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I would like to show that

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

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
up vote 31 down vote accepted

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). $

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+1, I really like this method! – Daniel Littlewood Dec 17 '12 at 19:45
(+1) nice technique. – Mhenni Benghorbal Jan 8 '13 at 7:30

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.

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It is the first I have heard about Frullani integral. nice answer (+1). – Mhenni Benghorbal Jan 8 '13 at 7:32

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