How to evaluate the infinite sum $\sum_{k=0}^{\infty}\frac{x^{2k+1}}{(2k+1)(k!)}$ Essentially what the title asks. For an argument $x$, how can I analytically acquire values for the function:
$$
f(x)=\sum_{k=0}^{\infty}\frac{x^{2k+1}}{(2k+1)(k!)}
$$
Again, it is important that I know how to do this analytically, as there are other series comparable to this one that I also wish to evaluate.
 A: $$f(x)=\int_{0}^{x}e^{t^2}\,dt =\int_{0}^{x}\exp\left(x^2+t^2-2tx\right)\,dt=xe^{x^2}\int_{0}^{1}e^{-t^2 x^2}\,dt = \frac{\sqrt{\pi}}{2} e^{x^2}\,\text{Erf}(x)$$
has the following continued fraction representation:
$$ f(z)=\sum_{n\geq 0}\frac{z^{2n+1}}{(2n+1)n!}=\frac{z}{1-\frac{2z^2}{3+\frac{4z^2}{5-\frac{6z^2}{7+\ldots}}}}. $$
For any $x\in(-1,1)$ the approximation $f(x)\approx e^{x^2}\arctan(x)$ is quite accurate.
A: Hint:
$$\sum_{k\ge0}\frac{x^{2k}}{k!}=\mathrm e^{x^2}.$$
A: You ask for specific values. By noting that this is simply
$$\sum_{k \ge 0} \int_0^x \frac{t^{2k}}{k!}dx= \int_0^x\sum_{k \ge 0} \frac{t^{2k}}{k!}dx= \int_0^xe^{t^2}dt = \frac{-i\sqrt{\pi}}{2}\operatorname{Erf}(ix)$$
All we have to do is look up known values of the Error Function. What we end up finding is this MSE question on the topic and this list of known values, which shows that there are no known, non-trivial, exact values of the Error Function.
Thus the only known values of $\operatorname{Erf}(x)$ are $x \in\{0,\pm\infty, \pm i\infty\}$
