Evaluate the improper integral $\int_0^\infty \ln(1-e^{-x})e^{-ax}x^bdx$

Question

Evaluate the integral $$\int_0^\infty \ln(1-e^{-x})e^{-ax}x^bdx, \quad a,b>0.$$

I'm trying to use substitution $t=1-e^{-x}$ and integration by parts. Also I've tried with Gamma function in parts. But I don't get anywhere. Any help will be appreciated.

Answer 1

Using the power series expansion of the natural logarithm yields;

$$\int_{0}^{\infty}\ln(1-e^{-x})e^{-ax}x^bdx=-\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{\infty}x^be^{-(n+a)x}dx$$

Applying the substitution $u=(n+a)x$ gives us;

$$-\sum_{n=1}^{\infty}\frac{1}{n\cdot (n+a)^{b+1}}\int_{0}^{\infty}u^be^{-u}du$$

$$=-\sum_{n=1}^{\infty}\frac{\Gamma(b+1)}{n\cdot (n+a)^{b+1}}$$

Answer 2

Here's a closed-form solution for nonnegative integers $b$, in terms of the gamma and polygamma functions, that uses differentiating under the integral sign twice.

First consider the case $b = 0$. Substituting $u := e^{-x}, du = -e^{-x} \,dx$, transforms the integral to $$\int_0^\infty e^{-a x} \log(1 - e^{-x}) \,dx = \int_0^1 u^{a - 1} \log(1 - u) \,du .$$ To evaluate this integral, we differentiate under the integral sign. Define $$I(t) := \int_0^1 u^{a - 1} (1 - u)^t \,du ,$$ so that $$I'(t) := \int_0^1 u^{a - 1} (1 - u)^t \log(1 - u) \,du,$$ and in particular so that the desired integral is $I'(0)$. The usual definition of the beta function and the formula for expressing the beta function, $\mathrm{B}$, in terms of the gamma function, $\Gamma$, gives $$I(t) = \mathrm{B}(a, t + 1) = \frac{\Gamma(a) \Gamma(t + 1)}{\Gamma(a + t + 1)} .$$ Differentiating with respect to $t$ gives $$(\psi(t + 1) - \psi(a + t + 1)) \frac{\Gamma(a) \Gamma(t + 1)}{\Gamma(a + t + 1)} ,$$ where $\psi$ is the digamma function, and evaluating at $t = 0$ yields $$\phantom{(ast)} \qquad \int_0^\infty e^{-a x} \log(1 - e^{-x}) \,dx = I'(0) = - \frac{\gamma + \psi(1 + a)}{a} , \qquad (\ast)$$ where $\gamma := -\psi(1)$ is the Euler-Mascheroni constant. For positive integers $a$ this integral is just $$-\frac{H_a}{a} ,$$ where $H_a := 1 + \frac12 + \frac13 + \cdots \frac1a$ is the $a$th harmonic number.)

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With $(\ast)$ in hand we now generalize to the case of a nonnegative integer $b$: Differentiating $(\ast)$ $b$ times with respect to $a$ yields $$\boxed{\int_0^\infty x^b e^{-a x} \log(1 - e^{-x}) \,dx = \frac{b!}{a^{b + 1}} \left(-\gamma - \sum_{k = 0}^b \frac{(-1)^k}{k!} a^k \psi^{(k)}(1 + a)\right)} ,$$ where $\psi^{(m)}$ is the polygamma function of order $m$.

For example, for $b = 1$, $$\int_0^\infty x e^{-a x} \log(1 - e^{-x}) \,dx = -\frac{\gamma + \psi(1 + a)}{a^2} + \frac{\psi^{(1)}(1 + a)}{a},$$ which for positive integers $a$ simplifies to $$ \int_0^\infty x e^{-a x} \log(1 - e^{-x}) \,dx = -\frac{H_a}{a^2} + \frac1a\left(\frac{\pi^2}{6} - \sum_{k = 1}^a \frac{1}{k^2}\right) .$$ I didn't compute them, but analogous formulae for positive integers $a$ should also exist for integers $b > 1$.

Computing using the asymptotics of the digamma function gives that for fixed $b$, as $a \to \infty$, $$\int_0^\infty x^b e^{-a x} \log(1 - e^{-x}) \,dx = b!\left(-\log a + (H_b - \gamma) - \frac{b + 1}{2 a} + R(a)\right) ,$$ where the remainder satisfies $R(a) \in O\left(\frac{1}{a^2}\right)$.