Here I have a generalization of the Fresnel integral in two variables, and I'd like to check that it converges:
$$ \frac{1}{2} \int_{\mathbb{R}^2 \times \mathbb{R}^2} \frac{d^2x}{(2\pi)^2} \frac{d^2y}{(2\pi)^2} \frac{\big[2\sinh \big(\frac{x_1 - x_2}{2}\big)\big]^2\big[2\sinh \big(\frac{y_1 - y_2}{2}\big)\big]^2}{\prod_{i,j=1,2}\big[2 \cosh \big(\frac{x_i - y_j}{2}\big)\big]^2} \exp \left[- \frac{1}{4\pi i}\Big((x_1^2 - y_1^2)+(x_2^2- y_2^2)\Big)\right]$$
The domain of integration is that $(x_1, x_2)$ and $(y_1, y_2)$ vary all over $\mathbb{R}^2$. This is from a physics computation.
Or prove that it diverges. This integral might give two different answers depending on which subspace we integrate first:
$\{x_1 = x_2\} \cap \{y_1 = y_2\}$ so that $\sinh (x_1 - x_2) = \sinh 0 = 0$.
$\{ x_1 = y_1\} \cap \{x_2 = y_2\}$ so that $e^{x_1^2 - y_1^2} = e^{x_2^2 - y_2^2} = e^0 = 1$
$\{ x_1 = y_2\} \cap \{x_2 = y_1\}$ so that $e^{x_1^2 - y_2^2} = e^{x_2^2 - y_1^2}= e^0 = 1$.
In the first case - without integrating over the remaining subspace $\mathbb{R}^4 / \big(\mathbb{R}(1,-1) \oplus \mathbb{R}(1,-1)\big) \simeq \mathbb{R}^2$ - we'd have that $\int = 0$. My notes have $\int = \frac{1}{16}$ and I wonder what the other two iterated integrals have us.
This looks like it could be realted to the Fresnel integral.
$$ \int_0^\infty \sin x^2 \, dx = \int_0^\infty \cos x^2 \, dx = \frac{\sqrt{2\pi}}{4} $$
