How to calculate the value of the n-th derivative of this function at this point

Question

$$\forall n\in \mathbb{N},\left.\frac{d^{2n+1}}{dx^{2n+1}}(x^{\ln x})\right|_{x=e^{\frac{n}{2}}}=0$$ How to prove this equation? I find that if $n=0$ or $n=1$, the answer is $0$. But I can't prove that it is true for all the nonegative integer. My idea is, calculate the analytical expression of its $n$-th derivative and then substitute it into the specific value. I find that $$\frac{d^{2n+1}}{dx^{2n+1}}(x^{\ln x})=\frac{x^{\ln x}}{x^{2n+1}}P_{2n+1}(\ln x)$$ here $P_{2n+1}(x)$ is a polynomial and the its degree is $2n+1$. And I find for $n=0,1,2$, $P_{2n+1}(\frac{n}{2})=0$. But I don't know how to prove the proposition "$\forall n,P_{2n+1}(\frac{n}{2})=0$". This is my ideas and I can't solve this question. I hope someone can help me! Thanks!

Answer 1

You want the $(2n+1)$-st derivative of the function $x^{\log x}=\exp((\log x)^2)$ at $x=e^{n/2}$, i.e., by the residue theorem, $$\frac1{2\pi i}\oint\frac {e^{(\log x)^2}}{(x-e^{n/2})^{2n+2}}\,dx$$ where we integrate along a contour encircling $e^{n/2}$. Using the substitution $x=e^{t+n/2}$, this integral is $$ e^{n^2/4-n/2-1}\frac1{2\pi i}\oint\frac {e^{t^2}}{(e^{t/2}-e^{-t/2})^{2n+2}}\,dt$$ where we now integrate along a circle centered at $t=0$. Since the integrant is even, the integral is zero, as you wanted to show.

(You can likely get this without integrals/residues, just using the same substitution.)