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

This is from wiki:

$$\gamma = - \int_0^\infty e^{-x} \ln x d x .$$

It is interesting as initially $\gamma$ is defined as a summation. But how to prove it? in an elementary way?

• Please, be more precise what you want, and what you tried. "Elementary" is relative: most people don't think improper integrals are elementary, you do, obviously. Where's your borderline between "elementary" and "non-elementary"? – Professor Vector Sep 14 '17 at 7:56
• there is an answer in this post math.stackexchange.com/questions/300531/… – poisson Sep 14 '17 at 9:42

\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 $\gamma$

Hint:

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

• But this is just a detour. Could you do it without appeal to the Gamma function? in an elementary way? – poisson Sep 14 '17 at 7:18
• @poisson: you didn't mention this requirement in your post. – Yves Daoust Sep 14 '17 at 7:22

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 $s=1$.

• But this is just a detour. Could you do it without appeal to the Gamma function? in an elementary way? – poisson Sep 14 '17 at 7:18

We start by the fact

$$\gamma = \lim_{n\to \infty} H_n -\log(n)$$

Then one can use it to prove

$$\tag {1}\Gamma(z) = \int^\infty_0 x^{z-1} e^{-x} \,dx= \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1+\frac{z}{n} \right)^{-1}e^{z/n}$$

Then we can deduce by taking log and differentiation

$$\int^\infty_0 e^{-x} \log(x) dx = -\gamma$$