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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 en.wikipedia.org/wiki/Exponential_integral 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
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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
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This is an example of Frullani's integral : mathworld.wolfram.com/FrullanisIntegral.html –  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
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1 Answer

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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