Taylor series for $e^{-x \ln x}$

Question

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?

Answer 1

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.