I have found this formula and I am trying to prove it , but I have not any idea how to deal with it:
$$e^{ax}-e^{bx} = x(a-b)\exp\left[\frac{a+b}{2}x\right]\prod_{i=1}^{\infty}\left[1+\frac{(a-b)^2x^2}{2k^2\pi^{2}}\right] $$
It's too complicated for me, any suggestions ?
On
We have the following infinite product representation for $\sinh\,z$:
$$\frac{\sinh\,z}{z}=\prod_{k=1}^\infty\left(1+\frac{z^2}{k^2\pi^2}\right)$$
Comparing this with
$$e^{ax}-e^{bx} = x(a-b)\exp\left[\frac{a+b}{2}x\right]\prod_{i=1}^{\infty}\left[1+\frac{(a-b)^2x^2}{2k^2\pi^{2}}\right]$$
we find that
$$\prod_{i=1}^{\infty}\left[1+\frac{(a-b)^2x^2}{2k^2\pi^{2}}\right]=\frac{\sinh\frac{x(a-b)}{\sqrt 2}}{\frac{x(a-b)}{\sqrt 2}}$$
Use that $\sinh\,z=\frac{\exp\,z-\exp(-z)}{2}$ to prove/disprove your equation.
This is the famous sine product formula in disguise. See for example this writeup.