Computation of multiple improper integral.

Question

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.

Answer 1

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.