A parameterized elliptical integral (Legendre Elliptical Integral) $$
K(a,\theta)=\int_{0}^{\infty}\frac{t^{-a}}{1+2t\cos(\theta)+t^{2}}dt
$$
For $$ -1<a<1;$$  $$-\pi<\theta<\pi$$
I know this integral to be a known tabulated Legendre elliptic integral, however the very fact that the numerator is parameterized completely throws a curveball.
Using:
$$
 K(a,\theta)=\int_{0}^{\infty}\frac{t^{-a}}{(1+t^{2}) + 2t\cos(\theta)}dt
$$
letting $2\gamma$ = $\theta$ 
$$
\rightarrow K(a,\theta)=\int_{0}^{\infty}\frac{t^{-a}}{(1+t^{2}) + 2t \cos(2\gamma)}dt
$$
in which case the trig function can be later manipulated using the double angle identity, turning it into a sine function
$$
\rightarrow K(a,\theta)=\int_{0}^{\infty}\frac{t^{-a}}{(1+t^{2}) + 2t \cos(2\gamma)}dt
$$
$$
\rightarrow K(a,\theta)=\int_{0}^{\infty}\frac{t^{-a}}{(1+t^{2}) + 2t(1-(\sin(\gamma))^{2})}dt
$$
So it does have a sine in the denominator that is sufficient for a Legendre Elliptical integral.  The rest is just solving for $k$ and simplifying the expression.  Which leaves me with the parameter $a$.  I have no idea what to do there.
Any help is certainly appreciated.

This is not elliptic integral, this can be expressed in terms of elementary functions:
\begin{align}
K(a,\theta)=\int_0^{\infty}\frac{t^{-a}dt}{t^2+2\cos\theta\, t+1}=\frac{1}{2i\sin\theta}\int_0^{\infty}\left(\frac{t^{-a}}{t+e^{-i\theta}}-\frac{t^{-a}}{t+e^{i\theta}}\right)dt=\\
=\frac{1}{2i\sin\theta}\left(\frac{\pi e^{ia\theta}}{\sin\pi a}-\frac{\pi e^{-ia\theta}}{\sin\pi a}\right)=\frac{\pi\sin \theta a}{\sin\theta\sin\pi a}.
\end{align}
