Using the identity
$$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$
The integral can be written
$$
I=\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\left(\cos(x)\cos(y)-\sin(x)\sin(y)\right)\,dx\,dy $$
Now by splitting the integrals
$$\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\cos(x)\cos(y)\,dx\,dy-\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\sin(x)\sin(y)\,dx\,dy
$$
Notice by symmetry of the integrals we have
$$\left(\int^\infty_0 \frac{\log x }{\sqrt {x}}\cos(x)\,dx \right)^2-\left(\int^\infty_0 \frac{\log x }{\sqrt {x}}\sin(x)\,dx \right)^2
$$
Both inegrals are solvable by using the mellin transforms
$$\int^\infty_0 x^{s-1}\sin(x)\,dx = \Gamma (s) \sin\left( \frac{\pi s}{2} \right)$$
$$\int^\infty_0 x^{s-1}\cos(x)\,dx = \Gamma (s) \cos\left( \frac{\pi s}{2} \right)$$
By differentiation under the integral sign and using $s=\frac{1}{2}$.
$$\int^\infty_0 \frac{\log x }{\sqrt {x}}\cos(x)\,dx =-\frac{1}{2} \sqrt{\frac{π}{2}} \left(2 \gamma +π+\log(16) \right) $$
$$\int^\infty_0 \frac{\log x }{\sqrt {x}}\sin(x)\,dx=\frac{1}{2} \sqrt{\frac{π}{2}} (-2 \gamma +π- \log(16))
$$
Collecting the results together we have
$$I=(\gamma+2\log 2)\pi^2$$
Best Answer
Let the considered integral be denoted by $I$. We have $$\eqalign{I&=\int_0^{1}\frac{\ln x}{\cosh^2x}dx+\int_{1}^\infty\frac{\ln x}{\cosh^2x}dx\cr &=\Big[(\ln x)\tanh x \Big]_0^1-\int_0^1\frac{\tanh x}{x}dx+ \Big[(\ln x)(\tanh x-1) \Big]_1^\infty-\int_1^\infty\frac{\tanh x-1}{x}dx\cr &=2\int_1^\infty\frac{dx}{x(1+e^{2x})}-\int_0^1\frac{1-e^{-2x}}{x(1+e^{-2x})}dx\cr &=2\int_1^\infty\frac{dx}{x(1+e^{2x})}-\int_0^1\frac{1-e^{-2x}}{x}\left(\frac{1}{1+e^{-2x}}-1\right)dx- \int_0^1\frac{1-e^{-2x}}{x}dx\cr &=2\int_1^\infty\frac{dx}{x(1+e^{2x})}+\int_0^1\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx- \int_0^1\frac{1-e^{-2x}}{x}dx\cr &=2\int_1^\infty\frac{dx}{x(1+e^{2x})}-\int_1^\infty\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx+\int_0^\infty\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx- \int_0^1\frac{1-e^{-2x}}{x}dx\cr &=\int_1^\infty\frac{1+e^{-2x}}{x(1+e^{2x})}dx+\int_0^\infty\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx- \int_0^1\frac{1-e^{-2x}}{x}dx\cr &=\int_1^\infty\frac{e^{-2x}}{x}dx- \int_0^1\frac{1-e^{-2x}}{x}dx+\int_0^\infty\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx\cr &=\underbrace{\int_2^\infty\frac{e^{-t}}{t}dt- \int_0^2\frac{1-e^{-t}}{t}dt}_A+\underbrace{\int_0^\infty\frac{e^{-t}-e^{-2t}}{t(1+e^{-t})}dt}_B} $$ Now, note that $$\eqalign{ A&=\int_2^\infty\frac{e^{-t}}{t}dt-\int_0^1\frac{1-e^{-t}}{t}dt- \int_1^2\frac{1-e^{-t}}{t}dt\cr &=\int_2^\infty\frac{e^{-t}}{t}dt-\int_0^1\frac{1-e^{-t}}{t}dt-\ln2 +\int_1^2\frac{e^{-t}}{t}dt\cr &=\int_1^\infty\frac{e^{-t}}{t}dt-\int_0^1\frac{1-e^{-t}}{t}dt-\ln2\cr &=-\gamma-\ln2 } $$ To calculate $B$ we can do the following $$\eqalign{ B&=\int_0^\infty\frac{e^{-t}-e^{-2t}}{t}\left(\sum_{n=0}^\infty(-1)^ne^{-nt}\right)dt\cr &=\sum_{n=0}^\infty(-1)^n\int_0^\infty\frac{e^{-(n+1)t}-e^{-(n+2)t}}{t}dt\cr &=\sum_{n=0}^\infty(-1)^n\ln\left(\frac{n+2}{n+1}\right)\cr &=\lim_{m\to\infty}\sum_{n=0}^{2m-2}(-1)^n\ln\left(\frac{n+2}{n+1}\right)\cr &=\lim_{m\to\infty}\ln\left(\prod_{k=1}^m\frac{2k}{2k-1}\prod_{k=1}^{m-1}\frac{2k}{2k+1}\right)\cr &=\lim_{m\to\infty}\ln\left(\frac{(2m)!!(2m-2)!!}{((2m-1)!!)^2}\right)=\ln\left(\frac{\pi}{2}\right)\cr } $$ By Stirling's formula. This yields $I=\ln \pi-2\ln 2-\gamma$. $\qquad\square$