Is $f_{n,k}(x)= \left| \frac{d^n}{dx^n} e^{-|x|^k} \right|$ integrable How to show that
\begin{align}
f_{n,k}(x)= \left|  \frac{d^n}{dx^n}  e^{-|x|^k} \right|
\end{align}
From working out a number of examples by hand I have the following conjecture about $f_{n,k}(x)$:


*

*if $k\ge 1$,  $f_{n,k}(x)$ is integrable for all $n\ge 1$,

*if $0<k<1$,  then there exist an $n$ such that  $f_{n,k}(x)$ is not integrable.


The key to this question is to work out behaviour at $x \to 0$. 
This question, is related to something I asked here. However, it is a little different as the behaviour at zero was not actually investigate there. 
 A: As you mentioned already, the key is the behavior of $f$ around the origin $x= 0$. $f_{n,k}$ should be integrable 


*

*for all $k,n\in\mathbb N$.

*for $k\in\mathbb R\setminus\mathbb N$ and $n<k+1$, $n\in\mathbb N$.


For all other $n,k$, the singularity at the origin becomes non-integrable.
Since the function is symmetric around $x=0$, it is sufficient to investigate integrability for $x\in(0,1)$ (the 1 being an arbitrary upper bound). Hence, we can drop the modulus in the exponent. Now we can show by induction and the chain rule that 
$$\frac{d^n}{dx^n} e^{-x^k} = e^{-x^k}\sum_{i=I(n,k)}^n a_i x^{k-i},$$
for appropriate coefficients $a_i$, and a possibly negative $I(n,k)$. The factor $ e^{-x^k}$ in front of the sum does not affect the integrability at $x= 0$, since $ e^{-0^k} = 1$ for all $k\in\mathbb N$.
The critical situation arises once the sum contains negative exponents: $x^{k-i}$ is integrable if and only if $k-i > -1$. If $k\in\mathbb N$, this can never happen, since once $k=i$, further derivatives cannot lower the exponent in $x^{k-i}$ any further (e.g., the third, fourth, ..., derivative of $x^2$ is zero). If $k\notin \mathbb N$, negative exponents in the sum will occur eventually. The smallest exponent in the sum is $k-n$. In this case, $f_{n,k}$ is integrable if and only if $k-n>-1$. Once $n$ exceeds this bound, the singularity stemming from $x^{k-n}$ is non-integrable.
