Let $f(x)=\int_0^x \sum_{i=0}^{n-1} \frac{(x-t)^i}{i!}dt.$ Find the nth derivative $f^{(n)}x.$ Let $f(x)=\displaystyle\int_0^x \displaystyle\sum_{i=0}^{n-1} \dfrac{(x-t)^i}{i!}dt.$ Find the nth derivative $f^{(n)}x.$
Note that $x$ appears both sides in the integrand and in the limits, so we need to work carefully. Write $g_n (x,t) = \displaystyle\sum_{i=0}^{n-1}\dfrac{(x-i)^i}{i!}$ so that $f(x) = \displaystyle\int_0^x g_n (x,t)e^{nt}dt.$ By definition, $f'(x) = \lim\limits_{h\to 0} \dfrac{\displaystyle\int_0^{x+h}g_n(x+h,t)e^{nt}dt - \displaystyle\int_0^x g_n(x,t) e^{nt}dt}{h}.$ Could anyone help me simplify this?
 A: Substitute $t =x-s$ in the integral:
$$
f(x)=\int_0^x \sum_{i=0}^{n-1} \dfrac{(x-t)^i}{i!}dt
= \int_0^x \sum_{i=0}^{n-1} \dfrac{s^i}{i!}ds 
$$
so that the integrand does not depend on $x$ anymore. Now differentiation becomes simple, e.g.:
$$
f'(x) = \sum_{i=0}^{n-1} \dfrac{x^i}{i!}
$$
etc.
A: In general
$$\int_{0}^{x} (x-t)^n \, dt = \left[ - \frac{(x-t)^{n+1}}{n+1} \right]_{0}^{x} =  \frac{x^{n+1}}{n+1}$$
which can be used in the following way.
\begin{align}
f(x) &= \int_{0}^{x} \sum_{k=0}^{n-1} \frac{(x-t)^k}{k!} \, dt \\ 
&= \sum_{k=0}^{n-1} \frac{x^{k+1}}{(k+1)!} \\
&= \sum_{k=1}^{n} \frac{x^k}{k!} = e_{n}(x) - 1,
\end{align}
where $e_{n}(x)$ is the finite exponential function. This also shows that $f(0) = e_{n}(0) - 1 = 0$.
Now, using
$$D^{(m)} ( x^n ) = \frac{n!}{(n-m)!} \, x^{n-m}$$
then
\begin{align}
f^{(m)}(x) &= \sum_{k=1}^{n} \frac{x^{k-m}}{(k-m)!} \\
&= \sum_{k=m}^{n} \frac{x^{k-m}}{(k-m)!} \\
&= \sum_{k=0}^{n-m} \frac{x^k}{k!} = e_{n-m}(x).
\end{align}
In the case $m=n$ this becomes $f^{(n)}(x) = e_{0}(x) =1$.
