What's the (limit of the) n-th order derivative of this integral? I met a question that asks me to calculate the result of an integral first:
$$f(s)=\int_0^1e^{-sx}dx,s\geq0\tag1$$
It is easy, if I am not wrong, the answer should be $1$ when $s=0$, and $\frac{1-e^{-s}}s$ for $s>0$.
Then it asks me to calculate the limit:
$$\lim_{s\rightarrow0}\frac{df(s)}{ds}\tag2$$
It is also easy, the derivative is:
$$\lim_{s\rightarrow0}\frac{df(s)}{ds}=\lim_{s\rightarrow0}\frac{e^{-s}s-(1-e^{-s})}{s^2}\tag3$$
because it is a $\frac00$ form limit, by using the L'Hôpital law, we can quickly calculate that the limit is $-\frac12$.
However, I got stuck in the final question, it asks me to calculate
$$\lim_{s\rightarrow0}\frac{d^nf(s)}{ds^n}\tag4$$
I tried to calculate the n-th derivative directly first, hoping to find some laws. But as all you can see from (3), it will be only more and more complex as I differentiate (3).
I also tried to use the Leibniz formula, like let $u=1-e^{-s}$ and $v=s^{-1}$, so $f(s)=(1-e^{-s})s^{-1}=uv$, but I still could not get any idea after just writing a long formula. So how can I calculate (4)? Could you give me some hints? Thank you very much!
 A: $f^{(n)}(s)=\int_0^{1}(-x)^{n}e^{-sx}dx$. So $\lim_{s \to 0} f^{(n)}(s) =\int_0^{1}(-x)^{n}dx=\frac {(-1)^{n}} {n+1}$
A: Alternatively, a series expansion and term-wise differentiation of $f(s)$ can be performed:
$$f(s) = \frac{1 - e^{-s}}{s} = \sum_{k=1}^\infty \frac{(-s)^{k-1}}{k!} = \sum_{k=0}^\infty \frac{(-1)^k}{k+1} \frac{s^k}{k!}.$$  Therefore, $$f^{(n)}(s) = \sum_{k=n}^\infty \frac{(-1)^k}{k+1} \frac{s^{k-n}}{(k-n)!}.$$  In all cases, the first term is constant with respect to $s$.  Now the result is trivial:  $$\lim_{s \to 0} f^{(n)}(s) = \frac{(-1)^n}{n+1} + \lim_{s \to 0} \sum_{k=n+1}^\infty \frac{(-1)^k}{k+1} \frac{s^{k-n}}{(k-n)!} = \frac{(-1)^n}{n+1}.$$

Step by step:  note $$e^s = \sum_{k=0}^\infty \frac{s^k}{k!},$$ hence $$e^{-s} = \sum_{k=0}^\infty \frac{(-s)^k}{k!},$$ and $$1 - e^{-s} = 1 - \left(\frac{(-s)^0}{0!} + \sum_{k=1}^\infty \frac{(-s)^k}{k!}\right) = - \sum_{k=1}^\infty \frac{(-s)^k}{k!},$$ hence $$\frac{1-e^{-s}}{s} = (-s)^{-1} \sum_{k=1}^\infty \frac{(-s)^k}{k!} = \sum_{k=1}^\infty \frac{(-s)^{k-1}}{k!}.$$
