Second moment bound from a clipped Gaussian tail ($\mathbb{E}{Z}^2\le (1+\log A)/b$)

If $\mathbb{P}(|Z|\ge t)\le A{e}^{-b{t}^2}$ for $A\ge 1$ and $b>0$, then $\mathbb{E}{Z}^2\le (1+\log A)/b$. Indeed the tail integral formula for moments gives $\mathbb{E}{Z}^2\le {\int }_0^{\infty }\min (1,A{e}^{-bu})du$; split at $u=(\log A)/b$. Clipping at one improves the constant compared with integrating the exponential bound alone.