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How can we prove this identity? (Taken from Wikipedia's List of Definite Integrals)

$$ \int_0^\infty \frac { e^{-ax} - e^{-bx} } x dx = \ln \frac b a $$

It seems that the integral is not elementary and

$$ \int_0^\infty \frac { e^{-ax} - 1 } x dx $$

does not converge.

Blue
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Adrian Savage
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4 Answers4

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This one has a really cute trick, namely that the integrand is equal to

$$\int_a^b e^{-xt} \, dt$$

as you can verify by a direct computation. This means we can write a double integral and change the order of integration, yielding

\begin{align*} \int_a^b \int_0^{\infty} e^{-xt} \, dx \, dt &= \int_a^b -\frac{e^{-xt}}{t}\big|_{x = 0}^{x = \infty} \, dt \\ &= \int_a^b \frac 1 t \, dt \\ &= \ln \frac b a \end{align*}

By the way, the integral you mention does converge. It's bounded by $e^{-ax}$ as $x \to \infty$, and bounded at the origin because $e^{-ax} - 1 = -ax + O(x^2)$.

Dave
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  • When I substitute $$ u = ax $$ in $$ \int_0^\infty \frac { e^{-ax} - 1 } x dx $$ I just get the original formula with a=1, as if it does not depend on $ a $. So it would be zero. But the integrand is positive, so this would not make sense. This is why I thought it did not converge. I'm not sure where my mistake is. – Adrian Savage Aug 14 '17 at 00:05
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    @AdrianSavage: The integral in your comment diverges at $\infty$. – robjohn Aug 14 '17 at 00:15
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    Just after you introduce the double integral, the final $dx$ should be a $dt$. It wouldn't allow me to make the edit, due to not enough difference from original post. – CDCM Aug 14 '17 at 00:35
  • @CDCM I just fixed that. – Dave Aug 14 '17 at 01:13
  • @CDCM Yes, thanks; also thanks for the fix, Dave. –  Aug 14 '17 at 02:16
5

Another approach to this integral is to write it as the limit $$ \begin{align} \int_0^\infty\frac{e^{-ax}-e^{-bx}}{x}\,\mathrm{d}x &=\lim_{\epsilon\to0^+}\int_\epsilon^\infty\frac{e^{-ax}-e^{-bx}}{x}\,\mathrm{d}x\\ &=\lim_{\epsilon\to0^+}\left(\int_\epsilon^\infty\frac{e^{-ax}}{x}\,\mathrm{d}x-\int_\epsilon^\infty\frac{e^{-bx}}{x}\,\mathrm{d}x\right)\\ &=\lim_{\epsilon\to0^+}\left(\int_a^\infty\frac{e^{-\epsilon x}}{x}\,\mathrm{d}x-\int_b^\infty\frac{e^{-\epsilon x}}{x}\,\mathrm{d}x\right)\\ &=\lim_{\epsilon\to0^+}\int_a^b\frac{e^{-\epsilon x}}{x}\,\mathrm{d}x\\ &=\int_a^b\frac1{x}\,\mathrm{d}x\\[9pt] &=\log(b/a) \end{align} $$

robjohn
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A more tricky solution.

$$I_a=\int \frac{e^{-ax}}x\,dx=\text{Ei}(-a x)\implies J_a=\int_k^\infty \frac{e^{-ax}}x\,dx=\Gamma (0,a k)+\log(k)$$ Using Taylor expansion around $k=0$ $$J_a=-\log (a)-\gamma +a k+O\left(k^2\right)$$ and then $$J_a-J_b=\int_k^\infty \frac{e^{-ax}-e^{-bx}}x\,dx=\log\left(\frac b a \right)+(a-b)k+O\left(k^2\right)$$

Claude Leibovici
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1

Here's a solution that uses differentiation under the integral:

Introduce a parameter $s$ and define

$$ I(s) = \int_0^{\infty} \dfrac{e^{-ax} - e^{-bx}}{x} e^{-sx}dx$$

Differentiate with respect to the parameter to get

$$ I'(s) = \int_0^{\infty} [e^{-(s+b)x} - e^{-(s+a)x}]dx = \dfrac{1}{s+b} - \dfrac{1}{s+a}$$

Integrating both sides gives

$$I(s) = \ln \dfrac{s+b}{s+a}$$

Evaluate at $s=0$ to get the result.

BobaFret
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