Explicit formula of Fourier Transform of $e^{-\pi x^4}$.

Question

The Fourier Transform of $f(x)$ is defined by $$\hat{f}(\xi)=\int_{-\infty}^{\infty}f(x)e^{-2\pi ix\xi}dx.$$ Let $f(x)=e^{-\pi x^{2k}},k\in\Bbb{N}$, I can get the Fourier Transform of $e^{-\pi x^{2k}}$ with the following implicit formula: $$\int_{-\infty}^{\infty}e^{-\pi x^{2k}}e^{-2\pi ix\xi}dx= \frac{1}{k}\sum_{n=0}^{\infty}\frac{(-1)^n(2\pi \xi)^{2n}} {\pi^{\frac{2n+1}{2k}}(2n)!} \Gamma\left(\frac{2n+1}{2k}\right).$$ When take $k=1$, we know that $$\int_{-\infty}^{\infty}e^{-\pi x^{2}}e^{-2\pi ix\xi}dx= \sum_{n=0}^{\infty}\frac{(-1)^n(2\pi \xi)^{2n}} {\pi^{\frac{2n+1}{2}}(2n)!} \Gamma\left(\frac{2n+1}{2}\right).$$ Combining $\Gamma\left(\frac{2n+1}{2}\right)=\frac{(2n-1)!!}{2^n}\sqrt{\pi}$, where $(2n-1)!!=1\cdot3\cdot5\cdots (2n-1)$, we get $$\int_{-\infty}^{\infty}e^{-\pi x^{2}}e^{-2\pi ix\xi}dx= e^{-\pi \xi^{2}}.$$ And this is a well known result. Here I want to know for any $k\in\Bbb{N}$, does the formula $$\frac{1}{k}\sum_{n=0}^{\infty}\frac{(-1)^n(2\pi \xi)^{2n}} {\pi^{\frac{2n+1}{2k}}(2n)!} \Gamma\left(\frac{2n+1}{2k}\right)$$ has an explicit expression as the case $k=1$? Any help will welcome, or some $k$, such as $k=2$, can be give an explicit expression. Thanks a lot!