Evaluating the integral, $\int_{0}^{\infty} \ln\left(1 - e^{-x}\right) \,\mathrm dx $ I recently got stuck on evaluating the following integral. I do not know an effective substitution to use. Could you please help me evaluate:
$$\int_{0}^{\infty} \ln\left(1 - e^{-x}\right) \,\mathrm dx $$
 A: 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
$$
\int_{0}^{1} | \ln (1 - e^{2\pi i x})|^{2} dx  = \zeta(2).
$$
A: 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.)
A: 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
$$I'(\alpha )=-\int_{0}^{\infty} \frac{e^{-x}}{1-\alpha  e^{-x}} \,\mathrm dx=
\left[\frac{x}{\alpha }-\frac{\ln \left(e^x-\alpha \right)}{\alpha }\right]_0^{\infty}=\frac{\ln(1-\alpha )}{\alpha }$$
$$I(\alpha )=-\operatorname{Li}_2(\alpha )+c$$
$$I(0)=0\implies I(\alpha )=-\operatorname{Li}_2(\alpha )$$
$$I(1)=-\operatorname{Li}_2(1)=-\zeta(2)=-\frac{\pi^2}{6}$$

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

A: Maybe this answer is too late. Let $t=1-e^{-x}$. Then
$$ \int_0^\infty\ln(1-e^{-x})dx=\int_0^1\frac{\ln t}{1-t}dt=-\frac{\pi^2}{6}. $$
A: break $\ln(1-e^{-x})$ like $\ln(1-t) = y-y^2/2-y^3/3- \cdots$ where $y = e^{-x}$
then break at the addition/subtraction point in integration.
then evaluate.
A: Integrating by parts it is easily reducible to integral representation of $\Gamma(s) \zeta(s)$ at $s=2$:
$$\begin{eqnarray*}
   \int_0^\infty \log\left(1-\mathrm{e}^{-x}\right) \, \mathrm{d} x &=& \lim  x \log\left(1-\mathrm{e}^{-x}\right) \vert_{0^+}^\infty  - \int_0^\infty x \frac{\mathrm{e}^{-x}}{1 - \mathrm{e}^{-x}} \mathrm{d} x \\ &=& - \int_0^\infty x \frac{\mathrm{e}^{-x}}{1 - \mathrm{e}^{-x}} \mathrm{d} x = -\Gamma(2) \zeta(2).
\end{eqnarray*}
$$
A: 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.
