Series including the Gamma funtion

Question

In one of my calculation, I came across the series: $$\sum_{n=0}^{\infty} \frac{x^n}{\Gamma(n+\alpha)}.$$ When $\alpha=1,$ this equals $e^x$ and when $\alpha=\frac{1}{2},$ this equals $\frac{1}{2\sqrt{x}}erf(\sqrt{x})e^x+\frac{1}{\sqrt{\pi}}.$ How can the general case be calculated in a closed form?

Answer 1

In 1903, the Swedish mathematician Gösta Mittag-Leffler introduced the following special function: $$ E_\alpha(z)=\sum_{n=0}^{\infty}\frac{z^n}{\Gamma(\alpha n+1)}, $$ where $z$ is a complex number, $\Gamma$ is the gamma function and $\alpha \geq 0$. The Mittag-Leffler function is a direct generalization of the exponential function to which it reduces for $\alpha = 1$.

In 1905, the Swedish mathematician Anders Wiman generalized Mittag-Leffler's function as the following: $$ E_{\alpha,\beta}(z)=\sum_{n=0}^{\infty}\frac{z^n}{\Gamma(\alpha n+\beta)}, $$ where $\alpha, \beta > 0$, and everything else is the same as in the definition of $E_\alpha(z)$.

You can find more details about these function in this survey article. I've also written a related answer yesterday.

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Update. We can derive a simple form in your special case. The upper incomplete gamma function $\Gamma(a,z)$ and the lower incomplete gamma function $\gamma(a,z)$ is defined as the following: $$ \Gamma(a,z) = \int_z^\infty t^{a-1}e^{-t}\,dt $$ and $$ \gamma(a,z) = \int_0^{z} t^{a-1}e^{-t}\,dt, $$ for $a > 0, z \geq 0$. Note that $\Gamma(a,0) = \lim_{z\to\infty} \gamma(a,z) = \Gamma(a)$ and $\gamma(a,z) + \Gamma(a,z) = \Gamma(a)$.

Using the series expansion of $\Gamma(a,z)$ given in 8.7.3 here, and the property given in 8.2.3 here, and changing the variable $a \mapsto a-1$, then using the recurrence relation given here, we find that

$$ \sum_{n=0}^{\infty} \frac{z^n}{\Gamma(n+a)} = \frac{1}{\Gamma(a)}\left(e^z\,z^{1-a}\,\gamma(a,z) + 1\right). $$