Let $\alpha \in (0,1)$. Is there a closed form for the sum $\sum_{n=1}^{\infty}x^{n\alpha}/\Gamma{(n \alpha)} $ ?
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$\begingroup$ Does it even converge for $x \neq 0$? Stirling's approximation shows that $$|\Gamma(n\alpha)x^{n\alpha}|^{1/n} \sim \left(\frac{n\alpha|x|}{e}\right)^{\alpha} \to \infty$$ as $n\to\infty$. $\endgroup$– Sangchul LeeCommented Jul 14, 2022 at 23:58
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$\begingroup$ Sorry. I mistyped it. $\endgroup$– MedoCommented Jul 15, 2022 at 0:10
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$\begingroup$ If $\alpha$ is a rational number then Mathematica provides some closed form. For small fractions like $\alpha=\frac12$ it's manageable but when the numbers increase, the result starts to be quite hairy. $\endgroup$– PC1Commented Jul 15, 2022 at 0:28
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5$\begingroup$ Essentially it is the Mittag-Leffler function $x^\alpha E_{\alpha,\alpha}(x^\alpha)$. $\endgroup$– metamorphyCommented Jul 15, 2022 at 5:55
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$\begingroup$ @metamorphy Indeed, It is the Mittag Leffler function (+1). Thanks $\endgroup$– MedoCommented Jul 15, 2022 at 19:14
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1 Answer
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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 $$G_{6}(x)=5+\frac{1}{6} \left(-\frac{6 \Gamma \left(-\frac{5}{6},x\right)}{\Gamma \left(-\frac{5}{6}\right)}+\frac{4 \Gamma \left(-\frac{4}{6},x\right)}{\Gamma \left(\frac{2}{6}\right)}+\frac{2 \Gamma \left(-\frac{2}{6},x\right)}{\Gamma \left(\frac{4}{6}\right)}+\frac{\Gamma \left(-\frac{1}{6},x\right)}{\Gamma \left(\frac{5}{6}\right)}\right)$$
answered Jul 15, 2022 at 3:03
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$\begingroup$ Would you like to take a look at this question: math.stackexchange.com/questions/4494002/… $\endgroup$– MedoCommented Jul 17, 2022 at 20:21