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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.

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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.

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  • $\begingroup$ For the convergence it should be ok because the LHS is $s^{-2}/n+O(\int_0^\infty (x^2/n^2)e^{-\sigma x}dx) = s^{-2} /n + O(\sigma^{-3} / n^2)$ $\endgroup$ – reuns Mar 25 '19 at 0:40
  • $\begingroup$ Is there a way to evaluate the resulting limit with the exponential integrals? $\endgroup$ – aleden Mar 25 '19 at 16:12

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