# 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?

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.
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: $$\gamma (n,x) = (n - 1)! \left (1 - e^{-x} \sum_{k = 0}^{n - 1} \frac{x^k}{k!} \right ).$$