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Prove \begin{equation} \int_0^1 \frac{x - 1 - x\ln(x)}{x\ln(x) - x\ln^2(x)}dx= \gamma \end{equation}

where $\gamma$ is Euler's constant

I don't know where to begin with this, a hint would be greatly appreciated!

Math_Day
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    The substitution $u=\ln x$ might be helpful because you can factor an $x$ in the denominator. –  Jun 30 '22 at 23:01

1 Answers1

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In similar spirit to this other answer of mine.

Enforce the substitution $x=e^{-u}$ to give

$$I=\int_{0}^{1} \frac{x-1-x \ln(x)}{x \ln(x)-x \ln^2 (x)}\,dx=\int_{0}^{\infty} \frac{1}{u} \left(\frac{1}{u+1}-e^{-u}\right)\,du$$

Consider the Mellin transform $$\int_{0}^{\infty} u^{s-1} \left(\frac{1}{u+1}-e^{-u}\right)\,du=\pi \csc (\pi s)-\Gamma (s)$$ Which follows from the beta function (see here) and definition of the gamma function respectively. Taking the limit as $s\mapsto 0$ gives the desired result of $\gamma$.

KStarGamer
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    ... or, as an option, integrating by part: $$\int_{0}^{\infty} \frac{1}{u} \left(\frac{1}{u+1}-e^{-u}\right)du=\big(\ln u-\ln (1+u)-\ln u ,e^{-u}\big)\Big|_0^\infty-\int_0^\infty\ln u,e^{-u}du$$ – Svyatoslav Jul 01 '22 at 01:40
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    @Svyatoslav Indeed, that is probably a more elementary option, but I personally like the Mellin transform method for the ability to easily generalise. – KStarGamer Jul 01 '22 at 01:57
  • Excellent approach. +1 I had first encountered this on Quora, and man I did a lot of non rigorous stuff in that answer. Still feels bad. -_- – Laxmi Narayan Bhandari Jul 01 '22 at 15:52