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Integration of exponential of inverse tangent

  1. Dec 14, 2011 #1
    Hi all, i need help with integration of exponential of inverse tangent, could not find it in table of integrals

    the whole equation is

    -am trying to integrate by parts but stuck at the arctan part

    Thank you.
  2. jcsd
  3. Dec 14, 2011 #2


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    You won't be able to find a closed form in terms of elementary functions. It looks like the answer will necessarily involve the special function [itex]\mbox{Ei}(z)[/itex], defined for real numbers as

    [tex]\mbox{Ei}(x) = \int_{-\infty}^x \frac{e^t}{t},[/tex]
    though your integral will require the extension of this function on the complex plane.

    Anyways, for your problem, I would use [itex]e^{-i\theta} = \cos\theta - i \sin\theta[/itex]. Since [itex]\theta = \tan^{-1}(z/\sqrt{a})[/itex] you can rewrite the sine and cosine in terms of only z, a, and (z^2+a)^(0.5). The last hint I will give is that the overall denominator you get can be factored (with imaginary roots) to cancel with part of the numerator.
  4. Dec 15, 2011 #3
    Thank you mute. I redid the initial equations and put the inverse tangent term as arctan (z/a) instead of arctan (z/a^1/2) as earlier.ignore the w as i already factor it out. so now i got

    I=∫A/[(z^2+a^2)^1/2]*exp(bz)exp(-i*arctan(z/a)) *dz

    =A∫1/[(z^2+a^2)^1/2]*exp(bz)*cos(theta)-i*sin(theta) *dz

    where theta=arctan(z/a)




    After this, let u=a+iz


    This is referring the exponential integral term rite. I hope this is what you meant.
    Thanks again.
  5. Dec 15, 2011 #4


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    I haven't checked it carefully, but yes, that looks about right. You can now identify

    [tex]\int du~\frac{e^{-ibu}}{u} = \mbox{Ei}(-ibu) + \mbox{const}[/tex]
    and put u back in terms of z.
  6. Dec 15, 2011 #5
    Thanks alot mute...
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