Let $X\sim\mathcal{N}(0,1)$ and $p>2$. I would like to prove that $$\left.\frac{\mathrm{d}}{\mathrm{d}t}\mathbb{E}[|X+t|^p]\right|_{t=0}=0,$$ but I don't see a (simple) way to do it.
Written explicitly, the derivative would be
$$\lim_{t\rightarrow 0}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{|x+t|^p-|x|^p}{t}\exp(-x^2/2)~\mathrm{d}x.$$
The denominator inside the integral doesn't generally converge to zero for fixed $x$, so it is not that simple. I also thought of using the bound $$\left| |x+t|^p-|x|^p\right|\leq p2^{p-1}|t|(|t|^{p-1}+|x|^{p-1}),$$ but I don't see this working either, the limit is strictly positive.
What can I do?
Hints: $\frac d {dt} |x+t|^{p}|_{t=0} =p|x+t|^{p-1} s(x)$ where $s(x)=1$ if $x \geq 0$ and $s(x)=-1$ otherwise. Using the inequality you have stated in combination with D CT show that you can differentiate under the expectation. Finally. $Ep|X+t|^{p-1} s(X)=0$ because the distribution of $X$ is symmetric.