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I would like to calculate the definite integral with K-Bessel funcitons and a and b complex (n and k integers):

$$\int_{0}^{\infty} x \;K_{a}(nx) \; K_{b}(kx) \; dx$$

I could not find it in litterature with a and b complex (We have $Re(a)<1$ and $Re(b)<1$ for convergence in zero!).

Any reference or help on this subject is welcome.

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Mathematica: $$\int_{0}^{\infty} x \;K_{a}(nx) \; K_{b}(kx) \; dx=\frac{1}{2n^2}(k n)^{-b} $$ $$\qquad\times \left[n^{2 b} \Gamma (b) \Gamma \left(-\frac{a}{2}-\frac{b}{2}+1\right) \Gamma \left(\tfrac{1}{2} (a-b+2)\right) \, _2F_1\left(\tfrac{1}{2} (-a-b+2),\tfrac{1}{2} (a-b+2);1-b;\frac{k^2}{n^2}\right)\right.$$ $$\qquad\left.+\,k^{2 b} \Gamma (-b) \Gamma \left(\tfrac{1}{2} (-a+b+2)\right) \Gamma \left(\tfrac{1}{2} (a+b+2)\right) \, _2F_1\left(\tfrac{1}{2} (-a+b+2),\tfrac{1}{2} (a+b+2);b+1;\frac{k^2}{n^2}\right)\right]$$

for $-2<{\rm Re}(a-b)<2$, $-2<{\rm Re}(a+b)<2$

if $k=n$ this simplifies to

$$\int_{0}^{\infty} x \;K_{a}(nx) \; K_{b}(nx)=\frac{\pi ^2 (a^2-b^2)}{4 n^2 (\cos \pi b-\cos \pi a)}$$

there do not seem to be further simplifications if $k\neq n$

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    $\begingroup$ See also entry 6.576.4 in Gradshteyn-Ryzhik: Tables of Integrals, Series, and Products (7th edition, 2007) for a somewhat different looking expression. $\endgroup$
    – GH from MO
    Commented Apr 3, 2016 at 17:46
  • $\begingroup$ In arxiv.org/abs/0801.0891 the following result is given $$\int\limits_0^\infty x\,K_0(ax)\,K_0(bx)\,dx=\frac{\ln{(a/b)}}{a^2-b^2},$$ which supplements the last equality (with $k=n$). $\endgroup$
    – Zurab Silagadze
    Commented Apr 3, 2016 at 23:36

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