Evaluate
$\displaystyle \int_0^1\int_0^1\frac{dx\,dy}{\ln^{3/2}\left(\dfrac1{xy}\right)}$
For the inner integral, substitute $\displaystyle u=xy,$
$\displaystyle \int_0^1 {\int_0^1 {\frac{{dx\,dy}}
{{\Big[ { - \ln (xy)} \Big]^{3/2} }}} } = \int_0^1 {\int_0^y {\frac{{du\,dy}}
{{y\left( { - \ln u} \right)^{3/2} }}} } = \int_0^1 {\int_u^1 {\frac{{dy\,du}}
{{y\left( { - \ln u} \right)^{3/2} }}} } .$
So
$\displaystyle \int_0^1 {\int_0^1 {\frac{{dx\,dy}}
{{\Big[ { - \ln (xy)} \Big]^{3/2} }}} } = \int_0^1 {\frac{{ - \ln u}}
{{\left( { - \ln u} \right)^{3/2} }}\,du} = \int_0^1 {\frac{1}
{{\sqrt { - \ln u} }}\,du} = \sqrt \pi .$