Integral involving $\ln$ and $\gamma$

Question

I want to know if its possible to have a closed form of this integral $$\int_0^\infty e^{-x}\ln(kx) dx$$ I know that if k = 1 then the integral is equal to $-\gamma$ but i want to find a generalized form

Answer 1

$$I(k)=\int_0^\infty e^{-x}\ln(kx) dx$$ $$I´(k)=\int_0^\infty e^{-x}\frac x{xk}dx$$ $$=\frac 1k\int_0^\infty e^{-x} dx$$ $$=\frac 1k$$ $$\implies I(k) = \ln(k)+c$$ $$I(1) = 0+c=-\gamma$$ $$\implies c=-\gamma$$ $$I(k) = \ln(k)-\gamma$$

Answer 2

$$\begin{align} \int_0^\infty e^{-x}\ln{(kx)}\mathrm{d}x &=\int_0^\infty e^{-x}(\ln{(k)}+\ln{(x)})\mathrm{d}x\\ &=\ln{(k)}\int_0^\infty e^{-x}\mathrm{d}x+\int_0^\infty e^{-x}\ln{(x)}\mathrm{d}x\\ &=\ln{(k)}-\gamma\\ \end{align}$$