A closed form for the sum of a series $\sum_{n=1}^{\infty}x^{n\alpha} /\Gamma{(n \alpha)}$

Question

Let $\alpha \in (0,1)$. Is there a closed form for the sum $\sum_{n=1}^{\infty}x^{n\alpha}/\Gamma{(n \alpha)} $ ?

Answer 1

Too long for a comment.

As @PC1 commented, CAS provide closed forms when $\alpha$ is the reciprocal of natural numbers.

The formulae are quite messy if $\alpha=\frac 1{2k}$ but interesting when $\alpha=\frac 1{2k+1}$. Let $$F_{2k+1}(x)=\sum _{n=1}^{\infty } \frac{x^{\frac{n}{2 k+1}}}{\Gamma \left(\frac{n}{2 k+1}\right)}$$ and look at the interesting simple patterns $$F_3(x)=\frac{1}{3} e^x x \left(9+\frac{2 \Gamma \left(-\frac{2}{3},x\right)}{\Gamma \left(\frac{1}{3}\right)}+\frac{\Gamma \left(-\frac{1}{3},x\right)}{\Gamma \left(\frac{2}{3}\right)}\right)$$ $$F_5(x)=\frac{1}{5} e^x x \left(25+\frac{4 \Gamma \left(-\frac{4}{5},x\right)}{\Gamma \left(\frac{1}{5}\right)}+\frac{3 \Gamma \left(-\frac{3}{5},x\right)}{\Gamma \left(\frac{2}{5}\right)}+\frac{2 \Gamma \left(-\frac{2}{5},x\right)}{\Gamma \left(\frac{3}{5}\right)}+\frac{\Gamma \left(-\frac{1}{5},x\right)}{\Gamma \left(\frac{4}{5}\right)}\right)$$ This is easy to generalize as a finite summation

Edit

If $\alpha=\frac 1{2k}$, we have in fact quite simple formulae writing

$$G_{2k}(x)=\frac{ F_{2k}(x)-\sqrt{\frac{x}{\pi}}}{x \,e^x}-\text{erf}\left(\sqrt{x}\right)$$

For example $$G_{6}(x)=5+\frac{1}{6} \left(-\frac{6 \Gamma \left(-\frac{5}{6},x\right)}{\Gamma \left(-\frac{5}{6}\right)}+\frac{4 \Gamma \left(-\frac{4}{6},x\right)}{\Gamma \left(\frac{2}{6}\right)}+\frac{2 \Gamma \left(-\frac{2}{6},x\right)}{\Gamma \left(\frac{4}{6}\right)}+\frac{\Gamma \left(-\frac{1}{6},x\right)}{\Gamma \left(\frac{5}{6}\right)}\right)$$