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I would like to know how to prove the following two definite integrals.

A: $$\int_{0}^1\frac{x-1}{\log x}dx=\log 2$$

B:$$\int_{0}^1\frac{\log x}{x-1}dx=\frac{\pi^2}{6}$$

I found these two by using wolfram-alpha, but I can't prove them. I suspect that the following relations might be used.

$$\log 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}$$and$$\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots=\sum_{n=1}^\infty\frac{1}{n^2}$$

I need your help.

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4 Answers 4

up vote 13 down vote accepted

Consider $$f(p)= \int_0^1\frac{x^p-1}{\log x}dx$$

Then $$f'(p)=\int_0^1x^pdx=\frac{1}{p+1}$$

Then integrate.

For the other, use that $$\int_0^1x^{n}\log x=-\frac{1}{(n+1)^2}$$

and that $$\frac{1}{1-x}=\sum_{n\geqslant 1}x^n$$

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Thank you very much. –  mathlove Aug 7 '13 at 6:20
    
Nice one on the Feynman trick. +1 –  Fixed Point Aug 20 '13 at 7:47

For A

Consider the following

$$I(a)=\int^1_0 \frac{x^a-1}{\log(x)}\, dx$$

Differentiate w.r.t to $a$ to get

$$I'(a)=\int^1_0 x^a dx=\frac{1}{a+1}$$

Hence

$$I(a) = \log(a+1)+C$$

By $a=0$ we have $C=0$ Hence

$$I(a) = \log(a+1)$$

Hence $$I(1) = \int^1_0 \frac{x-1}{\log(x)}\, dx = \log(2)$$

For part B

We know that

$$\int^1_0 \frac{\log(x)}{x-1}\, dx=-\int^1_0 \frac{\log(1-x)}{x}\, dx=\operatorname{Li}_2(1)=\zeta(2)$$

This can be proven through the power expansion

$$-\int^1_0 \frac{\log(1-x)}{x}=\sum_{k\geq 1}\frac{1}{k^2}=\zeta(2)$$

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Thank you very much. –  mathlove Aug 7 '13 at 6:32

\begin{eqnarray*} \int_{0}^{1}{x - 1 \over \ln\left(x\right)}\,{\rm d}x & = & \int_{0}^{\infty}{{\rm e}^{-x} - {\rm e}^{-2x} \over x}\,{\rm d}x = -\int_{0}^{\infty}\ln\left(x\right)\,\left(-{\rm e}^{-x} + 2{\rm e}^{-2x}\right)\,{\rm d}x \\ & = & \int_{0}^{\infty}\ln\left(x\right)\,{\rm e}^{-x}\,{\rm d}x - \int_{0}^{\infty}\ln\left(x \over 2\right)\,{\rm e}^{-x}\,{\rm d}x = \ln\left(2\right)\ \overbrace{\quad\int_{0}^{\infty}{\rm e}^{-x}\quad}^{1} = \ln\left(2\right) \\[1cm] &&\mbox{Or/and try this} \\ I\left(\mu\right) & \equiv & \int_{0}^{1}{x^{\mu} - 1 \over \ln\left(x\right)}\,{\rm d}x\,, \quad I'\left(\mu\right) = \int_{0}^{1}{x^{\mu}\ln\left(x\right) \over \ln\left(x\right)}\,{\rm d}x \\ I'\left(\mu\right) & = & \left.{x^{\mu + 1} \over \mu + 1}\right\vert_{0}^{1} = {1 \over \mu + 1} \quad\Longrightarrow\quad I\left(1\right) - I\left(0\right) = \int_{0}^{1}{{\rm d}\mu \over \mu + 1} = \ln\left(2\right) \end{eqnarray*}

\begin{eqnarray*} &&\\[1cm] \int_{0}^{1}{\ln\left(x\right) \over x - 1}\,{\rm d}x & = & \int_{0}^{\infty}x{\rm e}^{-x}\,{1 \over 1 - {\rm e}^{-x}}\,{\rm d}x = \int_{0}^{\infty}x{\rm e}^{-x}\,\sum_{n = 0}^{\infty}{\rm e}^{-nx}\,{\rm d}x = \sum_{n = 0}^{\infty} \int_{0}^{\infty}x\,{\rm e}^{-\left(n + 1\right)x}\,{\rm d}x \\ & = & \sum_{n = 0}^{\infty}{1 \over \left(n + 1\right)^{2}}\ \overbrace{\quad\int_{0}^{\infty}x\,{\rm e}^{-x}\,{\rm d}x\quad}^{1} = \sum_{n = 0}^{\infty}{1 \over \left(n + 1\right)^{2}} \\ & = & {\pi^{2} \over 6}\,,\quad\mbox{( Basel Problem )} \end{eqnarray*}

http://en.wikipedia.org/wiki/Basel_problem. See also: http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf

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Note your second equality is followed by a Frullani integral. –  Pedro Tamaroff Oct 31 '13 at 22:12
    
@PedroTamaroff I didn't know that. Thanks to your comment I found some interesting $\tt .pdf$ about it: ( 129.81.170.14/~vhm/papers_html/final15.pdf ). Thanks. –  Felix Marin Nov 2 '13 at 20:34

In what follows, we will use the Taylor series expansion for the natural logarithm: $$\quad\ln(1-x)\ =\ -\sum_{n=1}^\infty\frac{x^n}n\quad,\quad|x|\leqslant1,$$ coupled with the fact that the integral of a sum is the sum of integrals. $$\int_0^1\frac{\ln x}{x-1}dx\ =\ -\int_0^1\frac{\ln(1-x)}xdx\ =\ +\int_0^1\frac1x\cdot\sum_{n=1}^\infty\frac{x^n}ndx\ =\ \sum_{n=1}^\infty\int_0^1\frac{x^{n-1}}ndx=$$ $$=\ \sum_{n=1}^\infty\frac{x^n}{n^2}\Bigg|_{\ 0}^{\ 1}\ =\ \sum_{n=1}^\infty\frac1{n^2}\ =\ \zeta(2)\ =\ \frac{\pi^2}6.$$ As to why the sum in question equals $\frac{\pi^2}6$ , see Basel problem.

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