How to evaluate the integral $\int_{-r/2}^{r/2} \int_{-r/2}^{r/2} \frac{1}{x^2+y^2+r^2/4} dx dy$

Question

I came across this integral while trying to evaluate the electrical force exerted by a charged plate in the form of a square with side length $r$. I tried the usual method of first keeping $y$ constant and integrating with respect to $x$:

$$\int_{-r/2}^{r/2} \int_{-r/2}^{r/2} \frac{1}{x^2+y^2+\frac{r^2}{4}} dx dy = \int_{-r/2}^{r/2} \frac{\arctan\left(\frac{r}{\sqrt{r^2+4y^2}}\right)}{\sqrt{r^2+4y^2}} dy $$

However, I have no idea how to proceed from there. Even WolframAlpha times out. I'm suspecting this can be solved by some sort of coordinate change.

Answer 1

$$\int_{-r/2}^{r/2}\int_{-r/2}^{r/2}\frac{1}{x^{2}+y^{2}+\frac{r^{2}}{4}}dxdy$$ $$x\mapsto \frac{r}{2} x\quad y\mapsto \frac{r}{2} y$$ $$\int_{-1}^{1}\int_{-1}^{1}\frac{r^{2}}{r^{2}x^{2}+r^{2}y^{2}+r^{2}}dxdy$$ $$\int_{-1}^{1}\int_{-1}^{1}\frac{1}{x^{2}+y^{2}+1}dxdy$$ $$4\int_{0}^{1}\int_{0}^{1}\frac{1}{x^{2}+y^{2}+1}dxdy$$ $$8\int_{0}^{1}\frac{\arctan\left(\frac{y}{\sqrt{y^{2}+1}}\right)}{\sqrt{y^{2}+1}}dy$$ $$y=\sinh\left(t\right)\quad dy=\cosh\left(t\right)\quad t=\operatorname{arcsinh}\left(y\right)$$ $$8\int_{0}^{\operatorname{arcsinh}\left(1\right)}\arctan\left(\tanh\left(t\right)\right)dt$$

$$\int \arctan(\tanh(x))dx$$

$$2\pi\cdot \text{asinh}(1)-4 C +4\cdot\Im\left[\text{Li}_2\left((3-2\sqrt{2})i\right)\right]$$

Where:

• $C$ is the Catalan constant
• $\Im$ is the imaginary part
• $\text{Li}_{\nu}(z)$ is the polylogarithm function

In particular (Let $w:=3-2\sqrt{2}$ for brevity): $$\Im\left[\text{Li}_2\left(w i\right)\right]=\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{w^{2k+1}}{\left(2k+1\right)^{2}}=w\cdot{}_{3}F_2\left(\left.{\frac{1}{2},\frac{1}{2},1\atop\frac{3}{2},\frac{3}{2}}\right|-w^2\right)$$

Where:

• $\displaystyle{}_{3}F_2\left(\left.{a_1,a_2,a_3\atop b_1,b_2}\right|z\right)$ is the hypergeometric function

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In general:

$$\int_{-r}^{r}\int_{-r}^{r}\frac{1}{x^{2}+y^{2}+\frac{r^{2}}{a^{2}}}dxdy=-4C+4\cdot\Im\left[\text{Li}_2\left(e^{-2\text{arcsinh}(a)}i\right)\right]+2\pi\cdot\text{arcsinh}\left(a\right)$$