$$ K(a,\theta)=\int_{0}^{\infty}\frac{t^{-a}}{1+2t\cos(\theta)+t^{2}}dt $$
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}