Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

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

3 Answers 3

up vote 25 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 $J(\alpha)=\log(\alpha+1)$.

share|improve this answer
Thanks you very much! –  Chon Jan 19 '12 at 18:06
(+1) nice answer. –  Mhenni Benghorbal Jan 8 '13 at 7:29
add comment

$\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). $

share|improve this answer
+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
add comment

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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.