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I recall having done this integral a long time ago, but I can't remember how I actually did it. Does anyone have any ideas? Anything (for example contour integration) goes, but I'd prefer an elegant change-of-variables method. The integral in question is

$I(v,w) = \int_{\mathbb{R}^+} \frac{ds}{s} \left[e^{-vs} - e^{-ws} \right] = \int_{\mathbb{R}^+} \frac{ds}{s} \left[e^{-s} - e^{-(w/v)s} \right].$

for $v,w \geq 0.$

Wolfram Alpha doesn't immediately recognize it, but I believe that the answer is proportional to $\ln(w/v).$ Thanks in advance!

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    @Peter Bala: I'd never seen that result, interesting. Ross and Qiauchu: thanks for your comments.2011-08-08

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You've already shown that this only depends on $\frac{w}{v}$, so let's consider $ I(x)=\int_0^\infty\left(e^{-s}-e^{-sx}\right)\frac{ds}{s} $ Take the derivative with respect to $x$ $ \begin{align} \frac{d}{dx}I(x)&=\int_0^\infty e^{-sx}ds\\ &=\frac{1}{x} \end{align} $ Since $I(1)=0$, we get by integrating $\frac{1}{x}$ that $ I(x) = \log(x) $

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    Thanks for the (neat) answer.2011-08-08