I am solving Laplace's equation $$\nabla^2 \Phi = 0$$ in the half space of $\mathbb{R}^3$ with $\Phi = f(y,z)$ at $x=0$.
As is given here http://www.damtp.cam.ac.uk/user/reh10/lectures/nst-mmii-handout10.pdf, the solution is $$\Phi(x,y,z) = \frac{x}{2\pi} \int_{\mathbb{R}^2} \frac{f(y',z')}{(x^2 + (y-y')^2 + (z-z')^2)^{3/2}} \; \textrm{d}y'\textrm{d}z'$$
However I want to calculate $\Phi_x|_{x=0}$, and I'm getting $$\Phi_x = \frac{1}{2\pi} \int_{\mathbb{R}^2} \frac{f(y',z') (-2x^2+(y-y')^2 + (z-z')^2)}{(x^2+(y-y')^2 + (z-z')^2)^{5/2}} \; \textrm{d}y'\textrm{d}z'$$ and $$\Phi_x|_{x=0} = \frac{1}{2\pi} \int_{\mathbb{R}^2} \frac{f(y',z')}{((y-y')^2 + (z-z')^2)^{3/2}} \; \textrm{d}y'\textrm{d}z'$$ but this doesn't make sense even in a principal value way since the denominator doesnt change sign! What am I doing wrong?
Thanks :)
It seems that you have only differentiated the $x$ before the integral. You have also to add a contribution coming from the differentiation of $x$ inside the integral. I haven't performed the calculation, but may be you get some cancellation.