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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 ?

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2 Answers 2

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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it's clear , thanks for your help . –  Prigmin Smith Sep 3 '11 at 12:06
    
Prigmin, there's a check mark next to Yuval's answer or mine that you can click if you're happy with any of them. –  J. M. Sep 3 '11 at 12:21
    
Yes,I've already do that . Thanks . –  Prigmin Smith Sep 4 '11 at 13:22
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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 step is the USE this formula to prove the sine product formula... –  GEdgar Sep 3 '11 at 14:24
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