# What is $\int_0^\infty \log\left(\Gamma(x)\right)e^{-sx}dx$?

I have been interested in the Laplace Transform of the log-Gamma function$$\int_0^\infty \log\left(\Gamma(x)\right)e^{-sx}dx$$ By expanding the Gamma Function into its Weierstrass Form we get

\begin{align*} &\int_0^\infty \left( -\log(x)-\gamma x+\sum_{n=1}^\infty \frac{x}{n}-\ln(1+\frac{x}{n})\right)e^{-sx}dx \\ &=-\int_0^\infty\log(x)e^{-sx}dx-\gamma\int_0^\infty xe^{-sx}dx+\lim_{k\to\infty}\sum_{n=1}^k \left( \frac{1}{n}\int_0^\infty xe^{-sx}dx-\int_0^\infty \ln\left(1+\frac{x}{n}\right)e^{-sx}dx \right) \\ &=\frac{\log(s)+\gamma}{s}-\frac{\gamma}{s^2}+\lim_{k\to\infty}\sum_{n=1}^k \left( \frac{1}{ns^2}-\int_0^\infty \ln\left(1+\frac{x}{n}\right)e^{-sx}dx \right) \end{align*}

However, I do not know of how to evaluate the last integral and the following limit. If anyone could be able to help me with this, I would greatly appreciate it.

Well, apart from elementary contributions you have already shown that the Laplace transform of $$\log\Gamma$$ only depends on the Laplace transforms of $$\log(1+x/n)$$, and
$$\int_{0}^{+\infty}\log(1+x/n)e^{-sx}\,dx =n\int_{0}^{+\infty}\log(1+x)e^{-snx}\,dx$$ only depends on the Laplace transform of $$\log(1+x)$$, which by integration by parts is given by $$-\frac{e^s}{s}\text{Ei}(-s)$$, with $$\text{Ei}$$ being the exponential integral.
• For the convergence it should be ok because the LHS is $s^{-2}/n+O(\int_0^\infty (x^2/n^2)e^{-\sigma x}dx) = s^{-2} /n + O(\sigma^{-3} / n^2)$ – reuns Mar 25 '19 at 0:40