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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.

Tulip
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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. – Tulip Feb 11 '13 at 20:41
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    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 01:38
  • @GerryMyerson : Hi Gerry. I am reading this book, it says "the study of $\Gamma (a,x)$ is postponed until Volume 2" on page 9. Do we have Volume 2 already? – Ryan Feb 13 '13 at 09:28
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    @Ryan, not that I know of, but surely Google would know. – Gerry Myerson Feb 13 '13 at 11:58

4 Answers4

18

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 $$ \gamma = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k} - \log{n}. $$

Chappers
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  • (+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 – Mark Viola Jan 03 '17 at 18:04
  • Well, this is a pretty old question! I thought this method should be here for completeness, at least. Happy New Year to you too. – Chappers Jan 03 '17 at 18:11
  • 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. – Mark Viola Jan 03 '17 at 18:58
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    Hi Chappers, could you explain the very first line, where you use the DCT to take a single limit for the limit of integration and integrand? can we take the limit of integration to be any function of n which diverges, like n^2 or 2n, or does DCT force us to use n?

    To clarify, I'm confused why we are allowed to pick the upper limit of integration to be exactly n, rather than n^2 or 2n: numerical integration easily shows that the integral from 0 to 2n of your integrand is not the same as the integral from 0 to n.

     – Ron Shvartsman Oct 18 '22 at 18:08
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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 $$ \int_0^1\log(\log(1/t))\,\mathrm{d}t=-\gamma\tag{5} $$

Community
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robjohn
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7

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

Noting that

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

we find that

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

Pedro
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Argon
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2

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 $$\gamma=\lim\limits_{n\to\infty}\left( H_n-\log n\right)$$

Community
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Pedro
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