Let $X$ be a r.v. on $(\mathbb{R},\mathcal{B}(\mathbb{R}), m)$ with $m$ the Lebesgue measure. Assume that $X$ has a law $\mu$ on $\mathbb{R}$ and assume that $X$ is both symmetric (also known as an even) and has finite polynomial moments of any order. Why is it then the case that $\mathbb{E}(X) = 0$? I am currently reading through a small note on probability theory which makes this claim, but it is not immediately obvious to me why $\mathbb{E}(X) = 0$. The only "work" that I can include to this post is that since $X(-c) = X(c)$, we have that
$$\mathbb{E}(X) = \int_{\mathbb{R}}X(x)dx = 2\int_0^\infty X(x)dx$$
and that for any $n\in\mathbb{N}_{\geq 0}$ we have
$$\mathbb{E}(X^n) = \int_{\mathbb{R}}X^n(x)dx < \infty$$
So if $\mathbb{E}(X) = c > 0$, maybe one could try to partition the level sets of $X$ with the $n$th root of $X$ to show some kind of contradiction with $\mathbb{E}(X) = c > 0$ and
$$\mathbb{E}(X^n) = \int_{\mathbb{R}}X^n(x)dx < \infty$$
But right now I do not see how to make the conclusion.