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Math Help - Evaluating Improper integral

  1. #1
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    Evaluating Improper integral

    1. The calculation of the probability of excitation of an atom originally in the
    ground state to an excited state, involves the contour integral
    INTEGRAL(-INF TO +INF) [s exp(iwt)/(t2 + s2)2dt]
    Evaluate the above integral.
    Please help me in evaluating this integral.
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  2. #2
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    Quote Originally Posted by GAGS View Post
    1. The calculation of the probability of excitation of an atom originally in the
    ground state to an excited state, involves the contour integral
    INTEGRAL(-INF TO +INF) [s exp(iwt)/(t2 + s2)2dt]
    Evaluate the above integral.
    Please help me in evaluating this integral.
     \int_{-\infty}^{\infty} \frac{s e^{(i \omega t)}}{(t^2 + s^2)^2} dt

    Is that the integral?
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  3. #3
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    REPLY

    Yes sir this is that integral
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     \int_{-\infty}^{\infty} \frac{s e^{(i \omega t)}}{(t^2 + s^2)^2} dt = \int_{-\infty}^{0}  \frac{s e^{(i \omega t)}}{(t^2 + s^2)^2} dt +  \int_0^{\infty} \frac{s e^{(i \omega t)}}{(t^2 + s^2)^2} dt

    = \lim_{n \to -\infty} \int_{n}^{0}  \frac{s e^{(i \omega t)}}{(t^2 + s^2)^2} dt+ \lim_{m \to \infty}  \int_{0}^{m} \frac{s e^{(i \omega t)}}{(t^2 + s^2)^2} dt

    Does this help?
    Last edited by Mush; December 15th 2008 at 05:29 AM.
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  5. #5
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    REPLY

    Yes upto a bit but we have just changed notations. How to solve this intergal?
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  6. #6
    Member Greengoblin's Avatar
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    First solve the integrals, as you would normally, and then take the limits of these functions, then the sum of the limits.
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  7. #7
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    I think what we're trying to say is that we don't know how to solve the integral (I don't think the function converges!), and that the only advice we can offer is general and vague advice on how to solve general improper integrals using limits...
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  8. #8
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    Quote Originally Posted by GAGS View Post
    1. The calculation of the probability of excitation of an atom originally in the ground state to an excited state, involves the contour integral
    \int_{-\infty}^{\infty}\frac{s \exp(i\omega t)}{(t^2 + s^2)^2}dt
    Evaluate the above integral.
    I'll assume that \omega>0. Integrate the function \frac{s \exp(i\omega z)}{(z^2 + s^2)^2} round the contour consisting of (a long interval on) the real axis plus a large semicircle in the upper half-plane. The result will be equal to 2πi times the residue at the double pole at z=is.

    The residue is \frac d{dz}\Bigl(\frac{se^{i\omega z}}{(z+is)^2}\Bigr)\Bigr|_{z=is}. If I have done the calculation correctly then \int_{-\infty}^{\infty}\frac{s \exp(i\omega t)}{(t^2 + s^2)^2}dt = \frac{\pi(s+\omega)e^{-\omega s}}{2s^2}.

    If \omega<0 then the answer will be the same except that you have to replace \omega by |\omega|.
    Last edited by Opalg; December 16th 2008 at 08:23 AM. Reason: corrected typo
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