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In my current research in electromagnetics I am encountering integrals of the form $$ \int_0^\infty dt J_0( r t) \frac{\exp(-h \sqrt{t^2 - a^2})}{\sqrt{t^2 - b^2}} t . $$ $a$ and $b$ are complex numbers, never purely real. If $a=b$, the integral can be done analytically (see, e. g. Watson, A Treatise on the Theory of Bessel Functions, pg. 416, Eqn 4). One just recognizes that the ratio $\exp(.)/\sqrt{.}$ is proportional to the Bessel function $$ K_{\frac{1}{2}} (.), $$ which appears in a doable integral of the correct form.

However, I haven’t been able to find any information on such integrals for $ a \ne b $. My current thought is to rewrite this as $$ const \int_0^\infty dt J_0(t r) K_{\mu}( h \sqrt{t^2 – a^2}) t \frac{\sqrt{t^2 - a^2}}{\sqrt{t^2 - b^2}}, $$ and to write the ratio of square roots as a hypergeometric function (to have proper analyticity to allow it to be integrated over the desired range) that I can expand and integrate term-by-term, or use Mellin transforms on, etc.

This gets me to my question – I know there are Mathematica packages (I don’t have Maple) for recognizing hypergeometric functions and using the relations between them, but I don’t really understand enough about them to know which package I could use. Can someone point me to the right package and some instructions on how to use it, or to some basic literature? I’m not at all educated on hypergeometric series and how to construct them for a given function. I looked at ”Special Functions” by Andrews, Askey and Roy, but I need a different introduction showing how to construct the hypergeometric series for a specified function.

Many thanks, Tom

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This isn't what you are looking for, but it doesn't keep it from being way entertaining, and on topic. math.upenn.edu/~wilf/AeqB.html You can get Mathematica code to go with it. –  Charlie Frohman Jun 26 '11 at 23:04
    
Charlie, thanks for the link. I had downloaded that book a while back, but had forgotten about it. It doesn't exactly solve my problem, but it is overall quite helpful! Tom –  Tom Dickens Jun 30 '11 at 0:20
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