I am trying to integrate a generating function.
On the Wikipedia page for Generating Functions, the section "Differentiation and integration of generating functions" states: $$ G^{\prime}(z) = \sum_{n \geq 0} (n+1) g_{n+1} z^n \\ z \cdot G^{\prime}(z) = \sum_{n \geq 0} n g_{n} z^n \\ \int_0^z G(t) dt = \sum_{n \geq 1} \frac{g_{n-1}}{n} z^n $$
However, it is unclear to me how to calculate $g_{n-1}$ or what $g_{n-1}$ is. I would like to be able to integrate the following:
$$ G(x)=\sum _{n=0}^{\infty} \frac{x^n}{n!} $$
But I am unclear on how to do so using the statements above. Thank you for your help.
$$G(x)=\sum _{n=0}^{\infty} \frac{x^n}{n!}$$ $$\begin{array}{ll} \int G(x) d x && = \int ( \sum _{n=0}^{\infty} \frac{x^n}{n!} ) dx \\ && = \sum _{n=0}^{\infty} (\int \frac{x^n}{n!} dx) \\ && = \sum _{n=0}^{\infty} (\frac {\int x^n dx} {n!}) && \text{because of linearity of integration} \\ && = \sum _{n=0}^{\infty} (\frac {x^{n+1}} {n!(n+1)}) && \text{by the definition of }\int x^n dx\\ && = \sum _{n=0}^{\infty} \frac {x^{n+1}} {(n+1)!} \\ && = \sum _{n=1}^{\infty} \frac {x^n} {n!} \\ \end{array} $$
Alternatively, write $$G(x)=\sum _{n=0}^{\infty} g_n {x^n}$$ by convention, we have $$ \sum _{n=0}^{\infty}\left( \frac{1}{n!}\times x^n \right) = \sum _{n=0}^{\infty} g_n {x^n} $$, thus $g_n = \frac 1 {n!}$ .
Applying to the formula, we have
$$ \int G(x) d x = \sum _{n=1}^{\infty} \frac{g_{n-1}}{n} = \sum _{n=1}^{\infty} \frac{\frac{1}{(n-1)!}}{n} = \sum _{n=1}^{\infty} \frac{1}{n(n-1)!}= \sum _{n=1}^{\infty} \frac{1}{n!}$$
which gives the same result.
In conclusion, integration of a generating function is not different from that of normal function.