What is the $n^{th}$ derivative of $\log_x(e)$? I happened to stumble upon the integral of $\log_x(e)$, finding it to apparently be non-elementary.
So I had to see if I could discern a pattern by differentiating, much like finding the integral of $W(x)$.
$$f(x)=\log_x(e)=\frac1{\ln(x)}$$$$f'(x)=-\frac 1{x\ln(x)^2}$$$$f^{\prime\prime}(x)=\dfrac1{(\ln x)^2}+\dfrac1{2x^2(\ln x)^3}$$
Hopefully, there is a pattern that can be used to find $f^{(n)}$.
So please find $f^{(n)}$ and if we can, substitute $n=-1$ to find the integral of the function.
 A: (BIG) HINT:
Notice:
$$f(x)=\log_x(e)=\frac{\ln(e)}{\ln(x)}=\frac{\log_e(e)}{\ln(x)}=\frac{\frac{\ln(e)}{\ln(e)}}{\ln(x)}=\frac{1}{\ln(x)}$$
So:
$$f^{(1)}(x)=-\frac{1}{x\ln^2(x)}$$
$$f^{(2)}(x)=\frac{2+\ln(x)}{x^2\ln^3(x)}$$
$$f^{(3)}(x)=-\frac{2\left(3\ln(x)\left(3+\ln(x)\right)\right)}{x^3\ln^4(x)}$$
So you've to show that:
$$\frac{\text{d}^n}{\text{d}x^n}\left[\frac{1}{\ln(x)}\right]=-\frac{n(1+n)\left(1+\lfloor{\frac{n-1}{2}}\rfloor\right)\ln^{(n)}(x)}{2\ln^2(x)}\space\space\space\space\space\space\text{for}\space n\in\mathbb{Z}\space\text{and}\space n\ge 0$$
And:
$$\frac{\text{d}^n}{\text{d}x^n}\left[\ln(x)\right]=\ln^{(n)}(x)=\lim_{\epsilon\to0}\epsilon^{-n}\sum_{k=0}^{n}\left(-1\right)^{n-k}\binom{n}{k}\ln\left(x+k\epsilon\right)\space\space\space\space\space\text{for}\space n\in\mathbb{Z}\space\text{and}\space n\ge 1$$
Notice for the last part that:
$$\frac{\text{d}^n}{\text{d}x^n}\left[\ln(x)\right]=\ln^{(n)}(x)=(-1)^{1+n}x^{-n}\Gamma(n)\space\space\space\space\space\text{for}\space n\in\mathbb{Z}\space\text{and}\space n> 0$$
A: The answer is simple: It doesn't work like that.
You cannot substitute $n=-1$ to get the integral because when finding the pattern, you take into account only derivatives. Nevertheless there exist a special function that is defined exactly as the anti-derivative of $1/\ln(x)$, namely $$\operatorname{li}(x)=\int_0^x\frac{1}{\ln(t)}\,dt$$
A: The idea doesn't work, and you have an example at hand. The derivatives of $x^{-1}$ are all rational functions but it's integral is the logarithm, which cannot fit in this pattern. 
A: Write $s(n,k)$
for Stirling numbers of the first kind.
Then
$$
f^{(n)}(x) =
\sum _{k=1}^{n}\frac{s\left( n,k \right) k!\, \left( 
-1 \right) ^{k}}{ \left( \ln  \left( x \right)  \right) ^{k+1}{x}^{n}}
$$
But you won't get any informaion trying to put $n=-1$ there.
