Suppose I had a function that satisfied the property $f(x)=f(x-1)+g(x)$. For any $x\in\mathbb N$, it is easy enough to see that this boils down to the statement
$$f(x)=f(0)+\sum_{k=1}^xg(k)$$
If we return to our functional equation and differentiate it $n$ times, we get
$$f^{(n)}(x)=f^{(n)}(x-1)+g^{(n)}(x)$$
Once again, for any $x\in\mathbb N$, we have
$$f^{(n)}(x)=f^{(n)}(0)+\sum_{k=1}^xg^{(n)}(k)$$
One could then integrate both sides. For example, this asserts that
$$\sum_{k=1}^xg(k)=xf'(0)+\int_0^x\sum_{k=1}^tg'(k)dt$$
If the sum inside the intagral is generalized to any upper bound.
How would I justify this for discrete functions?
An example of this is letting $f_p(x)=\sum_{k=1}^xk^p$, as it seems to follow the above result. We can see from the above that
$$f_p(x)=a_px+p\int_0^xf_{p-1}(t)dt$$
$$a_p=1-p\int_0^1f_{p-1}(t)dt$$
which appears to be a true recursive formula.
Under what conditions can I extend discrete functions $f(x)$ and $g(x)$ into continuous differentiable functions and apply this method?
