how we can integrate $\int_0^\infty x^{n-1}\log(x)\exp(-ax)\exp(-bx)\,dx$ and $\int_0^{\infty} x^{n-1} \log(1+x)\exp(-ax)\exp(-bx)\, dx$

Question

I have tried to solve the given integrals, but could not find any solution in closed form $$\int_0^\infty x^{n-1}\log(x)\exp(-ax)\exp(-bx)\,dx$$ and $$\int_0^{\infty} x^{n-1} \log(1+x)\exp(-ax)\exp(-bx)\, dx$$

Answer 1

Consider that for any $n>-1$ and $k>0$: $$ \int_{0}^{+\infty}x^n e^{-x}\,dx = \Gamma(n+1),\qquad \int_{0}^{+\infty}x^n e^{-kx}\,dx = \frac{\Gamma(n+1)}{k^{n+1}}\tag{1}$$ and differentiate with respect to $n$. That leads to: $$\int_{0}^{+\infty}x^n\log(x)e^{-kx} = \frac{d}{dn}\left(\frac{\Gamma(n+1)}{k^{n+1}}\right) = \color{red}{\frac{n!}{k^{n+1}}\left(H_n-\gamma-\log k\right)}.\tag{2} $$ The other integral (depending on $\log(x+1)$) can be dealt with in a similar way, but has a less elementary form.