Gravity of a Circular Ring at a co-planar external point. I have a circular ring of unit mass and fixed radius R which lies in the XY plane at point $O$ with coordinates $O:(0,0)$.
I wish to find a formula for the gravitational force at a point $P: (D,0)$ which lies in the same plane as the ring and is at some variable distance D from the ring centre O.
(Note: There are many treatments of the case for a target lying on the axis of the ring.  The  reference so far found nearest to this co-planar case is Problems 5-12, 5-13, (no solutions given) p.127 in Classical Dynamics of Particles and Systems by Jerry B. Marion.
I expect that the formula should be of the form $F = GM*f(D)$ where $G$ is the gravitational constant, $M$ is the mass   and $f$ is some function similar to the Newtonian spherical divergence function $f(D) = \frac{1}{ D^2}$ (where the factor $\frac{1}{4.\pi}$ is absorbed in the value of the constant $G$ ).
So far I have obtained an integral formula by initially modelling the ring as a series of $N$ small point masses of mass $\frac{1}{N}$ separated by angle $\delta\theta$, whose distance from target is $L$ where: 
$$L^2 = (D-a)^2+b^2 = D^2-2aD+R^2 = D^2\left(1 -\frac{2a}{D} +\frac{R^2}{D^2}\right)$$ 
where $a (= R\cos\theta)$ and $b(=R\sin\theta)$ are the $x$ and $y$ coordinates of the point.  
Due to symmetry and vector addition of forces there is no net force in the y-direction and so the effective force contribution (along $x$) for a point is given by multiplying by the cosine factor $(D-a)/L$ thus:-
$$ F = \frac{-GM}{N}\frac{1}{4\pi.L^2}\frac{D-a}{L} 
= \frac{-GM}{ N}  \frac{D-a}{L^3} $$
$$ F = \frac{-GM}{ N}  \frac{D-R\cos\theta}{\left(D^2\left(1 -\frac{2a}{D} +\frac{R^2}{D^2}\right)\right)^{\frac{3}{2}}} $$
$$ F = \frac{-GM}{ N}  \frac{D-R\cos\theta}{D^3 \left(1 -\frac{2a}{D} +\frac{R^2}{D^2} \right)^{\frac{3}{2}}} $$
$$ F = \frac{-GM}{ N}  \frac{1-(R/D)\cos\theta}{D^2 \left(1 -\frac{2a}{D} +\frac{R^2}{D^2} \right)^{\frac{3}{2}}} $$
I then obtained the following integral formula for the force exerted on the target point by the ring:-
$$ F = \frac{-GM}{ D^2} \frac{1}{2\pi}\int_0^{2\pi}\frac{1-Q\cos\theta}{\left(1-2Q\cos\theta+Q^2\right)^{\frac{3}{2}}} \text{d}\theta$$
where $Q = R/D$.
$$ F = \frac{-GM}{ D^2} \frac{1}{2\pi}  \frac{1}{(2Q)^{3/2}}\int_0^{2\pi}\frac{1-Q\cos\theta}
{\left(\frac{Q^2+ 1}{2Q} - \cos\theta\right)^{\frac{3}{2}}} \text{d}\theta$$
Defining $A = \frac{Q^2+ 1}{2Q}$, Wolfram Alpha gives...
$$ \int_0^{2\pi}\frac{ 1 - Q \cos x}{(A -\cos x)^{3/2}} dx $$
$$=\left[\frac{2}{(A^2-1)\sqrt{A - \cos x}}\left(A^2-1\right)Q\sqrt{\frac{A - \cos x}{A-1}} \operatorname{F}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)-AQ\sin x- (A-1)(AQ-1)\sqrt{\frac{A-\cos x}{A-1}}\operatorname{E}\left(\frac{x}{2}~\big|~\frac{2}{1-A}\right)
+\sin x\right]_0^{2\pi}$$
Where $E(x|m)$ is an elliptic integral of the 2nd kind with parameter $m=k^2$, and $F(x|m)$ is an elliptic integral of the 1st kind with parameter $m=k^2$.
