Case 1: $B^{(0)}(\mathbf{x}) = (-x,0,z)$

 

Now there are two terms to be calculated. Firstly,

$$F_1(\mathbf{x}';\mathbf{x}) = \iiint (-x + x') \frac{\partial^2 G}{\partial x' \partial z'} \, dx' \, dy' \, dz'$$

$$= - \int x G \, dy' + \iint x' \frac{\partial G}{\partial x'} \, dx' \, dy'$$

$$= - \int x G \, dy' + \int x' G \, dy' - \iint G \, dx' \, dy'$$

$$= \frac{x - x'}{4\pi} \, \mathrm{sinh}^{-1}\left(\frac{y'}{\sqrt{{x'}^2 + {z'}^2}}\right)$$

$$\quad + \frac{1}{4\pi} \left( y' \, \mathrm{sinh}^{-1}\left(\frac... ...2 + {z'}^2}}\right) - z' \, \mathrm{atan}\left(\frac{x'y'}{z'r'}\right) \right)$$

$$= \frac{x}{4\pi} \, \mathrm{sinh}^{-1}\left(\frac{y'}{\sqrt{{x'}^2 ... ...z'}^2}}\right) - \frac{z'}{4\pi} \, \mathrm{atan}\left(\frac{x'y'}{z'r'}\right)$$ (32)

and secondly

$$F_2(\mathbf{x}';\mathbf{x}) = \iiint (z - z') \frac{\partial^2 G}{\partial {z'}^2} \, dx' \, dy' \, dz'$$

$$= z \iiint \frac{\partial^2 G}{\partial {z'}^2} \, dx' \, dy' \, dz' - \iiint z' \frac{\partial^2 G}{\partial {z'}^2} \, dx' \, dy' \, dz'$$

$$= \iint z \frac{\partial G}{\partial z'} \, dx' \, dy' - \iint z' \... ... z'} \, dx' \, dy' + \iiint \frac{\partial G}{\partial z'} \, dx' \, dy' \, dz'$$

$$= \frac{z}{4\pi} \mathrm{atan}\left(\frac{x'y'}{z'r'}\right) - \iint \frac{{z'}^2}{4\pi {r'}^3} \, dx' \, dy' + \iint G \, dx' \, dy'$$

$$= \frac{z}{4\pi} \mathrm{atan}\left(\frac{x'y'}{z'r'}\right) - \frac{z'}{4\pi} \mathrm{atan}\left(\frac{x'y'}{z'r'}\right)$$

$$\quad - \frac{1}{4\pi} \left( y' \, \mathrm{sinh}^{-1}\left(\frac... ...2 + {z'}^2}}\right) - z' \, \mathrm{atan}\left(\frac{x'y'}{z'r'}\right) \right)$$

$$= \frac{z}{4\pi} \mathrm{atan}\left(\frac{x'y'}{z'r'}\right) - \fra... ...c{x'}{4\pi} \, \mathrm{sinh}^{-1}\left(\frac{y'}{\sqrt{{x'}^2 + {z'}^2}}\right)$$ (33)

Due to the linearity of equations 14 and 21 these two terms can be combined at this stage to give

$$F(\mathbf{x}';\mathbf{x}) = F_1(\mathbf{x}';\mathbf{x}) + F_2(\mathbf{x}';\mathbf{x})$$

$$= \frac{x-x'}{4\pi} \, \mathrm{sinh}^{-1}\left(\frac{y'}{\sqrt{{x'}... ...{z'}^2}}\right) + \frac{z-z'}{4\pi} \mathrm{atan}\left(\frac{x'y'}{z'r'}\right)$$ (34)

which can be substituted directly to give $B^{(1)}_z$.

Note that for $F_1$ and $F_2$ here both primed and unprimed coordinates appear, whereas for the constant fields the unprimed coordinates did not appear in the expressions for $F$.