How to prove an equality involving an infinity series and the modified Bessel functions? Using a different approach I computed the Fourier expansion of a certain function whose Fourier expansion was already computed in a certain paper. If both results are correct this would imply a certain equality which I want to present here. My question here is if there are direct ways to show the truth of the equality.
$$\sum_{n=0}^\infty \frac{\Gamma(s+n) (4 \pi \alpha \beta)^{s+n}}{\Gamma(2s+n) n! (\alpha+\beta)^{s+n-1/2}}  K_{s+n-1/2}(2 \pi (\alpha+\beta))\\
=2 \pi (\alpha\beta)^{1/2} I_{s-1/2} (2 \pi  \beta) K_{s-1/2}(2\pi  \alpha)
$$
Here $s \in \mathbb C$ with $\Re(s)>1/2$ and $\alpha \ge \beta>0$.
The functions $I_s$ and $K_s$ are just the usual modified Bessel functions of the first and second kind.
 A: To evaluate the l.h.s. series
\begin{equation}
 S=\sum_{n=0}^\infty \frac{\Gamma(s+n) (4 \pi \alpha \beta)^{s+n}}{\Gamma(2s+n) n! (\alpha+\beta)^{s+n-1/2}}  K_{s+n-1/2}(2 \pi (\alpha+\beta))
\end{equation}
we use the integral representation for the modified Bessel function:
\begin{equation}
 K_{\nu}\left(z\right)=\frac{\pi^{\frac{1}{2}}(\frac{1}{2}z)^{\nu}}{\Gamma\left(\nu+\frac{1}{2}\right)}\int_{0}^{\infty}e^{-z\cosh t}(\sinh t)^{2\nu}\mathrm{d}t
\end{equation}
with $\nu=s+n-1/2$ and $z=2\pi(\alpha+\beta)$ to express
\begin{equation}
 S=\sum_{n=0}^\infty \frac{ (4 \pi^2 \alpha \beta)^{s+n}}{\Gamma(2s+n) n!}
 \int_{0}^{\infty}e^{-2\pi(\alpha+\beta)\cosh t}(\sinh t)^{2s+2n-1}\mathrm{d}t
\end{equation}
All the terms are positive, then changing the order of summation and integration gives
\begin{equation}
 S=\int_{0}^{\infty}\mathrm{d}t\,e^{-2\pi(\alpha+\beta)\cosh t}(\sinh t)^{2s-1}(4 \pi^2 \alpha \beta)^{s}\sum_{n=0}^\infty \frac{ (4 \pi^2 \alpha \beta)^{n}}{ n!\Gamma(2s+n)}
 (\sinh t)^{2n}
\end{equation}
With the series expansion of the modified Bessel function of the first kind:
\begin{equation}
 I_{\nu}\left(z\right)=(\tfrac{1}{2}z)^{\nu}\sum_{n=0}^{\infty}\frac{(\tfrac{1}{4}z^{2})^{n}}{n!\Gamma\left(\nu+n+1\right)}
\end{equation}
we deduce, by taking $\nu=2s-1,z=4\pi\sqrt{\alpha\beta}\sinh t$,
\begin{equation}
  S=2\pi\sqrt{\alpha\beta}\int_{0}^{\infty}e^{-2\pi(\alpha+\beta)\cosh t}I_{2s-1}\left( 4\pi\sqrt{\alpha\beta}\sinh t \right)\mathrm{d}t
\end{equation}
This integral is tabulated (G.& R. (6.669.4)) (Aaternatively it could be evaluated as a tabulated Laplace transform (Ederlyi TI (4.17.15))
by changing $u=\cosh t-1$),
\begin{align}
 \int_0^\infty e^{-\tfrac12(a_1+a_2)\cosh x}I_{2\mu}(\sqrt{a_1a_2}\sinh x)\mathrm dx=\frac{\Gamma(1/2+\mu)}{\sqrt{a_1a_2}\Gamma(1+2\mu)}W_{0,\mu}(a_1)M_{0,\mu}(a_2)
\end{align}
where $W_{0,\mu}(a_1)$ and $M_{0,\mu}(a_2)$ are Whittaker functions. This identity is valid for $\Re(\mu+1/2)>0,\Re \mu>0,a_1>a_2$. Here, with $\alpha>\beta$ and $s>1/2$, we take $a_1=4\pi\alpha,a_2=4\pi\beta,\mu=s-1/2$. Remarking that (DLMF1) and (DLMF2)
\begin{align}
 W_{0,\mu}\left(2z\right)&=\sqrt{\frac{2z}{\pi}}K_\mu\left(z\right)\\
 M_{0,\mu}\left(2z\right)&=2^{2\mu+\frac{1}{2}}\Gamma\left(1+\mu\right)\sqrt{z}I_{\mu}\left(z\right)
\end{align}
we obtain
\begin{equation}
 S=2\pi\sqrt{\alpha\beta}
 \frac{\Gamma(s)\Gamma\left(s+1/2\right)}{\sqrt{4\pi}\Gamma(2s)}2^{2s}I_{2s-1/2}\left(2\pi\beta\right)K_{s-1/2}\left(2\pi\alpha\right)
\end{equation}
Finally, using the Gamma duplication formula,
\begin{equation}
 S=2\pi\sqrt{\alpha\beta}
 I_{s-1/2}\left(2\pi\beta\right)K_{s-1/2}\left(2\pi\alpha\right)
\end{equation}
which is the expected result.
