Ask a about hard integral of $\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$ I want to evaluate the integral:
$$I(a,b)=\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$$
Attempt:$$\frac{\partial ^2I}{\partial a\partial b}=4ab\int_{0}^{\infty}\frac{\log x}{(a^2+x^2)(b^2+x^2)}dx=\frac{4ab}{b^2-a^2}\int_{0}^{\infty}\log x\left(\frac{1}{a^2+x^2}-\frac{1}{b^2+x^2}\right)dx$$
$$=\frac{4ab}{b^2-a^2}\frac{\pi}{2}\left(\frac{\log a}{a}-\frac{\log b}{b}\right)=\frac{2\pi(b\log a-a\log b)}{b^2-a^2}$$
Then $$I(a,b)=2\pi\int_{0}^{b}\int_{0}^{a}\frac{(y\log x-x\log y)}{y^2-x^2}dxdy$$
But this integral very hard to solve,can anyone help me,thank you! 

This derivation is slightly different from Maxim's, because I'm not fluent in Meijer G.  The beginning is the same, and assume $0<b<a.$  Then
$$ I(a,b)=b\,\int_0^\infty (\log{b} + \log{x} ) \log{(1+ \big(\frac{a/b}{x}\big)^2  ) }\log{(1+1/x^2)} dx $$
Let $r=b/a \le 1$ so that some series manipulations are permissible. The integral without the $\log{x}$ is easily performed in Mathematica:
$$  \int_0^\infty  \log{(1+ \big(r\,x\big)^{-2}  ) }\log{(1+1/x^2)} dx = \pi\Big( (1+\frac{1}{r})(\text{arctanh(r)} - \log{(1-r^2)})-2\log{r}\Big)$$
Define 
$$J(s;r)=\int_0^\infty x^s  \log{(1+ \big(r\,x\big)^{-2}  ) }\log{(1+1/x^2)} dx \,.$$
The objective is to find $$\frac{d}{ds}J(s;r)\Big|_{s=0} $$
Within Mathematica J(s;r) can be found in terms of elementary functions and Gauss's hypergeometric $F(a,b;c,x).$
$$\frac{J(s;r)}{\pi}=\sec{(\frac{\pi s}{2})} \Big\{\!\frac{2\,r^2}{
(s\!+\!1)(s\!\!+3)}F(1,\!\frac{s+3}{2};\! \frac{s+5}{2}, r^2)  -
\frac{2\,r^{1-s}}{
(s\!+\!1)(s\!-\!1)}F(1,\!\frac{1-s}{2};\! \frac{3-s}{2}, r^2)$$
$$+ \frac{1}{s\!+\!1}\Big[ r^{-1-s}\log{(1-r^2)} + \log{(-1+1/r^2)}+\frac{2}{s+1} - \pi\,\tan{(\pi\,s/2)}  \Big]
\Big\}$$
Do the derivative and take $s \to 0.$ $F(1,\frac{1}{2}; \frac{3}{2}, r^2)$ and 
 $F(1,\frac{3}{2}; \frac{5}{2}, r^2)$ evaluate to elementary functions.  However the derivatives with respect to $s$ of the hypergeometrics do not.  However, by using the series definition in terms of Pochhammer symbols, an easy calculation shows
$$ \frac{d}{ds} \frac{(3/2+s/2)_k}{(5/2+s/2)_k} \Big|_{s=0} = \frac{2k}{(2k+3)^2} \quad , \quad \frac{d}{ds} \frac{(1/2-s/2)_k}{(3/2-s/2)_k} \Big|_{s=0} = \frac{-2k}{(2k+1)^2}  $$  In detail,
$$\frac{d}{ds} F(1,\frac{1-s}{2}; \frac{3-s}{2}, r^2) \Big|_{s=0} 
= -\sum_{k=0}^\infty\frac{2k}{(2k+1)^2} r^{2k} = -\sum_{k=0}^\infty\frac{2k+1 -1}{(2k+1)^2} r^{2k} = $$
$$=-\frac{\text{arctanh(r)}}{r} + \sum_{k=0}^\infty \frac{r^{2k}}{(2k+1)^2}=
-\frac{\text{arctanh(r)}}{r} + \frac{1}{2r} \Big( \text{Li}_2(r) -  \text{Li}_2(-r) \Big)
$$
The series with the $(2k+3)^2$ in the denominator can be brought to this form with an index shift in the summation.  Collect all the results and you finally get
$$\frac{I(a,b)}{\pi\,b}= \log{b} \Big( (1+\frac{1}{r})(2\,\text{arctanh}(r) + \log{(1-r^2)} ) -2\log{r}  \Big)\, + 
\big(1-\frac{1}{r}\big) \big( \text{Li}_2(r) -  \text{Li}_2(-r) \big) $$
$$-\Big(\frac{\pi^2}{2} + \log{(r^{-2}-1)} +2\,\big(1+\log{r}+\frac{1}{r} \big)\,\text{arctanh}(r) + \frac{1+\log{r}}{r} \, \log{(1-r^2)} \Big) $$
