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.

  • $\begingroup$ 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 $$\gamma:= \lim_{N\rightarrow \infty}\sum_{k=1}^N \frac{1}{k} -\log N.$$ $\endgroup$ – Eric Naslund Feb 11 '13 at 20:23
  • $\begingroup$ 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. $\endgroup$ – aziiri Feb 11 '13 at 20:41
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    $\begingroup$ 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. $\endgroup$ – Eric Naslund Feb 11 '13 at 20:48
  • $\begingroup$ 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. $\endgroup$ – Gerry Myerson Feb 12 '13 at 1:38
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    $\begingroup$ @Ryan, not that I know of, but surely Google would know. $\endgroup$ – 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 $$ \int_0^1\log(\log(1/t))\,\mathrm{d}t=-\gamma\tag{5} $$


Another way, from another definition: by the Dominated Convergence Theorem, $$ \int_0^{\infty} e^{-u} \log{u} \, du = \lim_{n \to \infty} \int_0^n \left( 1 - \frac{u}{n} \right)^{n-1} \log{u} \, du. $$ Then, changing variables to $v=1-u/n$, $$ \begin{align} \int_0^n \left( 1 - \frac{u}{n} \right)^{n-1} \log{u} \, du &= n \int_0^1 v^{n-1} \log{( n(1-v))} \, dv \\ &= n\log{n} \int_0^1 v^{n-1} \, dv + (n+1) \int_0^1 v^{n-1} \log{(1-v)} \, dv \\ &= \log{n} - n \int_0^1 \sum_{k=1}^{\infty} \frac{v^{k+n-1}}{k} \, dv \\ &= \log{n} - n \sum_{k=1}^{\infty} \int_0^1 \frac{v^{k+n-1}}{k} \, dv \\ &= \log{n} - n \sum_{k=1}^{\infty} \frac{1}{k(k+n)} \\ &= \log{n} - \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{n+k} \right) \\ &= \log{n} - \sum_{k=1}^{n} \frac{1}{k}, \end{align} $$ using uniform convergence and partial fractions. But this is precisely the definition $$ \gamma = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k} - \log{n}. $$

  • $\begingroup$ (+1) This is the approach I would have taken. I'm shocked that it hadn't received an up vote. And Happy New Year! -Mark $\endgroup$ – Mark Viola Jan 3 '17 at 18:04
  • $\begingroup$ Well, this is a pretty old question! I thought this method should be here for completeness, at least. Happy New Year to you too. $\endgroup$ – Chappers Jan 3 '17 at 18:11
  • $\begingroup$ I see you just posted this. I used the same approach you did HERE to develop the limit definition of the Gamma function from the standard integral representation. The overarching purpose was different from the one herein, but the approach you used is quite elegant and straightforward. Anyway, I'm pleased to see you've returned. $\endgroup$ – Mark Viola Jan 3 '17 at 18:58

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

Noting that

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

we find that

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

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