I am not familiar with fractional calculus, so I want to know what I am doing wrong.
The convention I use
$$\int^\infty_{-\infty}e^{-\frac{x^2}{2}}e^{-i k x}dx=\sqrt{2 \pi}e^{-\frac{k^2}{2}}$$
I am assuming this should work
$$\frac{d^{0.5}}{dx^{0.5}}\rightarrow\sqrt{i k}=\frac{1+i}{\sqrt2}\sqrt k$$
And inverse transformation
$$\frac{1+i}{\sqrt2}\sqrt k\sqrt{2 \pi}e^{-\frac{k^2}{2}}\rightarrow\frac{1}{2 \pi}\frac{1+i}{\sqrt2}\sqrt{2 \pi}\int^\infty_{-\infty}\sqrt k\,e^{-\frac{k^2}{2}}e^{i k x}dk=\frac{1+i}{2\sqrt \pi}\bigg(\int ^\infty_0 \sqrt k e^{-\frac{k^2}{2}}e^{i k x}dk-i\int ^\infty_0 \sqrt k e^{-\frac{k^2}{2}}e^{-i k x}dk\bigg)$$
Since I am not an integration mastermind, mathematica evaluates this to be
$$\frac{\sqrt{2} \Gamma \left(\frac{3}{4}\right) \, _1F_1\left(\frac{3}{4};\frac{1}{2};-\frac{x^2}{2}\right)-2 x \Gamma \left(\frac{5}{4}\right) \, _1F_1\left(\frac{5}{4};\frac{3}{2};-\frac{x^2}{2}\right)}{2^{3/4} \sqrt{\pi }}$$
And mathematica input would be (should you want to look at it yourself)
1/(2^(3/4)
Sqrt[\[Pi]]) (Sqrt[2]
Gamma[3/4] Hypergeometric1F1[3/4, 1/2, -(x^2/2)] -
2 x Gamma[5/4] Hypergeometric1F1[5/4, 3/2, -(x^2/2)])
Which looks like
This looks about right to me.
But I cannot achieve the same result using the taylor series.
$$e^{-\frac{x^2}{2}}=\sum _{n=0}^{20} \frac{(-1)^n }{2^n n!}x^{2n}$$
After partially differentiating
$$\sum _{n=0}^{20} \frac{(-1)^n x^{2 n-0.5} \Gamma (2 n+1)}{\left(2^n n!\right) \Gamma (2 n+0.5)}$$
Which doesn't quite make sense
