I looking to evaluate $$\int_{0}^1\ln(\ln(\frac{1}{x})) dx$$

I know the answer is the Euler-Mascheroni constant, $\gamma$ but how do I get that result?

I've tried differentiating under the integral, but that didn't seem to work. Maybe series could work since $\gamma$ is connected to a lot of series and is defined by using harmonic series.

  • 2
    $\begingroup$ Sub $x=e^{-u}$ and use the fact that $$\int_0^{\infty} du \, e^{-u} \log{u} = -\gamma$$ $\endgroup$ – Ron Gordon Feb 5 '18 at 15:56
  • $\begingroup$ this integral can not expressed by the known elementary functions $\endgroup$ – Dr. Sonnhard Graubner Feb 5 '18 at 15:56

Enforcing $x= e^{-u}$ $$\int_{0}^1\ln\left(\ln\left(\frac{1}{x}\right)\right) dx =\int_{0}^\infty\ln(u)e^{-u} du =-\gamma $$

Where $\gamma$ is the Euler Mascheroni constant

  • 2
    $\begingroup$ The answer is a bit circular (how one knows that the last integral is the Euler constant). In principle, the OP should tell what the definition of $\gamma$ is. Conventionally, it is defined as the difference of the harmonic series and the logarithm function for large $n$. But then there is still some work to be done. $\endgroup$ – Fabian Feb 5 '18 at 16:11
  • $\begingroup$ This is actually helpful thanks. The second integral is just the digamma function evaluated at 1. I for some reason didn't even try substitution whoops! $\endgroup$ – Tom Himler Feb 5 '18 at 17:33

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