I found the following bivariate integral and I do not know how to attack it. $$ \int_0^\infty \int_0^\infty e^{-f(x,y)} \frac{\partial^2 f(x,y)}{\partial x \, \partial y} \, \mathrm{d}x \, \mathrm{d}y,$$ where $f:\mathbb{R^{+}}\times\mathbb{R^{+}} \mapsto \mathbb{R^{+}}$ is differentiable infinitely many times. Moreover I know that for any $y \in [0,\infty)$, $\lim_{x\mapsto \infty} f(x,y) = 0$, for any $x \in [0,\infty)$, $\lim_{y\mapsto \infty} f(x,y) = 0$ and that the integral exists finite.
Do you have some technique to suggest to attack this kind of integral or similar bivariate integrals?
Thanks a lot!
$$ \int_0^\infty \int_0^\infty e^{-f(x,y)} \frac{\partial^2 f(x,y)}{\partial x \, \partial y} \, \mathrm{d}x \, \mathrm{d}y = \int_0^\infty \left( \frac \partial {\partial y} \int_0^\infty e^{-f(x,y)} \frac{\partial f(x,y)}{\partial x} \, \mathrm{d}x \right) \, \mathrm{d}y $$ According to Theorem 2 on this page, the equality above is valid if $\dfrac {\partial f} {\partial x}$ and $\dfrac{\partial^2 f}{\partial y\, \partial x}$ are both continous. Or at least that is true if one integrates over a bounded interval, so you may want to do a bit of work with limits as something approaches $+\infty.$
This makes the inner integral into $\displaystyle\int_{a(y)}^{b(y)} e^{-u} \, du $ where $a(y) = f(0,y)$ and $b(y)=f(\infty,y) = \lim\limits_{u\,\to\,+\infty} f(u,y).$ Then think about Leibniz's rule.