Show that $ \int_0^xe^{x-t}f(t)dt=\sum_{n=0}^\infty( f^{(n)}(0)(e^x-\sum_{i=0}^n \frac{x^i}{i!}))$ ,$-R\lt x\lt R$. For a function $f$ that has a power series representation centered at the origin, with radius R>0, show that :
$F(x)=\int_0^xe^{x-t}f(t)dt=\sum_{n=0}^\infty( f^{(n)}(0)(e^x-\sum_{i=0}^n \frac{x^i}{i!}))$ ,$-R\lt x\lt R$.
I'm learning about power series and this problem has me stumped, I don't know how to approach it. I tried fiddling around with the taylor representation for $e^x$ , but whatever I obtain does not even look close to what I am supposed to get.
How should I approach this type of problem?
 A: Since $f$ has a power series representation centred about the origin, it can be expressed as
$$f(x) = \sum{n = 0}^\infty \frac{f^{(n)}(0)}{n!} t^n, \quad |t| < R,$$
where $R > 0$. Now for $x$ within the interval of convergence we have
$$F(x) = e^x \sum_{n = 0}^\infty \frac{f^{(n)}(0)}{n!} \int_0^x t^n e^{-t} \, dt,$$
after the summation has been interchanged with the integration.
For the remaining integral, $I_n$, since $n = 0,1,2,\ldots$ integrating by parts $n$ times it can be seen that
$$I_n = n! \left (1 - e^{-x} \sum_{k = 0}^n \frac{x^k}{k!} \right ).$$ 
Hence
$$F(x) = \sum_{n = 0}^\infty f^{(n)} (0) \left (e^x - \sum_{k = 0}^n \frac{x^k}{k!} \right ),$$
as required. 

Comment
The integral $I_n$ is actually equal to the lower incomplete gamma function $\gamma (n + 1,x)$. With this identification, for $n \in \mathbb{N}$, the follow known result (see Eq. (8) in the link) can then be directly used:
$$\gamma (n,x) = (n - 1)! \left (1 - e^{-x} \sum_{k = 0}^{n - 1} \frac{x^k}{k!} \right ).$$
