Evaluate the double integral over the square from -R to R in both x and y of x divided by the square root of (R squared plus y squared) times (R squared minus x squared).

Step-by-step solution

1. 1

   Split the integrand into an x-part and a y-part. The square root factors, so $\frac{x}{\sqrt{(R^2+y^2)(R^2-x^2)}} = \frac{x}{\sqrt{R^2-x^2}}\cdot\frac{1}{\sqrt{R^2+y^2}}.$ Nothing mixes the two variables.

2. 2

   Use separability to split the double integral into a product. Over the rectangle $[-R,R]\times[-R,R]$, $I = \left(\int_{-R}^{R}\frac{x\,dx}{\sqrt{R^2-x^2}}\right)\left(\int_{-R}^{R}\frac{dy}{\sqrt{R^2+y^2}}\right).$ If either factor is zero, the whole integral is zero - so examine the simpler one first.

3. 3

   Check that the x-integral is improper, and that it converges. The integrand blows up as $x\to\pm R$, so convergence must be established before invoking symmetry. On $[0,R]$: $\int_0^{R}\frac{x\,dx}{\sqrt{R^2-x^2}} = \left[-\sqrt{R^2-x^2}\right]_0^{R} = 0+R = R,$ a finite value. The mirror half converges to $-R$ by the same computation.

4. 4

   Apply odd symmetry. $f(x)=\dfrac{x}{\sqrt{R^2-x^2}}$ satisfies $f(-x)=-f(x)$, and $[-R,R]$ is symmetric about $0$. Since both halves converge absolutely, $\int_{-R}^{R}f(x)\,dx = -R+R = 0.$

5. 5

   Conclude, noting the other factor is finite. $\displaystyle\int_{-R}^{R}\frac{dy}{\sqrt{R^2+y^2}} = 2\operatorname{arsinh}(1) = 2\ln\!\left(1+\sqrt2\right)$, a positive finite number. A finite number times zero is zero: $I = 0.$

6. 6

   Note why the convergence check mattered. Had the $x$-integral diverged (for example with $\tfrac{x}{R^2-x^2}$ instead), the two halves would be $+\infty$ and $-\infty$, and 'odd function on a symmetric interval' would not justify the value $0$ - the integral would simply fail to exist.