I am struggling to reproduce the following result. Given-
$$I_1 = \int_{y=-\infty}^\infty \int_{y'=-\infty}^\infty\int_{x=-\infty}^\infty \int_{x'=-\infty}^\infty dy dy' dx dx' \frac{n(x) n(x')}{[(x-x')^2 + (y - y')^2]^{3/2}}$$ and
$$I_2 = \int_{y=-\infty}^\infty \int_{y'=-\infty}^\infty\int_{x=-\infty}^\infty \int_{x'=-\infty}^\infty dy dy' dx dx' \frac{1}{[(x-x')^2 + (y - y')^2]^{3/2}},$$
where $n(x) = 1, -\infty < x \leq 0$ and $n(x) = -1$ for $0 \leq x < \infty$ I need to show- $$I_1 - I_2 = -4\int_{y=-\infty}^\infty \int_{y'=-\infty}^\infty\int_{x=-\infty}^0\int_{x'=0}^\infty dy dy' dx dx' \frac{1}{[(x-x')^2 + (y - y')^2]^{3/2}}.$$
I tried expressing $I_1$ as
$$I_1 = \int_{x=-\infty}^0\int_{x'=-\infty}^0 (+A) + \int_{x=-\infty}^0\int_{x'=0}^\infty (-A)+\int_{x=0}^\infty\int_{x'=-\infty}^0 (-A)+\int_{x=0}^\infty \int_{x'=0}^\infty (+A)=2\biggl[\int_{x=-\infty}^0\int_{x'=-\infty}^0 (+A) + \int_{x=-\infty}^0\int_{x'=0}^\infty (-A)\biggr] $$ and $I_2$ as $$ I_2 =4\int_{x=-\infty}^0\int_{x'=-\infty}^0 (+A),$$ where I absorbed the integrals over y and y' using the factor A, which can be calculated later. However, it does not look like I'm getting the correct result. It would be great if someone could help.
