Long story short, I was bored and wanted to find the Fourier Transform of the three basic Jacobi Elliptic Functions $\mathbf{sn}(x\mid k)$, $\mathbf{cn}(x\mid k)$, and $\mathbf{dn}(x\mid k)$. However, when I went searching for an answer on the internet, no one really seemed to have one. Therefore, I attempted to solve it by hand.
Because the proof is a bit long, I will place in a few basics:
(1) I used the Fourier Series Representations of these functions
(2) I changed the functions from $\mathbf{pn}(x\mid k)$ to $\mathbf{pn}(2\pi t \mid k)$ ($\mathbf{p}$ can be $\mathbf{s,c,d}$) so that it could be much simpler to do the transform.
(3) I evaluated $\mathbf{sn}$ to be:
$$\mathcal{F\{\mathbf{sn}(2\pi t\mid k)\}(s)}=\frac{\pi}{ikK} \sum^\infty_{n=0}\frac{q^{n+\frac{1}{2}}(\delta(s-\frac{\pi(2n+1)}{2K})-\delta(s + \frac{\pi(2n+1)}{2K}))}{1-q^{2n+1}}$$
where $K\equiv K(k)=\int_0^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1-k^2\sin^2\theta}}$, $k$ is the elliptic modulus (eccentricity), $k'$ is the complimentary modulus ($k'=\sqrt{1-k^2}$), $q=e^{\frac{-\pi K(k')}{K(k)}}$ (the nome)
I wish to know if this is the right solution to the problem. Also, here is a thing that I made in Desmos for the Elliptic Functions. I will hope to add more to this, but I may have to type it up on a word doc to have it explained thoroughly.