I'm trying to solve
\begin{equation} S_{p,q}:= \int_{-\infty}^\infty \text da \ \int_{-\infty}^\infty \text db \ \exp \Big[i u (a^2-b^2) - v(a^2-b^2)^2\Big] a^p b^q \end{equation} with positive real valued constants $u$ and $v$ for all cases where $p+q\le4$. Its easy to show that $S$ vanishes if either $p$ or $q$ is odd. Lets also notice, that for $v \rightarrow 0$ the integral can be written as the product of two integrals which both converge. So i would expect that for $v>0$ the integral also converges, because a nonvanishing $v$ makes the integrand smaller.
Because Mathematica can not solve this integral, I tried the following transformations. First we substitute $a \rightarrow x(a):=a^2$ and $b \rightarrow y(b):=b^2$ which leads to \begin{equation} S_{p,q} = \int_0^\infty \text dx \ \int_0^\infty \text dy \ \exp\Big[iu(x-y) -v (x-y)^2\Big] x^{(p-1)/2} y^{(q-1)/2}. \end{equation}
My next transformation was setting \begin{equation} \begin{pmatrix}g\\h\end{pmatrix} := \begin{pmatrix}1&1\\1&-1\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} \end{equation} which gives \begin{equation} S_{p,q} = \frac 1 {|-2|} \int_{-\infty}^\infty \text d h \ \exp \left(iuh-vh^2\right) \int_{|h|}^\infty \text d g \ \left(\frac {g+h}2\right)^{(p-1)/2} \left(\frac {g-h}2\right)^{(q-1)/2}. \end{equation} The transformation of the limits can easily be seen, when you note, that in the $x-y-$ System, the integral goes over the upper right quadrant. The $g-h-$ System is rotated by 45° relative to the $x-y-$ System. So if you let $h$ run over all real numbers, $g$ has to be greater or equal to $|h|$ for $(x,y)$ being in the upper right quadrant.
Now for all interesting cases (where $p+q \le 4$) the second integral diverges. My question is: Does this also mean, that $S$ diverges? If so, why does $S$ for $v \rightarrow 0$ converge? And if not, do you have an idea how to solve this integral?