asymptotic approximation when $a\to 0^+$ of $I(a):=\int_0^\infty \int_0^{a/x}e^{-x-y}\ dy\ dx.$ I want to find an asymptotic approximation when $a\to 0^+$ for the integral 
$$I(a):=\int_0^\infty \int_0^{a/x}e^{-x-y}\ dy\ dx.$$
I found the following approximation:
$$C_1\, a\, \mathrm{ln}(1/a) < I(a)< C_2 \sqrt{a},$$
where $C_1$ and $C_2$ are constant terms.
I would like to have a better approximation, something of the type $I(a)=\Theta(\sqrt{a})$ or $I(a)=\Theta(a\, \mathrm{ln}(1/a))$ when $a\to 0^+$.
In the folowing I explain how I found those bounds.
First we split our integral:
$$I(a):=\int_0^\sqrt{a} \int_0^\sqrt{a} e^{-x-y}\ dy\ dx + \int_0^\sqrt{a} \int_\sqrt{a}^{a/x}e^{-x-y}\ dy\ dx +\int_\sqrt{a}^\infty \int_0^{a/x}e^{-x-y}\ dy\ dx.$$
But the two last integral are equal (clear if you draw the picture of the domain of integration and observe the symmetry), so we have: $I(a)=I_1(a)+2\, I_2(a)$ with
$$I_1(a):=\int_0^\sqrt{a} \int_0^\sqrt{a} e^{-x-y}\ dy\ dx\quad  \text{ and }\quad I_2(a):=\int_\sqrt{a}^\infty \int_0^{a/x}e^{-x-y}\ dy\ dx.$$
We consider that $a$ is small, so we have $0<a<1$ and in the domain of integration of $I_1(a)$, we have $e^{-2}<e^{-x-y}<1$. Hence
$e^{-2} a <I_1(a)< a$. That is
$$I_1(a)=\Theta(a).$$
This first integral will not play any role for our bounds, because the second integral will be much bigger (for small values of $a$).
Now we observe that if $x>\sqrt(a)$, then $a/x<a/\sqrt{a}=\sqrt{a}$. Hence
$$I_2(a)<\int_\sqrt{a}^\infty \int_0^\sqrt{a}e^{-x-y}\ dy\ dx
<\int_\sqrt{a}^\infty e^{-x} \int_0^\sqrt{a}1\ dy\ dx
=\sqrt{a}\,e^{-\sqrt{a}}<\sqrt{a}.$$
So we have proved the upper bound, and will show the above bound now.
We have for $x>\sqrt{a}>a$ and $a<1$,
$$\int_0^{a/x}e^{-y}\ dy> \frac{a}{x} e^{-a/x} >\frac{a}{x} e^{-1}.$$
So
$$I_2(a)> a\,e^{-1} \int_\sqrt{a}^\infty \frac{e^{-x}}{x}\ dx
>a\,e^{-1} \int_\sqrt{a}^1 \frac{e^{-x}}{x}\ dx
>a\,e^{-2} \int_\sqrt{a}^1 \frac{1}{x}\ dx
=\frac{e^{-2}}{2}\,a\,\textrm{ln}(1/a).$$
This gives us the lower bound.
 A: Your method will definitely work, we just need to keep track of the errors along the way.
You showed that
$$
\begin{align}
I(a) &= 2I_2(a) + \Theta(a) \\
&= 2\int_{\sqrt{a}}^\infty e^{-x} - e^{-a/x-x}\,dx + \Theta(a). \tag{1}
\end{align}
$$
As you observed, the tail of the integral is insignificant.  In fact we have
$$
\begin{align}
0 &\leq \int_{1}^\infty e^{-x} - e^{-a/x-x}\,dx \\
&\leq \int_{1}^\infty e^{-x} - e^{-a/1-x}\,dx \\
&= (1-e^{-a}) \int_1^\infty e^{-x}\,dx,
\end{align}
$$
so that
$$
\int_{1}^\infty e^{-x} - e^{-a/x-x}\,dx = O(a). \tag{2}
$$
In the remaining interval $x \in [\sqrt{a},1]$ we have $e^{-a/x} = 1 - a/x + O(a/x)^2$.  Substituting this into the integral yields
$$
\begin{align}
\int_\sqrt{a}^1 e^{-x} - e^{-a/x-x}\,dx &= \int_{\sqrt{a}}^1 e^{-x} - e^{-x}\Bigl[1 - a/x + O(a/x)^2\Bigr]\,dx \\
&= a \int_\sqrt{a}^1 \frac{e^{-x}}{x}\,dx + O\left(a^2\int_\sqrt{a}^1 \frac{e^{-x}}{x^2}\,dx\right). \tag{3}
\end{align}
$$
In the integrals we substitute $e^{-x} = 1 + O(x)$ to get
$$
\int_\sqrt{a}^1 \frac{e^{-x}}{x}\,dx = \int_\sqrt{a}^1 \frac{dx}{x} + O(1) = \frac{\log(1/a)}{2} + O(1)
$$
and
$$
\int_\sqrt{a}^1 \frac{e^{-x}}{x^2}\,dx = \frac{1}{\sqrt{a}} + O(\log a).
$$
Thus $(3)$ becomes
$$
\int_\sqrt{a}^1 e^{-x} - e^{-a/x-x}\,dx = \frac{a\log(1/a)}{2} + O(a).
$$
Combining this with $(1)$ and $(2)$, we conclude that
$$
I(a) = a\log\left(\frac{1}{a}\right) + O(a)
$$
as $a \to 0^+$.
