What is the closed form of this :$\int_{0}^{\infty}\frac{\log(x)\log^2(1+x)}{x^s}\,dx$ with $s$ is a complex variable?

Question

This question is related to this question , I want to know if it is possible to get a closed form of the below integral with $s$ is a complex variable $$\int_{0}^{\infty}\frac{\log(x)\log^2(1+x)}{x^s}\,dx$$ ,

Answer 1

Consider the integral

$$I(a,b)=\int_0^{\infty} \frac{x^a (1+x)^b}{x^s}\,dx$$ We are then looking for $$\frac{\partial^3 I}{\partial a \,\partial b^2 }\Big|_{(a,b)=(0,0)}$$ We can evaluate the first integral in terms of the beta function:

\begin{align} I(a,b)&=\int_0^{\infty} \frac{x^{a-s}}{(1+x)^{-b}}\,dx \\ \\ &=\mathcal{B}\left(a-s+1, -(b+a+1)+s\right) \\ \\ &=\frac{\Gamma(a-s+1)\Gamma(s-(b+a+1))}{\Gamma(-b)} \end{align} Letting $m=a-s+1$ and $n=s-(b+a+1)$, we find that

\begin{align} \\ \frac{\partial I}{\partial a}&=\frac{\Gamma(m)\Gamma(n)}{\Gamma(-b)}\left[\psi^{(0)}(m)-\psi^{(0)}\left(n\right)\right] \\ \\ \frac{\partial^2}{\partial a \partial b}&= \Gamma(n)\Gamma(m)\left[\frac{\psi^{(0)}(n)+\psi^{(0)}(-b)}{\Gamma(-b)} \left[\psi^{(0)}(n)-\psi^{(0)} (m) \right]+\frac{\psi^{(1)}(n)}{\Gamma(-b)} \right] \\ \\ \frac{\partial ^3}{\partial a \,\partial b^2}&=\,\,... \end{align}