Calculate $\frac{1}{H^2W^2}\int_0^H{\int_0^W{\int_0^H{\int_0^W{e^{-\alpha \sqrt{\left( x-z \right) ^2+\left( y-t \right) ^2}}}}dxdydzdt}}$

Question

I am stucked by the following problem in part of my current research process. Could you please help me?

(You can first jump to the Background part to see what I am calculating for.)

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Problem:

Given three constant real positive numbers $H,W,\alpha \in \mathbb{R}^+$, calculate the following quadruple integral:

$$ I(H,W,\alpha)=\frac{1}{H^2W^2}\int_0^H{\int_0^W{\int_0^H{\int_0^W{e^{-\alpha \sqrt{\left( x-z \right) ^2+\left( y-t \right) ^2}}}}dxdydzdt}}. $$

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Some of My Efforts:

The original problem is too difficult for me.

I just want the accurate answer and do not have to require the complete analysing process.

So I first tried to get the answer from WolframAlpha. But it seems to be failed to calculate the integral.

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Background:

I was trying to calculate the average "correlation" value of a two-dimensional area $[0,H]\times[0,W]$. In my task, the "correlation" $\rho$ is modeled as:

$$ \rho(d)=e^{-\alpha d}, $$

where $d\in \mathbb{R}^+$ is the distance of two points.

For a given point $(z,t)$ satisfying $0\le z\le H$ and $0\le t \le W$, its average "correlation" is:

$$ f(z,t)=\frac{1}{HW}\int_0^H{\int_0^W{\rho \left( \sqrt{\left( x-z \right) ^2+\left( y-t \right) ^2} \right)}dxdy}. $$

And for the whole area, the average "correlation" is:

$$ I=\frac{1}{HW}\int_0^H{\int_0^W{f\left( z,t \right)}dzdt}. $$

Finally I get the formula in the original problem, resulting in a quadruple integral.

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Some Other Aspects of Help That You May Provide:

I think that the original problem is too difficult. There may be some other operations helpful for me:

1. I can change the modeling of $\rho$. I model it from my observed data (see the figure below). Since when $d=0$, there is $\rho=1$ (the value of each point is strongly correlated to itself), and $\rho$ is approximately decreasing but always positive in $[0,+\infty)$. I guess the formula could be $\rho(d)=e^{-\alpha d}$. Is there a better modeling?

(The horizontal axis corresponds to $d$, and the vertical axis corresponds to $\rho$. The values of $\rho$ are more reliable when $0\le d\le 300$, since the data sample number of $0\le d\le 300$ are sufficiently large and number of $d>300$ are relatively small.)

2. I can change the calculation way of "average". Since the integration may be too difficult to calculate, I can use another way to measure the "average" of the area. Is there a better way? I try not to modify the original integral form, since it is easy to interpret and understand.

3. Instead of directly calculating $I(H,W)$, I can also change to calculate $I(H_1,W_1)-I(H_2,W_2)$, since I care the difference of "average correlations" of two areas. However, the difference seems to be more difficult to calculate.

Since $\rho$ is decreasing in $[0,+\infty)$, I intuitively think that a small area will own a larger "average correlation", i.e., $I(H_1,W_1)<I(H_2,W_2)$ when $H_1\times W_1 > H_2\times W_2$.

As you can see, I just have a vague and qualitative cognition of my above measuring and modeling now. So many things can be adjusted to make the problem easier. (I am now looking for them.)

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Could you please help me based on all the above descriptions?

(I would like to add more details if the statements are not clear enough.)

Answer 1

You can make the computation significantly easier if you change the correlation strength to

$$\rho(d)=e^{-ad^2}$$

which shouldn't affect the results qualitatively. This choice of function comes with the added advantage that the integrals over the coordinates factorize:

$$I=\frac{1}{H^2}\left(\int_0^H dx\int_0^H dz e~^{-a(x-z)^2}\right)\frac{1}{W^2}\left(\int_0^W dy\int_0^W dt e~^{-a(y-t)^2}\right):=F(W)F(H)$$

which according to Mathematica, evaluates to:

$$F(H)=\sqrt{\frac{\pi}{aH^2}}\text{erf}(H\sqrt{a})+\frac{e^{-aH^2}-1}{aH^2}$$

Answer 2

Too Long For Comment:

