Definite integral $\int_{-\infty}^\infty \int_{-\infty}^\infty \exp(-x^2 y^2) \; dx dy$?

Question

The deceptively simple-looking double integral $$ \int_{-\infty}^\infty \int_{-\infty}^\infty \exp(-x^2 y^2) \; dx dy $$ is clearly finite, but both maple and Wolphram alpha gives the value $\infty$! Any ideas how to calculate this symbolically?

Answer 1

Note for $y>0$ we have $$ \int_{-\infty}^{+\infty} e^{-x^2y^2}\;dx = \frac{\sqrt{\pi}}{y} $$ so $$ \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} e^{-x^2y^2}\;dx\;dy \ge \int_0^{+\infty}\int_{-\infty}^{+\infty} e^{-x^2y^2}\;dx\;dy =\sqrt{\pi}\int_0^{+\infty}\frac{dy}{y} = +\infty. $$

Wolfram Alpha is right for once!

Answer 2

$$I=\int_{-t}^{+t}\int_{-t}^{+t} \exp(-x^2 y^2) \, dx \, dy=4 t^2 \,\, _2F_2\left(\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2}; -t^4\right)$$

For large $t$ $$\large\color{blue}{I\sim \sqrt{\pi } \,\,(4 \log (t)+\gamma +2\log (2))}$$ which is very good for any $t>1$.

If $t=e$ the difference is $4.65\times 10^{-27}$.

If you want more terms, just add $$\frac{e^{-t^4}}{t^6} \left(1-\frac{2}{t^4}+\frac{23}{4 t^8}-\frac{22}{t^{12}}+\frac{1689}{16 t^{16}}+O\left(\frac{1}{t^{20}}\right)\right) $$

Edit

For those we want to play, give Wolfram Alpha the following

Series[4*t^2*HypergeometricPFQ[{1/2, 1/2}, {3/2, 3/2}, -t^4],{t,Infinity,6}]