Contour integral of $\int_0^\infty \log(x) e^{-x} dx$

Question

Is it possible to resolve this integral using integral contour? What should be the contour? \begin{equation} \int_0^\infty \log(x) e^{-x} dx = -\gamma \approx -0.577216 \end{equation} where $\gamma$ is the Euler‐Mascheroni constant.

Answer 1

We have: $$ \int_{0}^{+\infty} x^{\alpha}e^{-x}\,dx=\Gamma(\alpha+1) \tag{1}$$ and: $$\begin{eqnarray*}\frac{d}{d\alpha}\int_{0}^{+\infty} x^{\alpha}e^{-x}\,dx &=& \int_{0}^{+\infty}x^{\alpha} \log(x)\, e^{-x}\,dx = \Gamma'(\alpha+1) \\[0.2cm]&=& \Gamma(\alpha+1)\cdot\frac{d}{d\alpha}\log\Gamma(\alpha+1)\\[0.2cm] &=& \Gamma(\alpha+1)\cdot\psi(\alpha+1)\tag{2}\end{eqnarray*}$$ by the definition of $\Gamma$ and $\psi$, so $\int_{0}^{+\infty}\log(x)\,e^{-x}\,dx$ equals $\psi(1)$.

On the other hand, the Weierstrass product: $$ \Gamma(z+1) = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{\frac{z}{n}}\tag{3}$$ gives, through logarithmic differentiation, $$ \psi(z+1) = -\gamma+\sum_{n\geq 1}\frac{z}{n(n+z)}\tag{4}$$ so $\color{red}{\psi(1)=-\gamma}$ follows.