Evaluate the following improper integral $\int_0^\infty (\ln(1-e^{-x}))^ce^{-ax}x^bdx, \quad a>0, b,c \in \mathbb{N}_0$

Question

Evaluate the improper integral $$ \int_{0}^{\infty}\ln^{\large c}\left(1 - {\rm e}^{-x}\right) {\rm e}^{-ax}x^{b}\,{\rm d}x, \quad a > 0,\ b,c \in \mathbb{N}_{0} $$

• Is it possible to find closed form of this integral ?.
• It's a generalisation of the following problem: $$ \int_{0}^{\infty}\ln\left(1 - {\rm e}^{-x}\right) {\rm e}^{-ax}x^{b}\,{\rm d}x $$
• I'm not that familiar with Polygamma Functions, so this is not obvious for me.

Answer 1

As I wrote in my comment in the linked question, suppose that $b, c$ are nonnegative integers. Set $$J(a, m) := \int_0^\infty e^{-a x} (1 - e^{-x})^m \,dx.$$ Substituting $u := e^{-x}$ gives $$J(a, m) = \int_0^1 u^{a - 1} (1 - u)^m \,du = \frac{\Gamma(a) \Gamma(1 + m)}{\Gamma(a + m + 1)} .$$ Differentiating gives $$\int_0^\infty e^{-ax} x^b (1 - e^{-x})^m \log^c(1 - e^{-x}) \,dx = (-1)^b \frac{\partial^b}{\partial a^b} \frac{\partial^c}{\partial m^c} \left(\frac{\Gamma(a) \Gamma(1 + m)}{\Gamma(a + m + 1)}\right) .$$ Evaluating at $m = 0$ gives that the integral in the question statement is \begin{multline*} I(a, b, c) = \int_0^\infty e^{-ax} x^b \log^c(1 - e^{-x}) \,dx \\ = (-1)^b \left.\frac{\partial^b}{\partial a^b} \frac{\partial^c}{\partial m^c}\right\vert_{m = 0} \left(\frac{\Gamma(a) \Gamma(1 + m)}{\Gamma(a + m + 1)}\right) . \end{multline*} For any particular $c$ we can evaluate the derivative explicitly in terms of the polygamma functions, $\psi^{(n)}(x) = \frac{d^n}{dx^n} \log \Gamma(x)$. As usual $\psi := \psi^{(0)}$ denotes the digamma function.

For $c = 0$, $$I(a, b, 0) = \frac{b!}{a^{1 + b}} .$$

For $c = 1$, we recover the formula from the linked question, \begin{multline} I(a, b, 1) = \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). \end{multline}

For $c = 2$, \begin{multline} I(a, b, 2) = \int_0^\infty x^b e^{-a x} \log^2(1 - e^{-x}) \,dx \\ = \Gamma(1 + b) \sum_{k = 0}^b \frac{2 \gamma \psi^{(k)}(1 + a) + \frac{d^k}{da^k}(\psi^2(1 + a)) - \psi^{(1 + k)}(1 + a)}{(-a)^{1 + b - k} \Gamma(1 + k)} \end{multline} The term $\frac{d^k}{da^k}(\psi^2(1 + a))$ can be expanded as $$ \frac{d^k}{da^k}(\psi^2(1 + a)) = \sum_{l = 0}^k {k \choose l} \psi^{(l)}(1 + a) \psi^{(k - l)}(1 + a) . $$

Answer 2

We can write:

$$\int_0^{\infty } \ln ^c(1-\exp (-x)) \exp (-a x) x^b \, dx=\\\underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{\partial ^c}{\partial y^c}\frac{\partial ^b}{\partial x^b}\frac{(-1)^b \Gamma (a+x) \Gamma (1+y)}{\Gamma (1+a+x+y)}=\\\underset{m\to 0}{\text{lim}}\left(\sum _{\text{$\alpha $1}=0}^b \sum _{\text{$\alpha $2}=0}^c \frac{(-1)^b (-1)^{\text{$\alpha $1}+\text{$\alpha $2}} m^{-b-c} \binom{b}{\text{$\alpha $1}} \binom{c}{\text{$\alpha $2}} \Gamma (a-m \text{$\alpha $1}) \Gamma (1-m \text{$\alpha $2})}{\Gamma (1+a-m (\text{$\alpha $1}+\text{$\alpha $2}))}\right)$$

For: $ \quad a > 0,\ b,c \in \mathbb{N}_{0} $