Joanna Benson

2022-01-04

I would like to show that
${\int }_{0}^{1}\frac{x-1}{\mathrm{ln}\left(x\right)}dx=\mathrm{ln}2$

Mary Nicholson

This is a classic example of differentiating inside the integral sign.
In particular, let
$J\left(\alpha \right)={\int }_{0}^{1}\frac{{x}^{\alpha }-1}{\mathrm{log}\left(x\right)}dx$
Then one has that
$\frac{d}{d\alpha }J\left(\alpha \right)={\int }_{0}^{1}\frac{d}{da}\frac{{x}^{\alpha }-1}{\mathrm{log}\left(x\right)}dx={\int }_{0}^{1}{x}^{\alpha }dx=\frac{1}{\alpha +1}$
and so we know that $J\left(\alpha \right)=\mathrm{log}\left(\alpha +1\right)+C$. Noting that $J\left(0\right)=0$ tells us that $C=0$ and so $J\left(\alpha \right)=\mathrm{log}\left(\alpha +1\right)$

Jacob Homer

${\int }_{0}^{1}\frac{x-1}{\mathrm{log}x}dx={\int }_{0}^{1}{\int }_{0}^{1}{x}^{t}dtdx$
$={\int }_{0}^{1}{x}^{t}dxdt$
$={\int }_{0}^{1}\frac{1}{1+t}dt$
$=\mathrm{log}\left(2\right)$

karton

Making the substitution $u=\mathrm{ln}x$, we get
$I={\int }_{-\mathrm{\infty }}^{0}\frac{{e}^{u}-1}{u}{e}^{u}du\phantom{\rule{0ex}{0ex}}=-{\int }_{0}^{+\mathrm{\infty }}\frac{{e}^{-2s}-{e}^{-s}}{s}ds\phantom{\rule{0ex}{0ex}}=\mathrm{ln}\frac{2}{1}\phantom{\rule{0ex}{0ex}}=\mathrm{ln}2$
since we recognize a Frullani integral type.

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