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I Closed form for integral

  1. Jul 3, 2016 #1


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    Does anyone know whether the following integral has a closed-form solution? If not, is anything known about the asymptotic behavior?

    [itex]f(x) = \int_{-\infty}^{+\infty} \frac{e^{iux}}{\sqrt{u^2 + 1}} du[/itex]
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  3. Jul 3, 2016 #2

    Paul Colby

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    Does it help that ##f''(x)-f(x)=\delta(x)##?

    Oops missed the square root
  4. Jul 3, 2016 #3

    Charles Link

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    I don't know the complex variable residue theory as well as I should, but for x>0, I think the contour closes in the upper half plane and has a single pole at z=+i inside the contour so that ## f(x)=2 \pi i exp(-x) ##. For x<0, the contour would close in the lower half plane with a pole at z=-i and the counterclockwise path would give a minus sign, so that ## f(x)=-2 \pi i exp(+x) ##. I'm not sure about this function with the sqrt. in the denominator is considered to have simple poles. Perhaps @micromass can answer that. editing... and I missed the term of sqrt (2i) in the denominator for the part without the pole. Will need to research this further...editing ... multiplying numerator and denominator by sqrt (u^2+1) gives the function simple poles, but I'm still at the drawing board=perhaps there is a simple solution... editing... thought I had a possible answer, but still at the drawing board...
    Last edited: Jul 3, 2016
  5. Jul 3, 2016 #4
    It is modified Bessel functions of the second kind ##K_0(x)## (I assume ##z=x## in your integral), see, for example, here, equation (6). You should find a lot of information about Bessel functions (as well as about all other special functions) in Abramowitz and Stegun.
  6. Jul 4, 2016 #5


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    Thanks. I should have noticed that the imaginary part of my integral vanishes by symmetry.
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