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From: Bobbine on 4 Dec 2005 10:50
how to get integral of (1/x)[e^(-ax)-e^(-bx)] from 0 to infinity
From: C6L1V@shaw.ca on 4 Dec 2005 21:33
Bobbine wrote:
> how to get integral of (1/x)[e^(-ax)-e^(-bx)] from 0 to infinity

Let I(a,b) be the thing you want. For fixed b > 0 you can easily
evaluate dI(a,b)/da using a theorem about differentiation inside an
integral. It is easy to evaluate I(b,b), so you can get I(a,b) for all
a.

R.G. Vickson

From: Zdislav V. Kovarik on 5 Dec 2005 10:41


On Sun, 4 Dec 2005, Bobbine wrote:

> how to get integral of (1/x)[e^(-ax)-e^(-bx)] from 0 to infinity
>
(Assuming a>0, b>0)

The general result is in
http://mathworld.wolfram.com/FrullanisIntegral.html

The classical proof is rather delicate, but in your problem there is a
shortcut through double integrals:

observe that (exp(-a*x) - exp(-b*x))/x

is the integral from a to b of exp(-x*t) dt.

Set up the double inegral, interchange the order, ...

Cheers, ZVK(Slavek)
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