Let's assume $\xi$ is an integrable random variable with density $f(x)$, $\xi > 0$ almost surely. In other words, $\int\limits_0^{\infty}xf(x)dx<\infty$, $f(x)=0$ $\forall x<0$. Let F(x) be a CDF of $\xi$. In case all further integrals exist, we can write: $$ \int\limits_{0}^{\infty}(1-F(x))dx=\left.(1-F(x))x\right|_0^{\infty}+\int\limits_0^{\infty}xf(x)dx $$ So if $\lim\limits_{x\rightarrow\infty}(1-F(x))\cdot x=0$, we can safely assume that $\int\limits_{0}^{\infty}(1-F(x))dx=E(\xi)$
The question is: do we need any additional requirements (beside abovementioned integrability and non-negativity) to be sure that $\lim\limits_{x\rightarrow\infty}(1-F(x))\cdot x=0$? If we do, are there some more "natural" requirements than simply stating the value of this limit?