Is it possible to find a function $F:\mathbb R^+\times \mathbb R^+\to\mathbb R$ such that \begin{align} \int_0^\infty\int_0^\infty F(x,y)e^{-xq-yq}dxdy=\frac{\alpha}{q^4}+\frac{\beta}{q^3}+\frac{\gamma}{q^2}, \end{align} with $\alpha, \beta, \gamma$ constants.
@Edit It is a problem that comes from probability. I have two independent random variables $X$ and $Y$ with exponential distribution of parameter $\frac{1}{\lambda}$ and I would like to understand with function $F$ of the two random variables has an expectation that is quadratic in $\lambda$. This means that I need a function $F$ such that \begin{align} \mathbb E(F(X, Y))=\int_0^\infty\int_0^\infty F(x, y)\frac{e^{-\frac{x}{\lambda}-\frac{y}{\lambda}}}{\lambda^2}dx dy=\alpha\lambda^2+\beta\lambda+\gamma. \end{align} If I call $q=\frac{1}{\lambda}$ the problem is reduced to the one written above, that means solving a double Laplace transform.
We already know that for $F(x,y)=\gamma$
$$\int_0^\infty\int_0^\infty \gamma e^{-xq-yq}\:dx\:dy = \frac{\gamma}{q^2}$$
Each of the terms $q^{-3}$ and $q^{-4}$ is (a scalar multiple of) a derivative of $q^2$. By Laplace transform properties in $1$D, we know that a derivative in the frequency domain is a multiplication by a monomial power in the time domain
$$\int_0^\infty tf(t)e^{-st}\:dt = -\frac{d}{ds}F(s)$$
In $2$D we now have options. Both $x\cdot1$ and $y\cdot 1$ accomplish a single derivative
$$\int_0^\infty\int_0^\infty \beta x e^{-xq-yq}\:dx\:dy = \int_0^\infty\int_0^\infty \beta y e^{-xq-yq}\:dx\:dy = \frac{\beta}{q^3}$$
And a second derivative can be accomplished one of three ways. This means that all functions with that specific Laplace transform are of the form
$$F(x,y) = \gamma+\beta(ax+by)+\frac{\alpha}{2}(ux^2+2vxy+wy^2)$$
where $a+b=u+v+w=1$