<s>[INST] Taylor series for $e^{-x \ln x}$ What is a Taylor expansion for the following function?
$$ e^{-x \ln x} $$
I assume you can't do a Taylor expansion around $x=0$, since the function doesn't exist at that point.  The next best choice seems to be $x=1$, since that's when $\ln(x)=0$.  So, when I try to do an expansion around $x=1$, I get an answer that seems to have no clean closed-form (verified with wolfram alpha):
$$ 1 - (x - 1) + \frac{1}{2} (x - 1)^3 - \frac{1}{3} (x - 1)^4 + \frac{1}{12}(x - 1)^5 +  \frac{1}{120}(x - 1)^6 -  \dots$$
(Notice the above series has a zero coefficient for the $(x-1)^2$ term...which makes it even trickier to generate a closed-form expression for).
However, I believe this function should be easily expanded and neatly integrated term-by-term (as suggested by the solution to question 1A on this prelim exam question from UC Berkeley).
Where am I going wrong?
 [/INST] We have
\begin{align*}
\color{blue}{x^{-x}}&=1 - (x - 1) + \frac{1}{2} (x - 1)^3 - \frac{1}{3} (x - 1)^4 + \frac{1}{12}(x - 1)^5 +  \frac{1}{120}(x - 1)^6 -  \dots\\
&=\sum_{n=0}^\infty a_n\frac{(x-1)^n}{n!}
\end{align*}
when expanded at $x=1$,

where
\begin{align*}
\color{blue}{(a_n)_{n\geq 0}=(1, -1, 0, 3, -8, 10, 6, -42, -160,\ldots)}
\end{align*}
is stored as A176118 in OEIS.

</s>