Here I have the integral $$\int_0^{\infty}x^{\ell-1}K_m(x)K_n(\frac bx)\,dx$$ which is the integral of the multiplication of two modified bessel function of the second kind. I find that this integral is expressed as as a series of the form $$\sum_{v=0}^{\infty}c_vb^{\rho+v}$$ in the reference, i.e., "An Infinite Integral Involving a Product of two Modified Bessel functions of the Second Kind" by T. M. Macrobert. which is used to express the integral as generalized hypergeometric function. I do not know why the integral can be expressed as this series. Can you give me some advice?

  • $\begingroup$ Could you give the "reference" you use? Expression for $c_\nu$ and $\rho$ and/or the result as a generalized hypergeometric function may also be useful. $\endgroup$ – Paul Enta Jun 14 at 8:12
  • $\begingroup$ Thanks for your advice! The reference I use is "AN INFINITE INTEGRAL INVOLVING A PRODUCT OF TWO MODIFIED BESSEL FUNCTIONS OF THE SECOND KIND" by T. M. Macrobert. $\endgroup$ – Micheal.Andy Jun 14 at 14:05
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    $\begingroup$ Their notation $\sum_{n, -n} f(n)$ means $f(n) + f(-n)$, so the result is a sum of four hypergeometric functions. This is solvable with the general method described here. The Mellin transform of $K_m$ is known, and the convolution evaluates to $$2^{l - 3} G_{0, 4}^{4, 0} {\left( \frac {b^2} {16} \middle| {\frac {l - m} 2, \frac {l + m} 2, -\frac n 2, \frac n 2 \atop -} \right)}.$$ When all poles are simple, this expands into a sum of ${_0 F_3}$ functions by Slater's theorem. $\endgroup$ – Maxim Jun 14 at 16:37
  • $\begingroup$ @Maxim: Thank you! I follow the steps that you provide in the link and I have a problem about the variable $\omega$. When using mellin convolution, $\omega$ is the variable of integration, if $\omega$ is equal to a constant, how to integrate it? $\endgroup$ – Micheal.Andy 2 days ago
  • $\begingroup$ Let's make it explicit that $\mathcal M$ takes a function and returns a function and that $f * g$ is also a function: $$\mathcal M[\omega \mapsto (f * g)(\omega)](p) = \mathcal M[f](p) \mathcal M[g](p), \\ (f * g)(\omega) = \mathcal M^{-1}[p \mapsto \mathcal M[f](p) \mathcal M[g](p)](\omega).$$ $\endgroup$ – Maxim 2 days ago

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