# Showing that $\int_{0}^{1} \frac{x-1}{\ln(x)} \mathrm dt=\ln2$

I would like to show that

$$\int_{0}^{1} \frac{x-1}{\ln(x)} \mathrm dt=\ln2$$

What annoys me is that $x-1$ is the numerator so the geometric power series is useless.

Any idea?

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Inverse question: how to compute $\int_0^1 \frac{\ln x}{x-1} d x$? –  sdcvvc Jun 28 '13 at 13:32

This is a classic example of differentiating inside the integral sign.

In particular, let $$J(\alpha)=\int_0^1\frac{x^\alpha-1}{\log(x)}\;dx$$. Then one has that $$\frac{\partial}{\partial\alpha}J(\alpha)=\int_0^1\frac{\partial}{\partial\alpha}\frac{x^\alpha-1}{\log(x)}\;dx=\int_0^1x^\alpha\;dx=\frac{1}{\alpha+1}$$ and so we know that $\displaystyle J(\alpha)=\log(\alpha+1)+C$. Noting that $J(0)=0$ tells us that $C=0$ and so $J(\alpha)=\log(\alpha+1)$.

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Thanks you very much! –  Chon Jan 19 '12 at 18:06
(+1) nice answer. –  Mhenni Benghorbal Jan 8 '13 at 7:29
$\displaystyle \int_{0}^{1}\frac{x-1}{\log{x}}\;{dx} = \int_{0}^{1}\int_{0}^{1}x^{t}\;{dt}\;{dx} =\int_{0}^{1}\int_{0}^{1}x^{t}\;{dx}\;{dt} = \int_{0}^{1}\frac{1}{1+t}\;{dt} = \log(2).$
Making the substitution $u=\ln x$, we get $$I=\int_{-\infty}^0\frac{e^u-1}u e^udu=-\int_0^{+\infty}\frac{e^{-2s}-e^{-s}}sds=\ln\frac 21=\ln 2,$$ since we recognize a Frullani integral type.