Page:Philosophical Transactions of the Royal Society A - Volume 184.djvu/46

46 Similarly

$\sin n\phi\int_0^\pi\frac{\cos n\phi\;d\phi}{\sqrt{\left(z^2 + c^2 - 2c\varpi\cos\phi + \varpi^2\right)}}$|undefined

is a solution.

Let $$\mathrm{V}$$ stand for either of these integrals. Then

$\frac{d^{p + q}\mathrm{V}}{dc^p\;dz^q}$|undefined

is also a solution of 's equation.

For different values of $$p$$ and $$q$$ the solutions are not all independent: as it is easily seen that

$\int_0^\pi\frac{\cos\phi\;d\phi}{\sqrt{\left(z^2 + c^2 - 2c\varpi\cos\phi + \varpi^2\right)}}$|undefined

satisfies a linear partial differential equation of the second order in $$c$$ and $$z$$.

But two independent sets of solutions are obtained by giving different values to $$p$$ in $$\frac{d^p\mathrm{V}}{dc^p}$$ and $$\left(\frac{d}{dc}\right)^p\cdot\left(\frac{d\mathrm{V}}{dz}\right)\cdot$$

The only cases considered in this paper are when $$n = 0$$ or $$n = 1$$. That is the solutions of the forms

$\left(\frac{d}{dc}\right)^p\int_0^\pi\frac{d\phi}{\sqrt{\left(z^2 + c^2 - 2c\varpi\cos\phi + \varpi^2\right)}}$;|undefined

$\left(\frac{d}{dc}\right)^p\frac{d}{dz}\int_0^\pi\frac{d\phi}{\sqrt{\left(z^2 + c^2 - 2c\varpi\cos\phi + \varpi^2\right)}}$;|undefined

$\cos\phi\left(\frac{d}{dc}\right)^p\int_0^\pi\frac{\cos\phi\;d\phi}{\sqrt{\left(z^2 + c^2 - 2c\varpi\cos\phi + \varpi^2\right)}}$;|undefined

and

$\cos\phi\left(\frac{d}{dc}\right)^p\frac{d}{dz}\int_0^\pi\frac{\cos\phi\;d\phi}{\sqrt{\left(z^2 + c^2 - 2c\varpi\cos\phi + \varpi^2\right)}}$;|undefined

as these cover all the cases to which the functions are applied in this paper.

It is easily seen that

$\varpi\int_0^\pi\frac{\cos\phi\;d\phi}{\sqrt{\left(z^2 + c^2 - 2c\varpi\cos\phi + \varpi^2\right)}}$|undefined

and, consequently,

$\varpi\left(\frac{d}{dc}\right)^p\int_0^\pi\frac{\cos\phi\;d\phi}{\sqrt{\left(z^2 + c^2 - 2c\varpi\cos\phi + \varpi^2\right)}}$|undefined