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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 ?

2 Answers 2

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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.

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    Yes,I've already do that . Thanks .2011-09-04
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This is the famous sine product formula in disguise. See for example this writeup.

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    A good answer. Unless, of course, the next ste$p$ is the USE this formula to prove the sine product formula...2011-09-03