I am stuck in trying to compute or more realistically upper bound the following double integral
$$
\int_\mathbb{R}\int_\mathbb{R} \frac{\exp(-(x-a)^2)\exp(-(y-a)^2)}{1+\exp(-x^2)+\exp(-y^2)}dxdy
$$
as a function of $a$. Of course a trivial (and $a$-independent) bound can be obtained via
$$
\frac{1}{1+\exp(-x^2)+\exp(-y^2)}\leq 1
$$
and then solving the resulting double integral closed form, but I was hoping in something better than that. Any suggestion is appreciated!
Computing or upper bounding a complicated integral
calculusintegrationmultiple integral
Best Answer
Denote your integral by $I$. The described estimate gives $I\le \pi$. More accurately, we have $$ \pi - I = \int_{\mathbb{R}}\int_{\mathbb{R}}e^{-(x - a)^2 - (y - a)^2}\frac{e^{-x^2} + e^{-y^2}}{1 + e^{-x^2} + e^{-y^2}}\,dx\,dy. $$ Since $0 < e^{-x^2} + e^{-y^2}\le 2$ we can write $$ \iint_{\Pi}e^{-(x - a)^2 - (y - a)^2}(e^{-x^2} + e^{-y^2})\,dx\,dy\ge \pi - I \ge \iint_{\Pi}e^{-(x - a)^2 - (y - a)^2}\frac{e^{-x^2} + e^{-y^2}}{3}\,dx\,dy. $$ The integrals in the latter inequalities can be calculated explicitly giving $$ \pi - I \approx\exp(-a^2/2) $$ up to some multiplicative constant.