Fourier transform of Quartic and higher even power Gaussians

Question

Is it possible to calculate the Fourier transform quartic Gaussian function analytically? $$ F(k) = \int_{-\infty}^{\infty} e^{-a x^4} e^{ikx} dx $$

Answer 1

Assuming $a>0$, Mathematica gives $$F(k)=\Gamma \left(-\frac{1}{4}\right)\frac{k^2 }{16 a^{3/4}}\, _0F_2\left(;\frac{5}{4},\frac{3}{2};\frac{k^4}{256 a}\right)+\frac{2 \Gamma \left(\frac{5}{4}\right)}{a^{1/4}}\, _0F_2\left(;\frac{1}{2},\frac{3}{4};\frac{k^4}{256 a}\right)$$

In order you be able to play with it, I tabulate a few values of these functions which are very smooth $$\left( \begin{array}{ccc} x & \, _0F_2\left(;\frac{5}{4},\frac{3}{2};x\right) &\, _0F_2\left(;\frac{1}{2},\frac{3}{4};x\right)\\ 0 & 1.00000 & 1.00000 \\ 1 & 4.19970 & 1.58215 \\ 2 & 8.56974 & 2.26770\\ 3 & 14.2754 & 3.06568 \\ 4 & 21.4945 & 3.98554 \\ 5 & 30.4174 & 5.03726 \\ 6 & 41.2484 & 6.23128 \\ 7 & 54.2055 & 7.57858 \\ 8 & 69.5218 & 9.09067 \\ 9 & 87.4456 & 10.7796 \\ 10 & 108.241 & 12.6580 \\ 11 & 132.190 & 14.7391 \\ 12 & 159.590 & 17.0367 \\ 13 & 190.758 & 19.5653 \\ 14 & 226.029 & 22.3399 \\ 15 & 265.760 & 25.3763 \\ \end{array} \right)$$

Edit

I did not face any specific problem to compute with Mathematica $$F_n(k) = \int_{-\infty}^{+\infty} e^{-a x^{2n}} e^{ikx} dx$$ even if, for sure, the formulae are more and more complex.

For example, $$F_3(k) =-\frac{k^4 \Gamma \left(-\frac{1}{6}\right) \, _0F_4\left(;\frac{7}{6},\frac{4}{3},\frac{3}{2},\frac{5}{3};-\frac {k^6}{46656 a}\right)}{432 a^{5/6}}+$$ $$\frac{2 \Gamma \left(\frac{7}{6}\right) \, _0F_4\left(;\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{5}{6};-\frac {k^6}{46656 a}\right)}{\sqrt[6]{a}}-\frac{\sqrt{\pi } k^2 \, _0F_4\left(;\frac{2}{3},\frac{5}{6},\frac{7}{6},\frac{4}{3};-\frac {k^6}{46656 a}\right)}{6 \sqrt{a}}$$

Answer 2

There is no analytcal way to do that. We can however get some answer in terms of special functions (like Euler's Gamma and Hypergeometric functions). The Fourier transform in such way is given by this result:

$$\frac{\frac{2 \Gamma \left(\frac{5}{4}\right) \, _0F_2\left(;\frac{1}{2},\frac{3}{4};\frac{k^4}{256 a}\right)}{\sqrt[4]{a}}-\frac{k^2 \Gamma \left(\frac{3}{4}\right) \, _0F_2\left(;\frac{5}{4},\frac{3}{2};\frac{k^4}{256 a}\right)}{4 a^{3/4}}}{\sqrt{2 \pi }}$$

where $\Gamma(\cdot)$ denotes Euler's Gamma function and $_0F_2$ denotes the Hypergeometric function.