Taylor series of $f(x)=\int_0^1 \frac{1-e^{-sx}}{s}ds$ Taylor series of: 
$$f(x)=\int_0^1 \frac{1-e^{-sx}}{s}ds$$
at $x_0 = 0$.
I've done:
By fundamental theory of calculus:
$$f'(x)=1-e^{-1x}$$
Which is clearly differentiable by e.g. $n$ times.
What do I need to do to get the expression:
$$\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$
Perhaps use $x_0=0$?
 A: First note that
$$\frac{1-e^{-sx}}{s} = -\frac 1s \sum_{n=\color{red}1}^\infty \frac{(-1)^n}{n!} (sx)^n = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n!} x^n s^{n-1}$$
Now integrate term by term: $$\int_0^1 \frac{1-e^{-sx}}{s} \ ds  = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n!} x^n \int_0^1 s^{n-1} \ ds = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n \cdot n!} x^n$$
A: Given that
$$f(x) = \int_{0}^{1} \frac{1 - e^{-xs}}{s} \, ds$$
then
\begin{align}
f^{\prime}(x) &= \int_{0}^{1} \frac{0+s \, e^{-xs}}{s} \, ds = \int_{0}^{1} e^{-xs} \, ds = \left[ - \frac{e^{-xs}}{x} \right]_{0}^{1} = \frac{1 - e^{-x}}{x}
\end{align}
Now, 
\begin{align}
f^{\prime}(x) &= \frac{1}{x} \, \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \, x^{n}}{n!}  = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{n}}{(n+1)!}
\end{align}
and upon integration
$$ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{n+1}}{(n+1) \, (n+1)!} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n}}{n \cdot n!}$$
In terms of the desired expression:
$f(x)$ is seen in the form
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{n}(0) \, (x-0)^{n}}{n!}$$
where 
\begin{align}
f^{n}(0) = \begin{cases} 0 & n=0 \\ \frac{(-1)^{n-1}}{n} & n \geq 1 \end{cases}
\end{align}
A: Why not just integrate the power series:
$$
\begin{align}
\int_0^1\frac{1-e^{-sx}}s\,\mathrm{d}s
&=\int_0^1\left(\sum_{k=1}^\infty (-s)^{k-1}\frac{x^k}{k!}\right)\,\mathrm{d}s\\
&=\sum_{k=1}^\infty (-1)^{k-1}\frac{x^k}{k\,k!}
\end{align}
$$
A: First let's take care of $x=0$.
$$
f(0) = \int_0^1 \frac{1-e^{s0}}{s} ds = \int_0^1 \frac{1 - 1}{s} ds = 0 .
$$
(the integral looks improper, but isn't really).
Now you have to compute derivatives with respect to $x$:
$$
f'(x) = \frac{d}{dx}\int_0^1 \frac{1-e^{sx}}{s} ds = 
\int_0^1 \frac{d}{dx}\frac{1-e^{sx}}{s} ds = 
$$
Can you take it from there?
You might also be able to do this problem by expanding $e^{sx}$ in a power series inside the integral, and then integrating term by term.
