A double integral over square

Question

Let $a>0$ and define $$F(a)=\int_0^a\int_0^a\int_0^a\int_0^a\ln^2\Big[(x-s)^2+(y-t)^2\Big]dx~dy~ds~dt.$$ Can $F$ be "explicitly'' computed? or perhaps be expressed in terms of well-known integrals?

Answer 1

Disclaimer: I've worked without checking some assumptions, and I've used Mathematica to verify my work.

I've done the following: we can interpret the integral as $$\mathbb{E} \left[ \log\left( (X-S)^2 + (Y-T) \right)^2\right]$$ where $X, S, Y, T$ follow a uniform distribution on $\left[0,1\right]$. We can then find the PDF of the term within the logarithm. (See bottom of post for the PDFs.)

For a = 1, I think (seems to check out numerically) it equals (brace yourself) $$ \frac{1}{18} (-96 C-88 \pi +415+4 \log 2) (12 \pi -31+3\log 2) $$ where $C \approx 0.915966$ is Catalan's constant.

Now, for the general case, we can substitute $ x' = \frac{x}{a}$ and analogously for the other variables, to get: $$ \begin{aligned} F(a) & =\int_0^a\int_0^a\int_0^a\int_0^a\log\left[(x-s)^2+(y-t)^2\right]^2\mathrm{d}x~\mathrm{d}y~\mathrm{d}s~\mathrm{d}t \\ & = \int_0^1\int_0^1\int_0^1\int_0^1 a^4 \log\left[(ax'-as')^2+(ay'-at')^2\right]^2\mathrm{d}x'~\mathrm{d}y'~\mathrm{d}s'~\mathrm{d}t' \\ & = 4a^4 \log(a)^2 + 4a^4 \log(a) \mathbb{E} \left[ \log\left( (X-S)^2 + (Y-T)^2 \right)\right] + a^4\mathbb{E} \left[ \log\left( (X-S)^2 + (Y-T) \right)^2\right] \\ & = 4a^4 \log(a)^2 + 4a^4 \log(a)\left(\frac{1}{6} (4 \pi -25+4\log 2)\right) + a^4 \left(\frac{1}{18} (-96 C-88 \pi +415+4 \log 2) (12 \pi -31+3\log 2))\right) \\ \end{aligned} $$

where I got the last equality from Mathematica.

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The PDFs:

I did this piece by piece: first I determined the PDF of the difference, which turned out to be $$ \begin{cases} 1+x & -1\leq x \leq 0 \\ 1-x & 0\leq x\leq 1 \\ 0 & \text{elsewise.} \end{cases} $$ The PDF of the squared term turns out to be $$ \begin{cases} \frac{1}{\sqrt{x}}-1 & 0 < x \leq 1 \\ 0 & \text{elsewise.} \end{cases} $$ and eventually, the PDF of the sum of the two squared terms is $$ f(x) = \begin{cases} \pi - 4\sqrt{x}+x & 0\leq x \leq 1 \\ -2-x+4 \sqrt{x-1}-2 \tan ^{-1}\left(\sqrt{x-1}\right)+2 \csc ^{-1}\left(\sqrt{x}\right) & 1\leq x \leq 2 \\ 0 & \text{elsewise.} \end{cases} $$ So, the integral for general $a$ should be equal to $$ 4a^4\log(a)^2 + 4a^4 \log(a) \int_0^2 \log(x) f(x)\, \mathrm{d}x+ a^4 \int_0^2 \log(x)^2 f(x)\, \mathrm{d}x $$ which Mathematica claims to be the value I stated earlier.

(Going to try and stop editing now, sorry if there's some errors/badly formatted things left.)