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

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.

• welcome to MSE, why do you believe it has a closed form? Commented Mar 12 at 20:54
• Well, it does not need to have closed form, but it converges. How can I find approximation of it? Commented Mar 12 at 21:37

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.)

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 $$R(a) \in O\left(\frac{1}{a^2}\right)$$.

• is it possible to have similar result for generalisation $\int_0^\infty (\ln(1-e^{-x}))^ce^{-ax}x^bdx, \quad a,b,c>0$? Commented Jun 14 at 23:18
• @minimax It should be possible to proceed similarly, at least if you're happy with $b$, $c$ being integers. I'd start with setting $I(a, m) := \int_0^\infty e^{-a x} (1 - e^{-x})^m \,dx = \int_0^1 u^{a - 1} (1 - u)^m \,dx = \operatorname{B}(a, m + 1)$ and differentiating. Feel free to ask that as a separate question. If you do, you can drop a link here, and I'll take a closer look. Commented Jun 15 at 21:30
• @TravisWillse See Here: math.stackexchange.com/questions/4932832/… Commented Jun 15 at 21:47
• For $b=1$, I get this $\frac{1}{a} \left[(- \gamma - \psi(1 + a))^2 +\psi^{(1)}(1)- \psi^{(1)}(1 + a)\right]$. I'm not sure if it is equivalent to your result. Commented Jun 16 at 14:53

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;

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

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