The electric field at some point $(x,y,z)$ generated by a ring of radius $R$ and uniform linear density $\lambda$ is
$$\mathbf E=\frac{\lambda R}{4\pi\varepsilon_0}\int_0^{2\pi} \frac{[(x-R\cos\phi)\mathbf{\hat x}+(y-R\sin\phi)\mathbf{\hat y}+z\mathbf{\hat z}]\text d\phi}{[x^2+y^2+z^2+R^2-2R(x\cos\phi+y\sin\phi)]^{\frac{3}{2}}}.$$
Notice how the function $f(\cos \phi,\sin \phi)$ in the title would be $\cos\phi$, $\sin\phi$ or $1$ depending on each of the main types of integrals above.
So what I was wondering was whether there is some way to solve those integrals: perhaps some special function (maybe elliptical?)? Or maybe series expansion of root in denominator followed by binomial expansion? What about residue theorem doing the substitution $z=e^{i\phi}$?, etc.
Best Answer
The angle sum identity gives that $B \cos x + C \sin x = \sqrt{B^2 + C^2} \cos (x - x_0)$ for some $x_0$, and using perioidicity and double-angle identities gives that the substitution $u := x - x_0$ transforms the integral to one of the form (henceforth provided that $A^2 > B^2 + C^2$) $$\int_{-\pi}^\pi \frac{(a + b \sin u + c \cos u) \,du}{(1 - \lambda \cos u)^{3/2}}, \qquad \lambda := \frac{\sqrt{B^2 + C^2}}{|A|} \in [0, 1) .$$ The denominator is even, so $$\int_{-\pi}^\pi \frac{\sin u \,\,du}{(1 - \lambda \cos u)^{3/2}} = 0 .$$
Applying the tangent half-angle substitution, $u = 2 \arctan t$, yields $$\int_{-\pi}^\pi \frac{du}{(1 - \lambda \cos u)^{3/2}} = 2 \int_{-\infty}^\infty \frac{1}{(1 - \lambda) + (1 + \lambda) t^2} \sqrt{\frac{1 + t^2}{(1 - \lambda) + (1 + \lambda) t^2}} \,dt , $$ which can be written in the form $$\int_{-\infty}^\infty q(t) \sqrt{\frac{1 + t^2}{1 + \mu^2 t^2}} \,dt$$ for some rational function $q$ and constant $\mu$, which is already close to a [Legendre normal form for the complete elliptic integral function $E$ of the second kind][1]. Asking Maple to handle the details gives $$\boxed{\int_{-\pi}^\pi \frac{du}{(1 - \lambda \cos u)^{3/2}} = \frac{4}{(1 - \lambda) \sqrt{1 + \lambda}} E \left(\sqrt{\frac{2 \lambda}{1 + \lambda}}\right)} .$$
It is not too difficult to compute the asymptotic behavior of the integral in $\lambda$: When $\lambda \ll 1$, $$\int_{-\pi}^\pi \frac{du}{(1 - \lambda \cos u)^{3/2}} = 2 \pi \left(1 + \frac{15}{16} \lambda^2 + R_1(\lambda)\right) , \qquad R_1(\lambda) \in O (\lambda^4)$$
When $1 - \lambda \ll 1$, \begin{multline*}\int_{-\pi}^\pi \frac{du}{(1 - \lambda \cos u)^{3/2}} = 2 \sqrt 2 \left[ \frac1{1 - \lambda} - \frac18 \log(1 - \lambda) + \frac18 (1 + 5 \log 2) + R_2(\lambda) \right], \\ R_2(\lambda) \in O((1 - \lambda) \log (1 - \lambda)) . \end{multline*}
A similar computation gives $$\boxed{\int_{-\pi}^\pi \frac{\cos u \, du}{(1 - \lambda \cos u)^{3/2}} = \frac{4}{\lambda \sqrt{1 + \lambda}} \left[\frac{1}{1 - \lambda} E \left(\sqrt{\frac{2 \lambda}{1 + \lambda}}\right) - K \left(\sqrt{\frac{2 \lambda}{1 + \lambda}}\right) \right]} ,$$ where $K$ is the complete elliptic integral of the first kind. When $\lambda \ll 1$, $$\int_{-\pi}^\pi \frac{\cos u \,du}{(1 - \lambda \cos u)^{3/2}} = \frac{3 \pi}{2} \lambda \left(1 + \frac{35}{32} \lambda^2 + R_3(\lambda) \right), \qquad R_3(\lambda) \in O(\lambda^4).$$
When $1 - \lambda \ll 1$, \begin{multline*}\int_{-\pi}^\pi \frac{\cos u \,du}{(1 - \lambda \cos u)^{3/2}} = 2 \sqrt 2 \left[ \frac1{1 - \lambda} + \frac38 \log(1 - \lambda) + \frac18 (9 - 15 \log 2) + R_2(\lambda) \right], \\ R_4(\lambda) \in O((1 - \lambda) \log (1 - \lambda)) . \end{multline*} [1]: https://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of_the_second_kind