TITLE: How to prove $\int_0^1\frac{1-x}{(\ln x)(1+x)}\ dx=\ln\left(\frac2{\pi}\right)$?
QUESTION [9 upvotes]: A friend asked me to prove

$$\int_0^1\frac{1-x}{(\ln x)(1+x)}\ dx=\ln\left(\frac2{\pi}\right)$$

and here is my approach:
\begin{align}
I&=\int_0^1\frac{1-x}{\ln x}\frac1{1+x}\ dx\\
&=\int_0^1\left(-\int_0^1x^y\ dy\right)\frac1{1+x}\ dx\\
&=\int_0^1\left(-\int_0^1\frac{x^y}{1+x}\ dx\right)\ dy\\
&=\int_0^1\left((-1)^n\sum_{n=1}^\infty\int_0^1x^{y+n-1}\ dx\right)\ dy\\
&=\int_0^1\left(\sum_{n=1}^\infty\frac{(-1)^n}{y+n}\right)\ dy\\
&=\sum_{n=1}^\infty(-1)^n\int_0^1\frac1{y+n}\ dy\\
&=\sum_{n=1}^\infty(-1)^n\left[\ln(n+1)-\ln(n)\right]\tag{1}
\end{align}

Now how can we finish this alternating sum into $\ln\left(\frac2{\pi}\right)$?
My idea was to use 
$$\operatorname{Li}_a(-1)=(1-2^{-a})\zeta(a)=\sum_{n=1}^\infty\frac{(-1)^n}{n^a}$$
and if we differentiate both sides with respect to $a$ we get
$$2^{1-a}(\zeta^{'}(a)-\ln2\zeta(a))-\zeta^{'}(a)=\sum_{n=1}^\infty\frac{(-1)^{n-1}\ln(n)}{n^a}\tag{2}$$
wolfram says that $\sum_{n=1}^\infty(-1)^n\ln(n)$ is divergent which means we can not take the limit for (2) where $a\mapsto 0$ which means we can not break the sum in (1) into two sums. So any idea how to do (1)?

Other question is, I tried to simplify the sum in (1) as follows
\begin{align}
S&=\sum_{n=1}^\infty(-1)^n\left[\ln(n+1)-\ln(n)\right]\\
&=-\left[\ln(2)-\ln(1)\right]+\left[\ln(3)-\ln(2)\right]-\left[\ln(4)-\ln(3)\right]+\left[\ln(5)-\ln(4)\right]-...\\
&=-2\ln(2)+2\ln(3)-2\ln(4)+...\\
&=2\sum_{n=1}^\infty(-1)^n\ln(n+1)
\end{align}
which is divergent again. what went wrong in my steps?
Thanks.

REPLY [2 votes]: Let
$$I=\int_0^1\frac{1-x}{\ln x(1+x)}\ dx$$
and $$I(a)=\int_0^1\frac{1-x^a}{\ln x(1+x)}\ dx.$$
Note that $I(1)=I$ and $I(0)=0.$
$$I'(a)=\int_0^1\frac{-x^a}{1+x}\ dx=\frac12\psi\left(\frac{a+1}{2}\right)-\frac12\psi\left(\frac{a+2}{2}\right)$$
$$\Longrightarrow \int_0^1I'(a)da=I(1)-I(0)=I=\frac12\int_0^1 \left[\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a+2}{2}\right)\right]da $$
$$=\left[\ln\Gamma\left(\frac{a+1}{2}\right)-\ln\Gamma\left(\frac{a+2}{2}\right)\right]_0^1=\ln\frac{\Gamma(\frac{a+1}{2})}{\Gamma(\frac{a+2}{2})}\Bigg|_0^1$$
$$=-\ln\Gamma\left(\frac32\right)-\ln\Gamma\left(\frac12\right)=-\ln \left(\Gamma\left(\frac32\right)\Gamma\left(\frac12\right)\right)$$
$$=-\ln \left(\frac12\Gamma^2\left(\frac12\right)\right)=-\ln\left(\frac{\pi}{2}\right).$$