Monster double integral Are there methods to analytically compute this integral, composed of the maximum function, $max(\exp(x_{0})-\exp(z_{0}), 0)$, and exponential functions?
$$\iint_{-\infty}^{\infty}\tilde{A}\max(\exp(x_{0})-\exp(z_{0}), 0)\exp(A_{2}(x_{M}-x_{0})^2)\exp(2A_{2}\alpha_{1}(x_{M}-x_{0}))\exp(-2A_{2}\Omega(x_{M}-x_{0}+\alpha_{1})(z_{M}-z_{0}))\exp(A_{2}\Omega^2(z_{M}-z_{0})^{2})\exp(A_{5}(z_{M}-z_{0}))\exp(A_{6}(z_{M}-z_{0})^{2})dx_{0}dz_{0}$$
$\tilde{A},A_{2},\alpha_{1},\Omega,A_{5},A_{6},x_{M},z_{M}$ are constants.
Specifically, $\tilde{A},A_{2},A_{5},A_{6}$ are negative, while the others take on any real number.
So now let $x_{M}-x_{0}= {x}'$ and $z_{M}-z_{0}= {z}'$.
We transform the integral to:
$$\iint_{\infty}^{-\infty}\tilde{A}\max(\exp(x_{M}-{x}')-\exp(z_{M}-{z}'), 0)\exp(A_{2}({x}')^2)\exp(2A_{2}\alpha_{1}({x}'))\exp(-2A_{2}\Omega({x}')({z}'))\exp(A_{2}\Omega^2({z}')^{2})\exp((A_{5}-2A_{2}\Omega\alpha_{1})({z}'))\exp(A_{6}({z}')^{2})d{x}'d{z}'$$
where the limits have changed because of the substitution.
The exponent can now elegantly be written as:
$$\left ( {x}', {z}' \right )\begin{pmatrix} A_{2} & -A_{2}\Omega \\ -A_{2}\Omega & A_{6}+A_{2}\Omega^2 \end{pmatrix}\left ( {x}', {z}' \right )^{T} +\left ( 2A_{2}\alpha_{1},(A_{5}-2A_{2}\Omega\alpha_{1}) \right )\left ( {x}', {z}' \right )^{T}$$
and by letting $ \hat{A}=\begin{pmatrix} A_{2} & -A_{2}\Omega \\ -A_{2}\Omega & A_{6}+A_{2}\Omega^2 \end{pmatrix}$ and $ J=\left ( 2A_{2}\alpha_{1},(A_{5}-2A_{2}\Omega\alpha_{1})\right )$
The integral becomes
$$\iint_{\infty}^{-\infty}\tilde{A}\max(\exp(x_{M}-{x}')-\exp(z_{M}-{z}'), 0)\exp(\left ( {x}', {z}' \right )\hat{A}\left ( {x}', {z}' \right )^{T}+J\left ( {x}', {z}' \right )^{T})d{x}'d{z}'$$

I have tried doing a simpler integral:
$\iint_{-\infty}^{\infty}\tilde{A}\max(\exp(x_{0})-\exp(z_{0}), 0)\exp(A_{1}(x_{M}-x_{0}-\alpha_{1})^2)\exp(A_{6}(z_{M}-z_{0}-\alpha_{2})^{2})dx_{0}dz_{0}$
but still can not reach a closed-form solution involving error function, dilogarithms, etc ...
 A: HINT.
The first integral can be presented in the form of
$$I=\iint\limits_{-\infty}^{\infty}\tilde{A}\max\left(e^{\large x_0}-e^{\large z_0}, 0\right)e^{\large P_{20}(\vec v_0,x_0,z_0)}\,\text dz_0\,\text dx_0,$$
where
$$\vec v_0 = \{x_M,z_M,A_2,A_5,A_6,\alpha_1,\Omega\},$$
$$P_{20}(\vec v_0, x_0, z_0) = A_{2}\bigg((x_{M}-x_{0})^2+2\alpha_{1}(x_{M}-x_{0})-2\Omega(x_{M}-x_{0})(z_{M}-z_{0}))$$
$$+\Omega^2(z_{M}-z_{0})^{2}\bigg)
+(A_{5}-2\Omega\alpha_1)(z_{M}-z_{0})+A_{6}(z_{M}-z_{0})^{2},$$
or
$$I=\iint\limits_{-\infty}^{\infty}\tilde{A}\max\left(e^{\large x_C+x_M}-e^{\large z_C+z_M
}, 0\right)e^{\large P_{2C}(\vec v_C,x_C,z_C)}\,\text dz_C\,\text dx_C,\tag1$$
where
\begin{cases}
\vec v_C = \{A_2,A_5,A_6,\alpha_1,\Omega\}\\[4pt]
P_{2C}(\vec v_C, x_C, z_C) = A_{2}\big(x_C^2-2\alpha_1x_C-2\Omega x_C z_C+\Omega^2z_C^2\big)\\[4pt]
\qquad\qquad\qquad\;-(A_{5}-2\Omega\alpha_1)z_C+A_{6}z_C^2.\tag2
\end{cases}
Transformation of the coordinates in the form of
$$x_C+iz_C = (x_R+iz_R)e^{\large i\varphi} + x_D +iz_D\tag3$$
should lead from $(2)$ to the expression in the form of
$$P_{2R}(\vec p, x_R, z_R)= p_{xx} x_R^2+p_{zz} z_R^2+p_0.\tag4$$
This allows to get a system of equations for unknowns $\;x_D,z_D,\varphi.\;$

