Derivative of an integral (recursive formula) Let $a\in (0, 1)$. Is it possible to get a general formula for
$$\frac{d^k}{da^k}\int_0^\infty a^ndn?$$
I know that
$$\int_0^\infty a^ndn=-\frac{1}{\log a}.$$
If I compute
$$\frac{d}{da} \Big( \int_0^\infty a^ndn\Big)= \frac{d}{da} \Big(-\frac{1}{\log a}\Big)=\frac{1}{a(\log a)^2},$$
while
$$\frac{d^2}{da^2} \Big( \int_0^\infty a^ndn\Big)= \frac{d^2}{da^2} \Big(-\frac{1}{\log a}\Big)=-\frac{\log a+2}{a^2(\log a)^3}.$$
So if I am not wrong is not easy to get a recursive formula that instead is possible to get for
$$\frac{d^k}{da^k}\sum_0^\infty a^ndn=\frac{k!}{(1-a)^{k+1}.}$$
Am I wrong?
 A: We can prove that
$$
\frac{d^k}{da^k}\frac{-1}{\log a}=(-1)^{k+1}a^{-k}\sum_{i=0}^k{k\brack i}i!\cdot \left(\frac{1}{\log a}\right)^{i+1}
$$
where $\displaystyle{k\brack i}$ is a Stirling number of the first kind. This is indeed not as simple as the result for $\frac{d^k}{da^k}\sum_{n=0}^\infty a^n$.
Once we have found this complicated formula (which I found using Mathematica to compute many derivatives, and OEIS to match the coefficients that appeared), it is simple enough to prove it by induction on $k$.
Using $D$ as a shorthand for $\frac{d}{da}$,
\begin{align}
D^{k+1}\left[\frac{-1}{\log a}\right] &= D\left[D^k\frac{-1}{\log a}\right]
\\
&=(-1)^{k+1}\sum_{i=0}^k {k\brack i}i!\cdot D\left[a^{-k}\cdot (\log a)^{-(i+1)}\right]
\\
&=(-1)^{k+1}\sum_{i=0}^k {k\brack i}i!\cdot \left[(-k)a^{-(k+1)}\cdot (\log a)^{-(i+1)}-a^{-(k+1)}(i+1)(\log a)^{-(i+2)}\right]
\\
&=(-1)^{k+2}a^{-(k+1)}\left(\sum_{i=0}^k {k\brack i}i!\cdot k\cdot (\log a)^{-(i+1)}+\sum_{i=0}^k {k\brack i}(i+1)!(\log a)^{-(i+2)}\right)
\\
&=(-1)^{k+2}a^{-(k+1)}\left(\sum_{i=0}^k {k\brack i}i!\cdot k\cdot (\log a)^{-(i+1)}+\sum_{i=1}^{k+1} {k\brack i-1}i!\cdot (\log a)^{-(i+1)}\right)
\\
&=(-1)^{k+2}a^{-(k+1)}\sum_{i=0}^{k+1} \left(k{k\brack i}+{k\brack i-1}\right)i! (\log a)^{-(i+1)}
\\
&=(-1)^{k+2}a^{-(k+1)}\sum_{i=0}^{k+1} {k+1 \brack i}i! (\log a)^{-(i+1)}.
\end{align}
