Help with a Vardi Integral I ran into some trouble trying to evaluate this integral:
$$\mathcal{I}=\int_{0}^{1}\frac{x}{x^{2}+1}\log\log\left(\frac{1}{x}\right)\mathrm dx$$
Basically, what I have at the moment is a generalized expansion
$$\int_{0}^{1}\frac{x^{p-1}}{\delta + x^{w}}\log\log\left(\frac{1}{x}\right)\mathrm dx = -\gamma\int_{0}^{1}\frac{x^{p-1}}{\delta+x^{w}}\mathrm dx-\frac{1}{\delta}\sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{\log(wk+p)}{\delta^{k}(wk+p)}$$
Hence,
$$\mathcal{I} = -\gamma\int_{0}^{1}\frac{x}{1+x^{2}}\mathrm dx - \sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{\log(2k+2)}{(2k+2)}$$
I can't figure out how to solve the sum. I'm positively certain there's some trick with digamma functions or Hurwitz-Zeta, but I'm too sleep-deprived to make sense of it. Maybe I'm going about this all wrong and there exists a more straightforward method to solve $\mathcal{I}$? Please let me know your opinion!
 A: Using
$$\frac x{x^2+1}=\sum_{n=0}^\infty (-1)^n x^{2n+1}$$ and, after a simple change of variable
$$\int  x^{2n+1} \log (-\log (x))\,dx=-\frac{\text{Ei}(2 (n+1) \log (x))-x^{2 n+2} \log (-\log (x))}{2 (n+1)}$$
$$\int_0^1  x^{2n+1} \log (-\log (x))\,dx=-\frac{\log \left(4 (n+1)^2\right)-2 i \pi +2 \gamma }{4 (n+1)}$$ you should arrive at the result
$$\mathcal{I}=\int_{0}^{1}\frac{x}{x^{2}+1}\log\log\left(\frac{1}{x}\right)\mathrm dx=-\frac{3}{4} \log ^2(2)$$
A: In fact your formula is very close to the answer (provided by @Claude Leibovici).
The sum in your formula
$- \sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{\log(2k+2)}{(2k+2)}=-\frac{\log2}{2} \sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{1}{(k+1)}-\frac{1}{2}\sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{\log(k+1)}{(k+1)}$
a) $-\frac{\log2}{2} \sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{1}{(k+1)}=\frac{\log2}{2} \sum\limits_{k\geq 1}(-1)^{k}\cdot\frac{1}{k}=-\frac{\log^22}{2} $
b)$ -\frac{1}{2}\sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{\log(k+1)}{(k+1)}=-\frac{1}{2}\sum\limits_{k\geq 1}(-1)^{k+1}\cdot\frac{\log(k)}{k}=-\frac{1}{2}\Bigl(\frac{\log(1)}{1}-\frac{\log(2)}{2}+\frac{\log(3)}{3}-+...\Bigr)$
After regrouping the terms in the parentheses
$b) =\frac{1}{2}\lim_{s\to+0}\frac{\partial}{\partial{s}}\Bigl(\frac{1}{1^{s+1}}-\frac{1}{2^{s+1}}+\frac{1}{3^{s+1}}-\frac{1}{4^{s+1}}+-...\Bigr)=$$=\frac{1}{2}\lim_{s\to+0}\frac{\partial}{\partial{s}}\Bigl((\frac{1}{1^{s+1}}+\frac{1}{3^{s+1}}+...)-(\frac{1}{2^{s+1}}+\frac{1}{4^{s+1}}+...)\Bigr)=$$=\frac{1}{2}\lim_{s\to+0}\frac{\partial}{\partial{s}}\Bigl((\frac{1}{1^{s+1}}+\frac{1}{2^{s+1}}+\frac{1}{3^{s+1}}+...)-2(\frac{1}{2^{s+1}}+\frac{1}{4^{s+1}}+...)\Bigr)=$$=\frac{1}{2}\lim_{s\to +0}\frac{\partial}{\partial{s}}\Bigl(\zeta(1+s)-\frac{2}{2^{s+1}}\zeta(1+s)\Bigr)$
But at $s\to0$ $\zeta(1+s)=\frac{1}{s}+\gamma+O(s)$
Therefore b) $=-\frac{\log2}{2}(\frac{\log2}{2}-\gamma)$
