Evaluating $\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}$ Some time ago I came across  to the following integral:  
$$I=\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}$$ What are the hints on how to compute this integral?

You may like this method. Note
$$ \int_0^1x^ada=-\frac{1-x}{\ln x}, \text{ for }x>0, H_a=\sum_{k=1}^\infty\frac{a}{k(a+k)}, $$
and hence
\begin{eqnarray}
\int_0^1\frac{1-x}{1+x}\frac{dx}{\ln x}&=&-\int_0^1\int_0^1\frac{x^a}{1+x}dadx\\
&=&-\int_0^1\int_0^1\frac{x^a}{1+x}dxda\\
&=&-\frac{1}{2}\int_0^1(H_{\frac{a}{2}}-H_{\frac{a-1}{2}})da\\
&=&-\frac{1}{2}\int_0^1\sum_{k=1}^\infty\left(\frac{a}{k(a+2k)}-\frac{a-1}{k(a+2k-1)}\right)da\\
&=&-\frac{1}{2}\sum_{k=1}^\infty\left(4\ln k-2\ln(k^2-\frac{1}{4})\right)\\
&=&-\frac{1}{2}\ln\prod_{k=1}^\infty\frac{k^4}{(k^2-\frac{1}{4})^2}\\
&=&-\prod_{k=1}^\infty\frac{k^2}{k^2-\frac{1}{4}}\\
&=&-\ln\left(\frac{\pi}{2}\right).
\end{eqnarray}
In the last step, we used the Wallis Formula from here.
