The
Euler-Mascheroni constant denoted by the letter \(\gamma\) is defined as \[\gamma=\displaystyle\lim_{n \rightarrow \infty } \biggl( \sum_{k=1}^n \frac{1}{k} - \ln(n) \biggr)= \lim_{n\to \infty} ({\mathcal{H}_{n}-\log(n)})\,.\] So we must prove that \[\gamma=-\displaystyle\int_{0}^{\infty}e^{-t}\log(t)\;dt\,.\]
The following (beautiful) proof was given by
Sangchul Lee here
Proof: It is easy to prove that the sequence of functions
\[f_n(x) = \begin{cases}
\left( 1 - \frac{x}{n}\right)^n & 0 \leqslant x \leqslant n \\
0 & x > n \end{cases}\] satisfies \(0 \leq f_n(x) \uparrow e^{-x}\). Thus by dominated convergence theorem, we have
\[\int_{0}^{\infty} e^{-x} \log x \; dx = \lim_{n\to\infty} \int_{0}^{n} \left( 1 - \frac{x}{n}\right)^n \log x \; dx.\] Now by the substitution \(x=nu\), we have
\begin{align*}
\int_{0}^{n} \Bigl({1-\frac{x}{n}}\Bigr)^n\log x \; dx
&= n\int_{0}^{1} ({1 - u})^n (\log n + \log u) \; du \\
&= \frac{n}{n+1}\log n + n\int_{0}^{1} (1-u)^n \log u \; du \\
&= \frac{n}{n+1}\log n + n\int_{0}^{1} v^n \log (1-v) \; dv \\
&= \frac{n}{n+1}\log n - n\int_{0}^{1} v^n \biggl( \sum_{k=1}^{\infty} \frac{v^k}{k} \biggr) \; dv \\
&= \frac{n}{n+1}\log n - n \sum_{k=1}^{\infty} \frac{1}{k(n+k+1)} \\
&= \frac{n}{n+1}\log n - \frac{n}{n+1} \sum_{k=1}^{\infty} \Bigl( \frac{1}{k} - \frac{1}{n+k+1}\Bigr) \\
&= \frac{n}{n+1} \biggl( \log n - \sum_{k=1}^{n+1} \frac{1}{k} \biggr)\,.
\end{align*} Therefore taking \(n\to+\infty\) yields \(-\gamma\).