# Integral representation of Euler's constant

Prove that : $$\gamma=-\int_0^{1}\ln \ln \left ( \frac{1}{x} \right) \ \mathrm{d}x.$$

where $\gamma$ is Euler's constant ($\gamma \approx 0.57721$).

This integral was mentioned in Wikipedia as in Mathworld , but the solutions I've got uses corollaries from this theorem. Can you give me a simple solution (not using much advanced theorems) or at least some hints.

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How do you define $\gamma$? Your sentence "but the solutions I've got uses corollaries from this theorem" is not so clear. The standard definition is $$\gamma:= \lim_{N\rightarrow \infty}\sum_{k=1}^N \frac{1}{k} -\log N.$$ – Eric Naslund Feb 11 '13 at 20:23
I define $\gamma$ by the sum you mentioned, what I meant by that sentence is that the solutions applies a theorem that I don't know and it is a bit advanced. – aziiri Feb 11 '13 at 20:41
Here is the only way I can think of at the moment. By a substitution, the integral equals $$-\int_0^\infty e^{-x}\log x,$$ and can consider an exponential generating series to see that this is the negative derivative of the Gamma function at $z=1.$ Our goal is now to prove that $\Gamma'(1)=-\gamma$. One way to deduce this is from the functional equation for the zeta function using the Laurent expansion for zeta. This requires a proof that $\gamma$ is in fact the constant term in the Laurent expansion for $\zeta(s)$ around $s=1$, and that is not too hard. – Eric Naslund Feb 11 '13 at 20:48
It's done without using anything very advanced on pages 176-7 of Boros and Moll, Irresistible Integrals. It's a little longer than I'd want to write out. – Gerry Myerson Feb 12 '13 at 1:38
@Ryan, not that I know of, but surely Google would know. – Gerry Myerson Feb 13 '13 at 11:58

In this answer, it is shown that since $\Gamma$ is log-convex, $$\frac{\Gamma'(x)}{\Gamma(x)}=-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x-1}\right)\tag{1}$$ Setting $x=1$ yields $$\Gamma'(1)=-\gamma\tag{2}$$ The integral definition of $\Gamma$ says \begin{align} \Gamma(x)&=\int_0^\infty t^{x-1}\,e^{-t}\,\mathrm{d}t\\ \Gamma'(x)&=\int_0^\infty\log(t)\,t^{x-1}\,e^{-t}\,\mathrm{d}t\\ \Gamma'(1)&=\int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t\tag{3} \end{align} Putting together $(2)$ and $(3)$ gives $$\int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t=-\gamma\tag{4}$$ Substituting $t\mapsto\log(1/t)$ transforms $(4)$ to $$\int_0^1\log(\log(1/t))\,\mathrm{d}t=-\gamma\tag{5}$$

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$$I = \int_0^1 \log (-\log x)\,dx = \int_0^\infty e^{-x} \log(x)\,dx$$

Noting that

$$\Gamma(s) = \int_0^\infty e^{-x} x^{s-1}\, dx$$

we find that

$$\Gamma'(1) = I = -\gamma$$

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I took the liberty to change $n$ to $s$. Hope you don't mind. – Pedro Tamaroff Jun 28 '13 at 1:50
No problem, @Peter – Argon Jun 28 '13 at 1:51
Or directly from Euler's initial formula for the Gamma/factorial function: $$n!=\int_0^1(-\ln x)^ndx\iff\Gamma'(n+1)=\int_0^1(-\ln x)^n\ln(-\ln x)dx\iff\Gamma'(1)=I.$$ – Lucian Dec 13 '13 at 12:18
@Pedro why did you do that? – draks ... Mar 20 '14 at 4:46
@draks... The Leprechaun threatened to burn my house down. – Pedro Tamaroff Mar 20 '14 at 4:58

You can see a proof here where we use that $$\Gamma(z) = \frac{\exp{(-\gamma z)}}{z}\prod\limits_{n=1}^\infty\frac{\exp \left({\frac z n}\right)}{1+\dfrac z n }$$

There is another proof here where we use $$\gamma=\lim\limits_{n\to\infty}\left( H_n-\log n\right)$$

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