24
$\begingroup$

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?

$\endgroup$
30
$\begingroup$

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 $J(\alpha)=\log(\alpha+1)$.

$\endgroup$
  • 1
    $\begingroup$ 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?) Thanks $\endgroup$ – Mr Reality Oct 13 '17 at 18:06
19
$\begingroup$

$\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). $

$\endgroup$
17
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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