Really, from $(3)$ should

*

*$$\dbinom{x_C}{z_C}=\dbinom{x_R\cos\varphi-z_R\sin\varphi+x_D}
{x_R\sin\varphi+z_R\cos\varphi+z_D},$$

*$$P_{2C}(\vec v_C, x_C, z_C) = p_{xx}x_R^2+p_{xz}x_Rz_R+p_{zz}z_R^2+p_x x_R+p_z z_R+p_0,$$
where $p_{xx},p_{xz},p_{zz},p_{x},p_{z},p_{0}$ are functions of $v_C$.

Obtaining $x_D,z_D,\varphi$ from the system $p_{xz}=p_x=p_z=0,$ we should get $(4).$

At the same time, is known closed form of the integral
$$I=\int\limits_{-\infty}^\infty \int\limits_{ax+b}^\infty e^{\large -x^2-z^2} \, \text dx\, \text dz = \dfrac\pi2 \operatorname{erfc}\left(\dfrac b{\sqrt{1+a^2}}\right)\tag5$$
(pointed by DinosaurEgg), and this allows to get the closed form of the given integral.

Let $x=r\cos t,\,z=r \sin t.$
If $\;a>0,\,b>0,\;$ then the angle coordinate corresponds to line $\;z=ax+b,\;$ when $\;t\in(\arctan a,\pi+\arctan a).$
The polar radius can be obtained from the equation $\;a r \cos t +b = r \sin t,\;$
with the result $r=\dfrac b{|\sin t - a\cos t|}=\dfrac b{\sqrt{a^2+1}\cos(t-\arctan a)}.$
Therefore,
$$I=\int\limits_{\arctan a}^{\pi+\arctan a} \int\limits_{\large \frac b{\sqrt{a^2+1}\cos(t-\arctan a)}}^\infty e^{\large -r^2} \, \text dr\, \text dt 
=\dfrac12 \int\limits_{0}^{\pi} e^{\left({\large \frac b{\sqrt{a^2+1}\cos t}}\right)^2} \, \text dt,$$
$$I= \dfrac\pi2 \operatorname{erfc}\left(\dfrac b{\sqrt{1+a^2}}\right).$$


About the factor with $\;\max\;$ function:

*

*Equation $(3)$ defines a linear transformation.

*Exponential function is monotonic.

*$\max\left(e^{\large x_C+x_M}-e^{\large z_C+z_M}, 0\right)=\left(e^{\large x_C+x_M}-e^{\large z_C+z_M}\right)h\left(x_C+x_M-z_C-z_M\right).$

*$\;I=I^\star - I^\diamond,\;$ where
$$I^\star=\iint\limits_{-\infty}^{\infty}\tilde{A}h\left(x_C+x_M-z_C-z_M\right)e^{x_C+x_M+\large P_{2C}(\vec v_C,x_C,z_C)}\,\text dz_C\,\text dx_C,$$
$$I^\diamond =\iint\limits_{-\infty}^{\infty}\tilde{A}h\left(x_C+x_M-z_C-z_M\right)e^{z_C+z_M+\large P_{2C}(\vec v_C,x_C,z_C)}\,\text dz_C\,\text dx_C,$$
$\hspace{100mu}h(t)=\begin{cases}
1,\; \text{if} \;t>0\\
0,\; \text{otherwize}
\end{cases}\quad$ is the Heaviside step function.

Therefore, we have $\;P_{2C}^\star =P_{2C}+x_C+x_M,\;P_{2C}^\diamond =P_{2C}+z_C+z_M,\;$ with the same transformation technique.
On the other hand:

*

*Equation $(4)$ can define ellipse, parabola or hyperbola.

*Integral $(1)$ diverges in the hyperbolic case.

*Integral $(1)$ can be easily calculated in the parabolic case.

*Integral $(1)$ can be calculated via $(5)$ in the elliplic case.


