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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?

  • 0
    https://math.stackexchange.com/questions/154881/is-the-integral-int-1-infty-fracx-a-x-b-logx-dx-convergent?noredirect=1&lq=12017-12-19

3 Answers 3

30

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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    Alex, will differentiating outside and inside the integral sign give the same result? If not, why? (Can you give a *link* showing where I can learn the difference between them?) Thanks2017-10-13
19

$\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). $

17

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 h$a$ve he$a$rd about Frullani integral. nice answer (+1).2013-01-08