# Derivation of an Integral Expression of the Euler-Mascheroni Constant – Part 2

I derived the expression

\int\limits_{0}^{\infty} \mathrm{e}^{-x} \mathrm{ln}(x) \mathrm{d} x = -\gamma
\label{eq:1608161}
\tag{1}

for the Euler-Mascheroni constant here. However, there is a far easier method that was fully derived in Advanced Integration Techniques by Zaid Alyafeai. I recommend this book to readers of this blog. It is free and contains many useful and interesting results.

We start with

\int\limits_{0}^{\infty} \mathrm{e}^{-x} x^t \mathrm{d} x = \Gamma(t+1)
\label{eq:1608162}
\tag{2}

Differentiate with respect to $$t$$

\int\limits_{0}^{\infty} \mathrm{e}^{-x} x^t \mathrm{ln}(x) \mathrm{d} x = \frac{d\Gamma(t+1)}{dt} = \Gamma(t+1) \psi^{(0)}(t+1)
\label{eq:1608163}
\tag{3}

Taking the limit of equation \eqref{eq:1608163}, $$t \to 0$$ yields

\int\limits_{0}^{\infty} \mathrm{e}^{-x} \mathrm{ln}(x) \mathrm{d} x = \Gamma(1) \psi^{(0)}(1) = -\gamma
\label{eq:1608164}
\tag{4}

## 5 thoughts on “Derivation of an Integral Expression of the Euler-Mascheroni Constant – Part 2”

1. I was happy to do so and I look forward to the new addition that you mentioned on your website.

1. Yeah I added around 50 more pages. It should be out after maybe two weeks.

1. When it is ready, let me know at appliedclassicalanalysis AT gmail DOT com and I will create a blog post with a link.