Suppose $0 < a < b < \infty$. I am trying to find a nontrivial upper bound for:
$$ \int^\infty_0 \int^\infty_0 \frac{dx \, dy}{xy + \frac{e^{a(x+y)}}{b^2}}. \tag{1} $$
Now, I know that
$$ \int^\infty_0 \frac{dx}{x + \frac{e^{ax}}{b}} \leq 2 \log\left(1 + \frac{b}{a}\right). \tag{2} $$
Can I obtain something like this for (1), i.e. a logarithmic dependence of a similar sort?
I attempted to use (2), but this only gives me:
\begin{eqnarray} \int^\infty_0 \int^\infty_0 \frac{dx \, dy}{xy + \frac{e^{a(x+y)}}{b^2}} &\leq \int^\infty_0 \int^\infty_0 \frac{ds \, dt}{st + \frac{a^2}{b^2}e^se^t} \\ &= \int^\infty_0 \frac{1}{s} \int^\infty_0 \frac{dt}{t + (\frac{a^2}{b^2}\frac{e^s}{s})e^t}\,ds \\ &\leq \int^\infty_0 \frac{1}{s} \cdot2 \log\left(1 + \frac{b^2}{a^2}\frac{s}{e^s}\right) \, ds.\tag{3} \end{eqnarray}
Surely,
$$ \int^1_0 \frac{1}{s} \cdot \log\left(1 + \frac{s}{e^s}\right) \, ds \leq \int^1_0 \frac{\log\left(1 + s\right)}{s} = \frac{\pi^2}{12}. $$
But what else can I deduce?