Replacing $\cos x$ by $1$ and $\sin x$ by $0$...
$$=\frac{2}{(A^2-1)\sqrt{A -1}}*\left[(A^2-1)Q \operatorname{F}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)-(A-1)(AQ-1) \operatorname{E}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)\right]_0^{2\pi}$$
Cancelling $(A^2-1)$...
$$=\frac{2}{\sqrt{A -1}}\left[Q\operatorname{F}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right) 
- \frac{(AQ-1)}{A+1} \operatorname{E}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)  \right]_0^{2\pi}$$
Being unfamiliar with Elliptic Integrals, this is as far as I can comfortably go at present.

After reading the wikipedia article Elliptic Integral, proceeding tentatively, from the definitions of elliptic integrals I think that $E(x|k^2)$ and $F(x|k^2)$ both go to zero when $x$ is zero, thus...
$$=\frac{2Q}{\sqrt{A -1}}\left[\operatorname{F}\left(\pi~\big|~\frac{-2}{A-1}\right)-\frac{(AQ-1)}{AQ+Q} \operatorname{E}\left(\pi~\big|~\frac{-2}{A-1}\right)\right]$$
Next perhaps it would be helpful to reformulate the problem so that the amplitude(?) term in the elliptic integrals changes from $\pi$ to $\pi/2$, thereby making the elliptic integrals "complete" and permitting them to be expressed as power series.  This reformulation could be done by modelling the gravitational effect ($Fx$ component only) of two half rings (positive $y$ and negative $y$), independently, and using respectively the angles $\theta_1$ and $\theta_2$ which both range from $0$ to $\pi/2$ but in different directions.

In order to express the result in terms of complete elliptic integrals, it is easier to compute the gravitational potential $\phi(D)$ first. Then the (radial) field is given by $F(D) = - \phi'(D)$. Following your approach we find
$$ \phi(D) = - \frac{G M}{2 \pi D} \int \limits_0^{2\pi} \frac{\mathrm{d} \theta}{\sqrt{1 - 2 Q \cos(\theta) + Q^2}} = - \frac{G M}{\pi D} \int \limits_0^{\pi} \frac{\mathrm{d} \theta}{\sqrt{1 - 2 Q \cos(\theta) + Q^2}} \, . $$
In the last step we have used the fact that the integral from $0$ to $\pi$ and that from $\pi$ to $2\pi$ have the same value. Now we can write
$$ - \cos(\theta) = \cos(\pi - \theta) = 1 - 2 \sin^2\left(\frac{\pi - \theta}{2}\right) $$
and introduce the new integration variable $\alpha = \frac{\pi - \theta}{2}$ to obtain
$$ \phi(D) = -\frac{2 G M}{\pi D} \int \limits_0^{\pi/2} \frac{\mathrm{d} \alpha}{\sqrt{1 + 2 Q + Q^2 - 4 Q \sin^2(\alpha)}} = -\frac{2 G M}{\pi D} \frac{1}{1+Q} \int \limits_0^{\pi/2} \frac{\mathrm{d} \alpha}{\sqrt{1 - \frac{4 Q}{(1+Q)^2} \sin^2(\alpha)}} \, . $$
But this integral is just the definition of the complete elliptic integral of the first kind and (using the parameter $m = k^2$ as the argument)
$$ \phi(D) = - \frac{2 G M}{\pi D} \frac{1}{1+Q} \operatorname{K}\left(\frac{4 Q}{(1+Q)^2}\right) = - \frac{2 G M}{\pi D} \operatorname{K}(Q^2) = - \frac{2 G M}{\pi D} \operatorname{K}\left(\frac{R^2}{D^2}\right)$$
follows. The final simplification is an application of Gauss's transformation. Taking the derivative we find the field
$$ F(D) = - \frac{2 G M}{\pi(D^2 - R^2)} \operatorname{E}\left(\frac{R^2}{D^2}\right) $$
in terms of the complete elliptic integral of the second kind.
