# Line integral over two squares

I want to calculate the following integral

$$\oint\limits_{C_1} \oint\limits_{C_2} \frac{\vec {dl_1} \cdot \vec {dl_2}} r,$$ where

• The square paths $$C_1$$ and $$C_2$$ over which the integration should be carried are shown in this picture:

(the length of both sides of the squares is a given parameter $$a$$, and the orientations are the ones shown).

• $$r$$ represents the distance from each $$\vec {dl_1}$$ to $$\vec {dl_2}$$.

The approach I took in order to solve this problem is considering the total integral as the (finite) sum of integrals restricted to a single side from each square and performing the calculation of each of these integrals: obviously, the chosen sides being perpendicular to each other imply the vanishing of the dot product and thus zero contribution from that term to the total value of the integral.

My problem arises from the behaviour of $$r$$, due to its dependence on both $$\vec {dl_1}$$ and $$\vec {dl_2}$$. For example, the $$r$$ i get for the top of the left square and the bottom of the right square is: $$r = \sqrt{a^2 + (3a - x - y)^2}$$ where

• $$x$$ is the coordinate of the point $$\vec {dl_1}$$ on the left square and
• $$y$$ is the coordinate of $$\vec {dl_2}$$ on the right square (I would integrate from 0 to $$a$$ both times).

Can this issue be solved and consequently is my approach fruitful? And if it is not, is there something easier that I am missing?

First, I would make a change of variables $$l_i$$ to $$l_i/a$$. Second, just let the vary in the limits on the figure. So for top of the left square $$x$$ varies from $$1$$ to $$0$$, and the bottom right from $$2$$ to $$3$$. So you will need to calculate $$\int_1^0dx\int_2^3dy\frac1{\sqrt{1^2+(y-x)^2}}$$ This type of integrals can be done with hyperbolic substitution and integration by parts. Let start with integration over $$y$$, with $$z=y-x$$, then $$z=\sinh u$$, $$u=\mathrm{arcsinh} z$$, $$dz=\cosh u du$$ $$\int_{2+x}^{3+x}\frac{dz}{\sqrt{1+z^2}}=\int_{\mathrm{arcsinh}(2+x)}^{\mathrm{arcsinh}(3+x)}\frac{\cosh u du}{\sqrt{1+\sinh^2u}}=\mathrm{arcsinh}(3+x)-\mathrm{arcsinh}(2+x)$$ I've used $$1+\sinh^2u=\cosh^2u$$.
For the integration over $$x$$, after changing variables, you get integrals of the type $$\int \mathrm{arcsinh} u du=\int u'\mathrm{arcsinh} u du\\=u\mathrm{arcsinh} u-\int u(\mathrm{arcsinh} u)'du\\=u\mathrm{arcsinh} u-\int\frac u{\sqrt{u^2+1}}\\=u\mathrm{arcsinh} u-\sqrt{u^2+1}$$