Consider the function for $x>0$ and $y>0$ $$ f(x,y) := \frac{\pi x^2}{2y} \mathrm{csch}\left( \frac{\pi y}{2} \right) \left( I_{-1-i y/2}(x) I_{-1+iy/2}(x) - I_{1-i y/2}(x) I_{1+i y/2}(x) \right) $$ where $I_{\nu}(z)$ is a modified Bessel function. It turns out that $$ f(x,y) = 1 $$ which I have confirmed numerically.
How do you prove this? I have tried checking known cross-products of Bessel functions and cannot figure this out.
Take the Wronskian product $$ I_{\nu}(z) I_{-1-\nu}(z) - I_{1+\nu}(z) I_{-\nu}(z) = -\frac{2\sin \nu\pi}{\pi z} $$ and apply the recurrence relation $$ I_{\pm \nu}(z) = \pm \frac{2\nu}{z} ( I_{-1 \pm \nu}(z) + I_{1 \pm \nu}(z) ). $$ This gives, after some cancellation, $$ -\frac{2\sin \nu\pi}{\pi z} = \frac{z}{2\nu} ( -I_{-1+\nu}(z) I_{-1-\nu}(z) - I_{1+\nu}(z) I_{-1-\nu}(z) ) , $$ which rearranges itself into your formula.
