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Math Help - Complex Analysis (Residual Theorem).

  1. #1
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    Complex Analysis (Residual Theorem).

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    This is what I did... I'm not sure if I'm in the right direction about this problem.

    I am french so excuse me if my translation in english isn't accurate.

    I used the residual theorem.







    The integral of is 0. (Cauchy's Theorem).

    Poles of order 1 at

    Only the pole at z = ia is in the semi-circle drawn above.

    Residual at z = ia :





    What does cos(ia) represent ? Does that even exist?

    If this is totally wrong, then how do I solve this problem?
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  2. #2
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    Hello,

    \cos(ia)=\frac{e^{i(ia)}+e^{-i(ia)}}{2}=\frac{e^{a}+e^{-a}}{2}=\cosh(a)
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  3. #3
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    Quote Originally Posted by illusion419 View Post

    The integral of is 0. (Cauchy's Theorem).
    That's not Cauchy's Theorem. You'd need to show that as the radius goes to infinity, the integral over the half-circle arc goes to zero. But it's easier if you use:

    \mathop\oint\limits_{H_u}\frac{e^{iz}}{z^2+a^2}dz=  2\pi i\mathop\text{Res}\limits_{z=ia}\left\{\frac{e^{iz  }}{z^2+a^2}\right\}=f+gi

    where H_u is the upper half circle contour. I think it's not to hard to show:

    \lim_{R\to\infty}\mathop\int\limits_C \frac{e^{iz}}{z^2+a^2}dz\to 0

    where C is the half-arc contour and on the straight-line segment along the real axis R we have:

    \mathop\int\limits_{R} \frac{e^{iz}}{z^2+a^2}dz=\int_{-\infty}^{\infty} \frac{cos(x)}{x^2+a^2}dx+i\int_{-\infty}^{\infty} \frac{sin(x)}{x^2+a^2}dx=f+gi (since the part over C is zero).

    Then:

    \int_{-\infty}^{\infty} \frac{cos(x)}{x^2+a^2}dx=\textbf{Re}\left[2\pi i\mathop\text{Res}\limits_{z=ia}\left\{\frac{e^{iz  }}{z^2+a^2}\right\}\right]

    [edit] I made a silly mistake on this initially but corrected it here. Hope that didn't cause problems for you Illusion.
    Last edited by shawsend; December 5th 2008 at 12:06 PM. Reason: corrected formulas
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