The known magnetic field of a simple loop $(B_z,B_r)$ is expressed as a combination of constant current I and the 2 Elliptic integrals K(k), E(k), k being related to the cylindrical position $(z,r)$ and loop radius. I am not aware of any expression of the magnetic field for a large loop with a current variable upon the loop radius, except for $r=0$.
Therefore I have been integrating the known loop magnetic field across the radius, and got as a result the following sum with no obvious generating function :
$$\sum_{n=0}^\infty {(-1)^n\frac{(2n){!}{!}}{(2n+3){!}{!}}t^nP_n(x)}$$
I tried to combine the function $(1+1/t)^\frac{1}{2}Arctanh((1+1/t)^\frac{-1}{2})-1$ and the generating function of Legendre polynomial 1st kind $P_n(x)$, without success.
When I define the Polynomial $M_n(x)=(-1)^n\frac{(2n){!}{!}}{(2n+3){!}{!}}P_n(x)$
the recursion is $M_{n+1}(x)=-2x\frac{(2n+1)}{(2n+5)}M_n(x)-\frac{4n^2}{(2n+3)(2n+5)}M_{n-1}(x)$
then how to find out its' generating function ?
An (elliptic) integral representation easily follows from $$\frac{(2n)!!}{(2n+3)!!}=\frac{2^n\Gamma(n+1)\sqrt\pi}{2^{n+2}\Gamma(n+5/2)}=\frac12\mathrm{B}\left(n+1,\frac32\right)=\frac12\int_0^1 y^n\sqrt{1-y}\,{\rm d}y$$ and the generating function of the Legendre polynomials: $$\sum_{n=0}^\infty(-1)^n\frac{(2n)!!}{(2n+3)!!} t^n P_n(x)=\frac12\int_0^1\sqrt\frac{1-y}{1+2xty+t^2 y^2}\,{\rm d}y.$$