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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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If you choose a value for $w/v$, Alpha does give $\log w/v$. I think if you watch convergence you can use the expression in under convergent series to get the result by cancelling all terms except the log before you let the lower limit go to zero. – Ross Millikan Aug 8 '11 at 14:29
From $I(1, a) + I(1, b) = I(1, a) + I(a, ab) = I(1, ab)$ and the fact that $I(1, x)$ is continuous it follows already that $I(1, x) = c \ln x$, but to compute the proportionality constant actually requires computing an integral. – Qiaochu Yuan Aug 8 '11 at 14:44
This is an example of Frullani's integral : – Peter Bala Aug 8 '11 at 15:13
@Peter Bala: I'd never seen that result, interesting. Ross and Qiauchu: thanks for your comments. – Gerben Aug 8 '11 at 15:18
up vote 6 down vote accepted

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. – Gerben Aug 8 '11 at 15:16

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