# Closed form for $\sum_{k=0}^\infty \frac{x^{2 k+1}}{(2k+1) \Gamma (\frac{2k+1}{q})}$

I've been trying to answer another question, and as a part of the solution encountered the following series:

$$S_q(x)=\sum_{k=0}^\infty \frac{x^{2 k+1}}{(2k+1) \Gamma (\frac{2k+1}{q})}$$

Where $$q=1,2,3,4,\ldots$$.

For $$q=1$$ and $$q=2$$ the closed form is trivial, however, for $$q \geq 3$$ it becomes harder and harder to find.

I would like to know the general form, which should look something like this:

$$S_q(x)= \sum_{n=1}^q F_n(x)$$ Where $$F_n(x)$$ is a hypergeometric function, possibly multiplied by some factors.

Expressing:

$$2k+1=qm+n$$

We have:

$$\sum_{m=0}^\infty \frac{x^{qm+n}}{(qm+n) \Gamma (m+\frac{n}{q})}= \frac{x^n}{n \Gamma(\frac{n}{q})} {_1 F_1} \left(1; 1+ \frac{n}{q}; x^q \right)$$

And the series in general should reduce to a finite sum of such terms, but I don't know how to correctly describe all the possible choices of $$n$$ for a particular $$q$$.

• $S_q(x)=\frac{1}{q}\Im E_{\frac{1}{q}}(ix)$ please have a look at e.g. here . Maybe there is some literature about the Mittag-Leffler function. – user90369 Aug 8 '19 at 11:05
• @user90369, thank you for the reference, I didn't know about this function – Yuriy S Aug 8 '19 at 11:26

## 2 Answers

First, multiply by $$q$$ to simplify the denominator:

$$qS_q(x)=\sum_{k=0}^\infty\frac{x^{2k+1}}{\Gamma\left[\frac{2k+q+1}q\right]}$$

If all you want is hypergeometric functions, then just look at the ratio between terms. Let $$k=qm+n$$ and $$a_{m+1}/a_m$$:

$$\frac{x^{2qm+2n+2q+1}}{\Gamma\left[\frac{2qm+2n+3q+1}q\right]}\div\frac{\Gamma\left[\frac{2qm+2n+q+1}q\right]}{x^{2qm+2n+1}}=\frac{q^2x^{2q}}{(2qm+2n+q+1)(2qm+2n+2q+1)}$$

Hence we have

$$qS_q(x)=\sum_{n=0}^{q-1}\frac{x^{2n+1}}{\Gamma\left[\frac{2n+q+1}q\right]}{}_1F_2\left(1;\frac{2n+q+1}{2q},\frac{2n+2q+1}{2q};\frac14x^{2q}\right)$$

In the case that $$q$$ is even, less terms are needed:

$$2qS_{2q}(x)=\sum_{n=0}^{q-1}\frac{x^{2n+1}}{\Gamma\left[\frac{2n+2q+1}{2q}\right]}{}_1F_1\left(1;\frac{2n+2q+1}{2q};\frac12x^q\right)$$

• Oh, this is more simple than I thought. Somehow I didn't think that ratio of terms will help in this case, but it clearly does. Thank you – Yuriy S Aug 8 '19 at 16:31
• Made some fixes. – Simply Beautiful Art Aug 8 '19 at 16:38
• So, it is a sum of hypergeometric functions after all – Yuriy S Aug 8 '19 at 16:43
• Yeah $\hphantom{blank text here}$ – Simply Beautiful Art Aug 8 '19 at 16:43

Unfortunately, I don't understand the sense to make a sum of series out of one series. At least these sub-series should have something particular, not only another structure. Perhaps it's clearer to build a closed form using the incomplete gamma function.

$$x!:=\Gamma(1+x) ~ ; ~~z>0$$

$$\displaystyle E_z(x):=\sum\limits_{k=0}^\infty\frac{x^k}{(kz)!}\hspace{2cm}$$ Mittag-Leffler function

$$\displaystyle f_z(x):=\sum\limits_{k=0}^\infty\frac{x^{k+z}}{(k+z)!}=e^x\left(1-\frac{\Gamma(z,x)}{\Gamma(z)}\right)$$

$$\displaystyle z:=\frac{1}{q} , ~ q\in\mathbb{N}$$

$$\displaystyle E_{1/q}(x) = e^{x^q} + \sum\limits_{j=1}^{q-1}f_{j/q}(x^q) = e^{x^q}\left(q - \sum\limits_{j=1}^{q-1}\frac{\Gamma(j/q,x^q)}{\Gamma(j/q)}\right)$$

$$\displaystyle S_q(x)=\frac{1}{q}\Im E_{1/q}(ix)$$