I am trying to take the spherically-symmetric Fourier transform (i.e. Hankel transform) of a function. I have seen it stated in a paper that the FT of this function "doesn't exist", so it may not, but I want to give it a good crack anyway (the paper was an astronomy paper, not a maths paper, so there may be a chance they just didn't try hard enough).

Anyway, in the end the function to integrate is:

$$ \int_0^c x \sin (\kappa x) \exp\left(-\frac{2x^\alpha}{\alpha}\right)\ dx.$$

I've tried doing this by parts but I don't think it works. I also tried Wolfram|Alpha which instantly told me that there is no solution in terms of standard mathematical functions. SymPy also gave me nothing.

I was considering trying to use Meijer-G functions to do it. However, to do this I either need the whole integrand to be represented by a single Meijer-G function (which I'm not sure is possible), or I need it to be composed of 2 or 3 multiplied G functions, and have the limits of the integral be $(0,\infty)$.

Following the latter route, I did a substitution, $y = 1/(x-c) - 1/c$, which gives the correct limits:

$$ \int_0^\infty \frac{cy}{(y+1/c)^3}\exp\left(-\frac{2}{\alpha}\left(\frac{cy}{y+1/c}\right)^\alpha\right)\sin\left(\kappa \frac{cy}{y+1/c}\right)\ dy, $$

but makes my eyes hurt and I suspect that finding a reasonable G function for these terms will be difficult/impossible.

Does anyone have any ideas? Thanks!

  • $\begingroup$ As the upper bound is arbitrary, this is equivalent to an indefinte integral and no "definite integral trick" will work. If you integrate on $\kappa$ you get rid of the factor $x$ (and trade $\sin$ for $\cos$). Then by rescaling the variable you can remove the $\alpha$ denominator. But what remains still looks hopeless. $\endgroup$ – Yves Daoust Sep 1 '16 at 21:42
  • $\begingroup$ Thanks for your comment. I can't see any of the symbols in your answer for some reason. But nice to know it still seems hopeless :) $\endgroup$ – StevenMurray Sep 2 '16 at 1:48
  • $\begingroup$ You can read the simplified integrand as the complex exponential of a (fractional) polynomial, which is often bad news. $\endgroup$ – Yves Daoust Sep 2 '16 at 9:11

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