Computation of multiple improper integral. In my recent work, I need to the details of the computation of the following multiple improper integral:
$$\iint_{[0,1]^2}e^{-\pi x^2y^2}dxdy-\iint_{[1,\infty)^2}e^{-\pi x^2y^2}dxdy.$$
As you see, the first one is a proper integral and the second one is improper integral. It is easy to know the convergence of the second one. At the beginning, I just use Wolfram mathematica 9.0 to get the finally result:
$$\frac{1}{4}(\gamma+\log(4\pi)),$$ 
where $\gamma$ is a Euler-Gamma constant.
But now I need to give the details of the computation.
The wolfram mathematica tells us that:
$$\iint_{[0,1]^2}e^{-\pi x^2y^2}dxdy=HypergeometricPFQ[(1/2,1/2);(3/2,3/2);-\pi];$$
$$\iint_{[1,\infty)^2}e^{-\pi x^2y^2}dxdy
=HypergeometricPFQ[(1/2,1/2);(3/2,3/2)-\frac{1}{4}(\gamma+\log(4\pi)).$$
I do not know how to get the hyperbolic geometric function and  how to cancel it. 
Any help and hints will welcome. Thanks a lot.
 A: From the definition of Error Function: 
$$ 
\begin{align} 
I_1&=+\iint_{{[0,1]}^2}e^{-\pi\,x^2 y^2}\,{dx dy}\,=\,\int_{0}^{1}dx\,\int_{0}^{1}+\,e^{-\pi x^2 y^2}\,dy\,=\,\int_{0}^{1}\left[\frac{\text{erf}\left(\sqrt{\pi}\,x y\right)}{2x}\right]_{y=0}^{y=1}\,dx \\[2mm] 
&=\int_{0}^{1}\frac{\text{erf}\left(\sqrt{\pi}\,x\right)}{2x}\,dx\qquad\color{blue}{\{{\small u=\text{erf}\left(\sqrt{\pi}\,x\right)\,\Rightarrow\,du=2e^{-\pi\,x^2};\,\,dv=\frac{dx}{2x}\,\Rightarrow\,v=\frac{\log{x}}{2}}\}}\\[2mm] 
&=\left[\left(\,\text{erf}\left(\sqrt{\pi}\,x\right)\right)\frac{\log{x}}{2}\right]_{0}^{1}\,-\,\int_{0}^{1}e^{-\pi\,x^2}\,\log{x}\,dx\,=-\int_{0}^{1}e^{-\pi\,x^2}\,\log{x}\,dx\qquad\qquad\qquad\color{blue}{\{1\}} \\[6mm] 
I_2&=-\iint_{[1,\infty)^2}e^{-\pi\,x^2 y^2}\,{dx dy}\,=\,\int_{1}^{\infty}dx\,\int_{1}^{\infty}-\,e^{-\pi x^2 y^2}\,dy\,=\,\int_{1}^{\infty}\left[\frac{\text{erf}\left(\sqrt{\pi}\,x y\right)-1}{2x}\right]_{y=\infty}^{y=1}\,dx \\[2mm] 
&=\int_{1}^{\infty}\frac{\text{erf}\left(\sqrt{\pi}\,x\right)-1}{2x}\,dx\qquad\color{blue}{\{{\small u=\text{erf}\left(\sqrt{\pi}\,x\right)-1\,\Rightarrow\,du=2e^{-\pi\,x^2};\,\,dv=\frac{dx}{2x}\,\Rightarrow\,v=\frac{\log{x}}{2}}\}}\\[2mm] 
&=\left[\left(\,\text{erf}\left(\sqrt{\pi}\,x\right)-1\right)\frac{\log{x}}{2}\right]_{1}^{\infty}\,-\,\int_{1}^{\infty}e^{-\pi\,x^2}\,\log{x}\,dx\,=-\int_{1}^{\infty}e^{-\pi\,x^2}\,\log{x}\,dx\qquad\quad\color{blue}{\{2\}} \\[6mm] 
\color{red}{I}&=I_1+I_2=-\int_{0}^{\infty}e^{-\pi\,x^2}\,\log{x}\,dx=-\int_{0}^{\infty}e^{-\,x^2}\,\log{\frac{x}{\sqrt{\pi}}}\,\,\frac{dx}{\sqrt{\pi}}\qquad\color{blue}{\{{\small x=\frac{t}{\sqrt{\pi}}}\}}\\[2mm] 
&=-\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}e^{-\,x^2}\,\log{x}\,dx+\frac{\log{\pi}}{2\sqrt{\pi}}\int_{0}^{\infty}e^{-\,x^2}\,dx={\small \frac{\gamma+\log{4}}{4}+\frac{\log{\pi}}{4}}=\color{red}{\frac{\gamma+\log{4\pi}}{4}}\quad\color{blue}{\{3\}} 
\end{align} 
$$ 
 
$\,{\color{blue}{\{1\}}}\,$ Two times L'Hopital's Rule, starting by $\,{\displaystyle\small \lim_{x\rightarrow 0}\left[\text{erf}(x).\log{x}\right]=\lim_{x\rightarrow 0}\,\frac{\log{x}}{1/\text{erf}(x)}}\,$
$\,{\color{blue}{\{2\}}}\,$ Four times L'Hopital's Rule, starting by $\,{\displaystyle\small \lim_{x\rightarrow \infty}\left[\text{erfc}(x).\log{x}\right]=\lim_{x\rightarrow \infty}\,\frac{\text{erfc}(x)}{1/\log{x}}}\,$
$\,{\color{blue}{\{3\}}}\,$ Integral of Gamma Conatant.  
