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I recently got stuck on evaluating the following integral. I do not know an effective substitution to use. Could you please help me evaluate:

$$\int_{0}^{\infty} \ln\left(1 - e^{-x}\right) \,\mathrm dx $$

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3  
Use braces for longer exponents, so e^{-x}, not e^-x –  t.b. Aug 29 '11 at 7:54
    
I've edited the LaTeX so that $e^{-x}$ appears correctly (and also merged your accounts, Jack). –  Zev Chonoles Aug 29 '11 at 7:55
    
The integrand is undefined, since $e^{-x}-1$ is negative. Do you mean $\ln|e^{-x}-1|$? –  Hans Lundmark Aug 29 '11 at 7:55
    
Yes, I dropped a negative sign from in front of the integrand and have edited the question. –  Jack Rousseau Aug 29 '11 at 9:12

6 Answers 6

up vote 42 down vote accepted

One route to evaluating the integral is $$-\int_0^\infty \ln(1-e^{-x})dx=\int_0^\infty\left(e^{-x}+\frac{e^{-2x}}{2}+\frac{e^{-3x}}{3}+\cdots\right)dx $$ $$=\int_0^\infty e^{-x}dx+\frac{1}{2}\int_0^\infty e^{-2x}dx+\frac{1}{3}\int_0^\infty e^{-3x}dx+\cdots$$ $$=1+\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{4}\cdot\frac{1}{4}\cdots $$ $$=\zeta(2)=\frac{\pi^2}{6}.$$

I don't know if a straightforward substitution could get you the answer, what with this being the Riemann zeta function and all, but you can see the integrals on the linked page and try your own hand at finding one. (Or someone else can try their hand.)

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Thank you, this is what I was looking for! Another post observed that zeta(2) is the area under the curve e^-x + e^-y = 1 and now I guess I have more to think about. haha –  Jack Rousseau Aug 29 '11 at 9:22
1  
@Jack: Was it math.stackexchange.com/questions/8337/…, by any chance? –  Hans Lundmark Aug 29 '11 at 11:29
    
Yes it was. - I see now, that my question was answered in that thread. @Hans: I find your comment there very intriguing. –  Jack Rousseau Aug 29 '11 at 20:44

Consider following parametric integral$$I(\alpha )=\int_{0}^{\infty} \ln\left(1 - \alpha e^{-x}\right) \,\mathrm dx$$

We have $I(0)=0$ and $I(1)$ will yield required integral

$$I'(\alpha )=-\int_{0}^{\infty} \frac{e^{-x}}{1-\alpha e^{-x}} \,\mathrm dx= \left[\frac{x}{\alpha }-\frac{\ln \left(e^x-\alpha \right)}{\alpha }\right]_0^{\infty}=\frac{\ln(1-\alpha )}{\alpha }$$

$$I(\alpha )=-\operatorname{Li}_2(\alpha )+c$$

$$I(0)=0\implies I(\alpha )=-\operatorname{Li}_2(\alpha )$$

$$I(1)=-\operatorname{Li}_2(1)=-\zeta(2)=-\frac{\pi^2}{6}$$

$$\large\int_{0}^{\infty} \ln\left(1 - e^{-x}\right) \,\mathrm dx=-\frac{\pi^2}{6}$$

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Maybe this answer is too late. Let $t=1-e^{-x}$. Then $$ \int_0^\infty\ln(1-e^{-x})dx=\int_0^1\frac{\ln t}{1-t}dt=-\frac{\pi^2}{6}. $$

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+1 For nice answer! –  Integrator Dec 10 at 10:55
1  
@Integrator, your answer is nice too. Actually, both our answers are the same. –  xpaul Dec 10 at 13:51

A related family of integrals is the following. Consider the polylogarithm function, $\mathsf{Li}_{s}(x) = \sum_{k \geq 1} \frac{x^{k}}{k^{s}}$. By Parseval's Theorem, $$ \int_{0}^{1} | \mathsf{Li}_{s}(e^{2 \pi i x})|^{2} dx = \sum_{k \geq 1} \left| \frac{e^{2 \pi i k x}}{k^{s}} \right|^{2} = \sum_{k \geq 1} \frac{1}{k^{2s}} = \zeta(2s). $$ Since $\mathsf{Li}_{1}(x) = - \ln (1 - x)$, we have $$ \int_{0}^{1} | \ln (1 - e^{2\pi i x})|^{2} dx = \zeta(2). $$

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Integrating by parts it is easily reducible to integral representation of $\Gamma(s) \zeta(s)$ at $s=2$:

$$\begin{eqnarray*} \int_0^\infty \log\left(1-\mathrm{e}^{-x}\right) \, \mathrm{d} x &=& \lim x \log\left(1-\mathrm{e}^{-x}\right) \vert_{0^+}^\infty - \int_0^\infty x \frac{\mathrm{e}^{-x}}{1 - \mathrm{e}^{-x}} \mathrm{d} x \\ &=& - \int_0^\infty x \frac{\mathrm{e}^{-x}}{1 - \mathrm{e}^{-x}} \mathrm{d} x = -\Gamma(2) \zeta(2). \end{eqnarray*} $$

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Not as snazzy as anon's answer, but:

If you start with the usual series for the logarithm:

$$-\frac{\log(1-x)}{x}=\sum_{k=0}^\infty \frac{x^k}{k+1}$$

the dilogarithm function $\mathrm{Li}_2(x)$ is

$$\mathrm{Li}_2(x)=-\int_0^x\frac{\log(1-t)}{t}\mathrm dt=\sum_{k=1}^\infty \frac{x^k}{k^2}$$

From the series representation, we find that $\mathrm{Li}_2(1)=\dfrac{\pi^2}{6}$; it remains to manipulate the integral

$$-\int_0^1\frac{\log(1-t)}{t}\mathrm dt=\int_1^0\frac{\log(1-t)}{t}\mathrm dt$$

Letting $t=\exp(-u),\quad \mathrm dt=-\exp(-u)\mathrm du$, we get

$$-\int_0^\infty\frac{\log(1-\exp(-u))}{\exp(-u)}\exp(-u)\mathrm du$$

which is the negative of the OP's integral.

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