Let $\theta\left(z\right):=\sum_{n\geq1}e^{-n^{2}z},\,\mathrm{Re}\left(z\right)>0.$ It is well known that $$\theta\left(z\right)=\left(\frac{\pi}{z}\right)^{1/2}\theta\left(\frac{\pi^{2}}{z}\right).$$
I would like to know if it is possible to extend this formula to the function $$\theta_{\ell}\left(z\right):=\sum_{n\geq1}e^{-n^{2\ell}z},\,\mathrm{Re}\left(z\right)>0,\,\ell\in\mathbb{N}^{+}.$$
I tried to work with the Poisson summation formula but I get nothing.
Thank you.
Note that $$\int_{0}^{+\infty}e^{-t^{2\ell}}e^{i\omega t}dt=\sum_{k\geq0}\frac{\left(i\omega\right)^{k}}{k!}\int_{0}^{+\infty}e^{-t^{2\ell}}t^{k}dt=\frac{1}{2\ell}\sum_{k\geq0}\frac{\Gamma\left(\frac{k+1}{2\ell}\right)}{k!}\left(i\omega\right)^{k}$$ and, in a similar way, $$\int_{-\infty}^{0}e^{-t^{2\ell}}e^{i\omega t}dt=\frac{1}{2\ell}\sum_{k\geq0}\frac{\Gamma\left(\frac{k+1}{2\ell}\right)}{k!}\left(-i\omega\right)^{k}$$ hence the Fourier transform of the function $f(t):=e^{-t^{2\ell}}$ is $$\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-t^{2\ell}}e^{i\omega t}dt=\frac{1}{\ell\sqrt{2\pi}}\sum_{k\geq0}\frac{\Gamma\left(\frac{2k+1}{2\ell}\right)}{\left(2k\right)!}\left(-1\right)^{k}\omega^{2k}.\tag{1}$$ For $\ell=1$, by $(1)$, it is quite simple to get, using Gamma duplication's formula, the classical Jacobi's funcional identitiy but for $\ell>1$ we can see we have no simple expression (probably we can write $(1)$ in terms of some hypergeometric functions), as already mentioned by reuns.