{dz}{iz}=dx where i=\sqrt{-1} will be used.

    \displaystyle \cos(x)=\frac{e^{ix}+e^{-ix}}{2}=\frac{z+z^{-1}}{2}

    Putting all of this in gives

    \displaystyle \int_{0}^{2\pi}\frac{dx}{1+\epsilon\cos(x)}=\oint_  {|z|=1}\frac{dz}{\left[1+\epsilon(\frac{z+z^{-1}}{2}) \right]iz}=\frac{2}{i\epsilon}\oint_{|z|=1}\frac{dz}{(z+\  frac{1-\sqrt{1-\epsilon^2}}{\epsilon})(z+\frac{1+\sqrt{1-\epsilon^2}}{\epsilon})}

    As you noted the integral only converges when |\epsilon|<1

    So their is only one pole inside the unit disk and it occurs at

    \displaystyle z=\frac{-1+\sqrt{1-\epsilon^2}}{\epsilon}

    Now by the residue theorem we get

    \displaystyle \frac{2}{i\epsilon}\cdot 2\pi i\left( \frac{1}{\frac{-1+\sqrt{1-\epsilon^2}}{\epsilon}+\frac{1+\sqrt{1-\epsilon^2}}{\epsilon}}\right)=\frac{4\pi}{\epsilo  n} \left( \frac{1}{\frac{2\sqrt{1-\epsilon^2}}{\epsilon}}\right)=\frac{2 \pi}{\sqrt{1-\epsilon^2}}
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  3. #3
    Member Pranas's Avatar
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    TheEmptySet, thank you very much.
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