MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following series has had me held up for the past one week:

$$ \sum_{n=0}^\infty\frac{(2m)_n m^n}{(2m+1/2)_n n!}A^{3n/2} t^n K_{2m+n-1/2}(2\sqrt{A}t)~~~~ A>0, ~t>0, ~m\geq1/2 $$

where $K_{2m+n-1/2}( \cdot )$ is the Modified Bessel Function of the 2nd kind. I have looked into a number of handbooks on special functions, but to no avail.

Any ideas on how to obtain the sum?

share|cite|improve this question
What makes you think the sum can be expressed in closed form? – Andrey Rekalo Oct 16 '11 at 16:01
What sort of result are you looking for? Are you trying to find a closed form in terms of Bessel functions? I think the whole of point of things related to the Bessel function is that there IS no closed form, so instead mathematicians just gave it a new symbol! – Christopher A. Wong Oct 17 '11 at 4:03
Any result that includes the Bessel function, Hypergoemetric functions, Zeta functions, Psi function, Digamma function etc. would do. The point of having it in such a form is (1) Standard numerical packages like Mathematica have fast built in routines for these kind of functions and (2) these functions connect nicely with each other via other relation formulas. In this particular case, this expression is supposed to be a probability density function, so i could for example find its variance in some closed form. – Iconoclast Oct 17 '11 at 11:55
@Christopher: "I think the whole point of trigonometric functions is that there IS no closed form for the function that gives lengths corresponding to angles, so instead mathematicians just gave it a new symbol!" How does that sound? – J. M. Nov 26 '11 at 14:53
That sounds entirely correct. =P – Christopher A. Wong Nov 26 '11 at 20:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.