An integral representation of Euler's constant $\gamma$

Solution 1:

$$ \begin{align} \int_0^\infty\log(x)\,e^{-x}\,\mathrm{d}x &=\lim_{n\to\infty}\int_0^n\log(x)\,\left(1-\frac xn\right)^n\,\mathrm{d}x\tag{1}\\ &=\lim_{n\to\infty}n\int_0^1\log(nx)\,(1-x)^n\,\mathrm{d}x\tag{2}\\ &=\lim_{n\to\infty}\left(\frac{n\log(n)}{n+1}+n\int_0^1\log(x)\,(1-x)^n\,\mathrm{d}x\right)\tag{3}\\ &=\lim_{n\to\infty}\left(\frac{n\log(n)}{n+1}+n\int_0^1\log(1-x)\,x^n\,\mathrm{d}x\right)\tag{4}\\ &=\lim_{n\to\infty}\left(\frac{n\log(n)}{n+1}-n\int_0^1\sum_{k=1}^\infty\frac{x^k}k\,x^n\,\mathrm{d}x\right)\tag{5}\\ &=\lim_{n\to\infty}\left(\frac{n\log(n)}{n+1}-n\sum_{k=1}^\infty\frac1{k(n+k+1)}\right)\tag{6}\\ &=\lim_{n\to\infty}\left(\frac{n\log(n)}{n+1}-\frac{n}{n+1}\sum_{k=1}^\infty\left(\frac1k-\frac1{n+k+1}\right)\right)\tag{7}\\ &=\lim_{n\to\infty}\left(\frac{n\log(n)}{n+1}-\frac{n}{n+1}H_{n+1}\right)\tag{8}\\[9pt] &=-\gamma\tag{9} \end{align} $$ Explanation:
$(1)$: Monotone Convergence
$(2)$: substitute $x\mapsto nx$
$(3)$: $\log(nx)=\log(n)+\log(x)$
$(4)$: substitute $x\mapsto1-x$
$(5)$: use the series for $\log(1-x)$
$(6)$: integrate term by term
$(7)$: Partial Fractions
$(8)$: use formula for Extended Harmonic Numbers
$(9)$: definition of $\gamma$

Solution 2:

Hint:

By differentiating under the integral sign $$\int_0^\infty x^te^{-x}\,dx=\Gamma(t+1)$$ you get $$\int_0^\infty \log x\,e^{-x}\,dx=\Gamma'(1)=-\gamma.$$

Solution 3:

The Laplace transform of $x^p$ (for $p > -1$) is $\Gamma(p+1) s^{-p-1}$. Now $\ln x = \left. \dfrac{d}{dp} x^p \right|_{p=0}$, so (after justifying some interchange of limits) the Laplace transform of $\ln x$ is $$\eqalign{\left.\dfrac{d}{dp} \Gamma(p+1) s^{-p-1} \right|_{p=0} &= \left.\left(\Psi(p+1) \Gamma(p+1) s^{-p-1} -\Gamma(p+1) s^{-p-1} \ln(s)\right)\right|_{p=0}\cr =\frac{-\gamma - \ln(s)}{s} }$$ Now take $s=1$.