Monster double integral

Question

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 $$\iint_{\infty}^{-\infty}\tilde{A}\max(\exp(x_{M}-{x}')-\exp(z_{M}-{z}'), 0)\exp(\left ( {x}', {z}' \right )\hat{A}\left ( {x}', {z}' \right )^{T}+J\left ( {x}', {z}' \right )^{T})d{x}'d{z}'$$

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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 ...

Answer 1

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 $\;x_D,z_D,\varphi.\;$

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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 $(4).$

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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, $$I=\int\limits_{\arctan a}^{\pi+\arctan a} \int\limits_{\large \frac b{\sqrt{a^2+1}\cos(t-\arctan a)}}^\infty e^{\large -r^2} \, \text dr\, \text dt =\dfrac12 \int\limits_{0}^{\pi} e^{\left({\large \frac b{\sqrt{a^2+1}\cos t}}\right)^2} \, \text dt,$$ $$I= \dfrac\pi2 \operatorname{erfc}\left(\dfrac b{\sqrt{1+a^2}}\right).$$

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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.

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