Consider $X_{1}$ is a random variable that admits the symmetric distribution, that is $X_{1}=_{d}-X_{1}$ (or $\mathbb{P}(X_{1}>x)=\mathbb{P}(X_{1}<-x)\ \text{for all}\ x$) such that $\mathbb{P}(|X_{1}|>x)=x^{-2}$ for $x\geq 1$.
I'd like to show that $\mathbb{E}X_{1}=0$. I had some attempts but got stuck. To show this, recall that we have a lemma:
for $p>0$, and $Y\geq 0$, we have $$\mathbb{E}Y^{p}=\int_{0}^{\infty}py^{p-1}\mathbb{P}(Y>y)dy.$$
Thus, $$\mathbb{E}X_{1}\leq|\mathbb{E}X_{1}|\leq\mathbb{E}|X_{1}|=\int_{0}^{\infty}\mathbb{P}(|X_{1}|>x)dx.$$
However, the problem here is that $\mathbb{P}(|X_{1}|>x)=x^{-2}$ only for $x\geq 1$. So how could I compute the integral from $0$ to $\infty$?
Thank you!
Edit 1:
I have some progress as follows:
As $x\mapsto\mathbb{P}(|X_{1}|>x)$ is a decreasing function, we have $$\int_{0}^{\infty}\mathbb{P}(|X_{1}|>x)dx\leq\sum_{k=1}^{\infty}\mathbb{P}(|X_{1}|>k)=\sum_{k=1}^{\infty}\dfrac{1}{k^{2}}=\dfrac{\pi^{2}}{6},$$ which is not helpful here...
Since $X\overset{d}{=}-X$, $\mathsf{E}X=\mathsf{E}(-X)=-\mathsf{E}X$ (the tail condition ensures that $X\in L^1$).