Let's consider the sum $$\sum_{n=1}^{+\infty}nI_{n}(ax)K_{n}(bx)$$ Here $I_{n}$ and $K_{n}$ are the modified Bessel functions of the first and second kind respectively and $b>a>0$ are real constants.
Using the asymptotics for $n\to +\infty$ for the Bessel functions I proved that the sum converges $\forall x\in\mathbb{R}^{+}$. I would like to know if a closed form exists.
Without the $n$ I would have used the summation formula, but in this case I can't think of anything helpful. Thanks in advance.
Edit: Following Gary advice, I substituted the product $I_{n}K_{n}$ with the integral. Then interchanged the integral and sum obtaining $$\int_{0}^{+\infty}e^{-(b-a)x\cosh{t}}\sum_{n=0}^{+\infty}(n+1)J_{2n+2}(2\sqrt{ab}x\sinh{t})d t$$
If i am not mistaken if I have the sum $\sum_{n=0}^{+\infty}a_{n\nu}J_{\nu+2n+1}$ it holds that $$a_{n\nu}=2(\nu+2n+1)\int_{0}^{+\infty}u^{-1}f(u)J_{\nu+2n+1}(u)du.$$ Here $\nu=1$ and $a_{n}=n+1$ and the condition becomes $$\frac{1}{4}=\int_{0}^{+\infty}u^{-1}f(u)J_{2n+2}(u)du,$$ which is a bit weird, because I have to find a function whose integral against every $J_{2n+2}$ is independant on $n$. Maybe the answer is trivial.