Lemma: For any integrable function $f$, we have

$$\begin{align} \int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,f{\left((x-y)^{2}\right)} &=2\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,f{\left((x-y)^{2}\right)};~~~\small{symmetry}\\ &=2\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}u\,f{\left(u^{2}\right)};~~~\small{\left[y=x-u\right]}\\ &=2\int_{0}^{1}\mathrm{d}u\int_{u}^{1}\mathrm{d}x\,f{\left(u^{2}\right)}\\ &=2\int_{0}^{1}\mathrm{d}u\,\left(1-u\right)f{\left(u^{2}\right)}.\\ \end{align}$$

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Suppose $\left(H,W,\alpha\right)\in\mathbb{R}_{>0}^{3}$. Applying the above lemma to $\mathcal{I}$, we find

$$\begin{align} \mathcal{I}{\left(H,W,\alpha\right)} &=\frac{1}{H^{2}W^{2}}\int_{0}^{H}\mathrm{d}x\int_{0}^{W}\mathrm{d}y\int_{0}^{H}\mathrm{d}z\int_{0}^{W}\mathrm{d}t\,e^{-\alpha\sqrt{(x-z)^{2}+(y-t)^{2}}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}z\int_{0}^{1}\mathrm{d}t\,e^{-\alpha\sqrt{H^{2}(x-z)^{2}+W^{2}(y-t)^{2}}};~~~\small{(x,y,z,t)\mapsto(Hx,Wy,Hz,Wt)}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}z\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}t\,e^{-\alpha\sqrt{H^{2}(x-z)^{2}+W^{2}(y-t)^{2}}}\\ &=4\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\left(1-u\right)\left(1-v\right)e^{-\alpha\sqrt{H^{2}u^{2}+W^{2}v^{2}}}.\\ \end{align}$$

We can further reduce the integral as follows:

$$\begin{align} \mathcal{I}{\left(H,W,\alpha\right)} &=4\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\left(1-u\right)\left(1-v\right)e^{-\alpha\sqrt{H^{2}u^{2}+W^{2}v^{2}}}\\ &=4\int_{0}^{1}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\left(1-u\right)\left(1-v\right)e^{-\alpha\sqrt{H^{2}u^{2}+W^{2}v^{2}}}\\ &~~~~~+4\int_{0}^{1}\mathrm{d}u\int_{u}^{1}\mathrm{d}v\,\left(1-u\right)\left(1-v\right)e^{-\alpha\sqrt{H^{2}u^{2}+W^{2}v^{2}}}\\ &=4\int_{0}^{1}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\left(1-u\right)\left(1-v\right)e^{-\alpha\sqrt{H^{2}u^{2}+W^{2}v^{2}}}\\ &~~~~~+4\int_{0}^{1}\mathrm{d}v\int_{v}^{1}\mathrm{d}u\,\left(1-v\right)\left(1-u\right)e^{-\alpha\sqrt{H^{2}v^{2}+W^{2}u^{2}}};~~~\small{(u,v)\mapsto(v,u)}\\ &=4\int_{0}^{1}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\left(1-u\right)\left(1-v\right)e^{-\alpha\sqrt{H^{2}u^{2}+W^{2}v^{2}}}\\ &~~~~~+4\int_{0}^{1}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\left(1-v\right)\left(1-u\right)e^{-\alpha\sqrt{H^{2}v^{2}+W^{2}u^{2}}}\\ &=4\int_{0}^{1}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\left(1-u\right)\left(1-v\right)\left[e^{-\alpha\sqrt{H^{2}u^{2}+W^{2}v^{2}}}+e^{-\alpha\sqrt{W^{2}u^{2}+H^{2}v^{2}}}\right]\\ &=4\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}x\,u\left(1-u\right)\left(1-xu\right)\left[e^{-\alpha\sqrt{H^{2}u^{2}+W^{2}x^{2}u^{2}}}+e^{-\alpha\sqrt{W^{2}u^{2}+H^{2}x^{2}u^{2}}}\right];~~~\small{\left[v=xu\right]}\\ &=4\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}u\,u\left(1-u\right)\left(1-xu\right)\left[e^{-u\alpha\sqrt{H^{2}+W^{2}x^{2}}}+e^{-u\alpha\sqrt{W^{2}+H^{2}x^{2}}}\right].\\ \end{align}$$

The integral over $u$ in the last line above is elementary, so in principle we have successfully reduced $\mathcal{I}$ to a single variable integral. Unfortunately, this integral for $u$ is quite cumbersome and doesn't lead to an integral over $x$ that is any easier, so I won't bother reproducing it here. It could help with numerical procedures though.

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