Laplace transform I

What is the Laplace transform of $e^{-x}\ln(1-e^{-x})$

We repost here the solution of this problem posted by @SrinivasR1729 on Twitter.
\[ \int_{0}^{\infty} e^{-ax}e^{-x}\ln(1-e^{-x}) dx = -\frac{H_{a+1}}{a+1}\]

Proof \begin{align*} \int_{0}^{\infty} e^{-ax}e^{-x}\ln(1-e^{-x}) dx = & \int_{0}^{1} (1-w)^a \ln(w) dw \quad (w \mapsto 1-e^{-x})\\ =& \int_{0}^{1} (1-w)^a \left(\frac{d}{dt}\Big|_{t=0+} w^t\right) dw\\ =& \frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} w^t(1-w)^a dw\\ =& \frac{d}{dt}\Big|_{t=0+} B\left(t+1,a+1\right)\\ \end{align*} Recall that the derivative of the complete beta function: \[ \frac{d}{da} B(a,b) = B(a,b)\left(\psi^{(0)}(a) - \psi^{(0)}(a+b)\right)\] Therefore: \begin{align*} \int_{0}^{\infty} e^{-ax}e^{-x}\ln(1-e^{-x}) dx = &\frac{d}{dt}\Big|_{t=0+} B\left(t+1,a+1\right)\\ =&\lim_{t \to 0+} \left(\psi^{0}(t+1)-\psi^{0}(a+t+2)\right)B(t+1,a+1) \\ =& \left(\psi^{0}(1)-\psi^{0}(a+2)\right)B(1,a+1)\\ =& \left(-\gamma-\psi^{0}(a+2)\right)B(1,a+1)\\ \end{align*} where we used the fact that $\psi^{(0)}(1) = -\gamma$, the Euler constant.

Recall that a hamonic number for real and complex values can be expressed with the following relation: \[H_{x} = \psi^{(0)}(x+1) + \gamma\] Also note that \[ B(1,a+1) = \frac{\Gamma(1)\Gamma(a+1)}{\Gamma(a+2)} = \frac{\Gamma(a+1)}{\Gamma(a+1)(a+1)} = \frac{1}{a+1}\] Hence \begin{align*} \int_{0}^{\infty} e^{-ax}e^{-x}\ln(1-e^{-x}) dx = & \left(\gamma-\psi^{0}(a+2)\right)B(1,a+1)\\ =& -\frac{H_{a+1}}{a+1} \end{align*} \[\boxed{ \int_{0}^{\infty} e^{-ax}e^{-x}\ln(1-e^{-x}) dx = -\frac{H_{a+1}}{a+1}}\]