Proving $\int_{0}^1\frac{x-1}{\log x}dx=\log 2$ and $\int_{0}^1\frac{\log x}{x-1}dx=\frac{\pi^2}{6}$ 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.
A: 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 
$$-\int^1_0 \frac{\log(1-x)}{x}=\sum_{k\geq 1}\frac{1}{k^2}=\zeta(2)$$
A: \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
A: 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 $$\frac{1}{1-x}=\sum_{n\geqslant 1}x^n$$
A: 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.
