Bernstein tail bound to expectation Let $Z$ be a nonnegative r.v. satisfying
\begin{align*}
P(Z \geq t) \leq C e^{-\frac{t^2}{2(v^2+bt)}}
\end{align*}
where $(v,b)$ are positive constants and $C \geq 1$. Show that
\begin{align*}
E(Z) \leq 2v(\sqrt{\pi}+\sqrt{\log C} ) + 4b(1+\log C).
\end{align*}
I've tried expanding $EZ = \int_0^\infty P(Z \geq t) dt$ and bounding the exponential on different subsets of the positive real line but the end expression still seems mysterious to me.
 A: To see where this type of bound comes from, start with the simpler case when $v=0$. Let $C'=\log C$, so that the hypotheses reads
$$
\mathbb P(Z\geq t)\leq \exp\left[C'-\frac{t}{2b}\right].
$$
The exponent is zero when $t=2bC'$, so this is a natural place to split the integral. Thus,
$$
\mathbb EZ\leq \int_0^{2bC'}1\ dt+\int_{2bC'}^{\infty}e^{C'-t/2b}\ dt=2bC'+2b = 2b(1+\log C).
$$

Returning to the problem at hand, we start using a crude union-type bound on the tail probability itself by observing that either $v^2\leq bt$ or $v^2\geq bt$, and thus
$$
\begin{align*}
\mathbb P(Z\geq t)&\leq C\exp\left[-\frac{t^2}{2(v^2+bt)}\right]\\
&\leq C\max \left\{\exp\left(-\frac{t}{4b}\right),C\exp\left(-\frac{t^2}{4v^2}\right)\right\}\\
&\leq C\exp\left(-\frac{t}{4b}\right)+C\exp\left(-\frac{t^2}{4v^2}\right).
\end{align*}
$$
Combining this inequality with the trivial one $\mathbb P(Z\geq t)\leq 1$ yields that
$$
\mathbb P(Z\geq t)\leq 1\wedge C\exp\left(-\frac{t}{4b}\right)+1\wedge C\exp\left(-\frac{t^2}{4v^2}\right),
$$
where $1\wedge A$ is a shorthand for $\min(1,A)$.
Our expectation bound will be obtained by writing
$$
\mathbb EZ=\int_0^{\infty}\mathbb P(Z\geq t)\ dt\leq I_1+I_2,
$$
where $$I_1=\int_0^\infty 1\wedge C\exp\left(-\frac{t}{4b}\right)\ dt,\qquad I_2=\int_0^\infty 1\wedge C\exp\left(-\frac{t^2}{4v^2}\right)\ dt.$$
To evaluate $I_1$, solve the equation $Ce^{-t_1/4b}=1$ to obtain $t_1=4b\log C$. Thus
$$
I_1=\int_0^{t_1}1\ dt+\int_{t_1}^{\infty}C\exp\left(-\frac{t}{4b}\right)\ dt=t_1+4b=4b(1+\log C).
$$
To evaluate $I_2$, solve the equation $Ce^{-(t_2/2v)^2}=1$ to obtain $t_2=2v\sqrt{\log C}$. Thus
$$
I_2=\int_0^{t_2}1\ dt+\int_{t_2}^{\infty}C\exp\left(-\frac{t^2}{4v^2}\right)\ dt=t_2+2v\int_{\sqrt{\log C}}^{\infty}Ce^{-t^2}\ dt,
$$
after a change of variables $t\mapsto t/2v$ in the second term. Putting together everything so far, we see that the desired bound reduces to showing that 
$$
\int_{\sqrt{\log C}}^{\infty}Ce^{-t^2}\ dt\leq \sqrt{\pi}.
$$
To obtain this final bound, we will use the Gaussian integral $\int_{-\infty}^\infty e^{-x^2}\ dx=\sqrt{\pi}.$ In fact, we will choose the change of variables $x=t-\sqrt{\log C}$ and observe that $$t^2-x^2=(x+\sqrt{\log C})^2-x^2=\log C + 2x\sqrt{\log C}\geq \log C,\qquad x\geq 0.$$
Equivalently, this reads $Ce^{-t^2}\leq e^{-x^2}$ for all $t\geq \sqrt{\log C}$, and therefore we have shown that
$$
\int_{\sqrt{\log C}}^{\infty}Ce^{-t^2}\ dt\leq \int_0^{\infty}e^{-x^2}\ dx=\frac{\sqrt{\pi}}{2}.
$$
Putting it all together, we have obtained the bound
$$
\mathbb EZ\leq 2v\left(\frac{\sqrt{\pi}}{2}+\sqrt{\log C}\right)+4b(1+\log C),
$$
which is slightly better than the desired bound of
$$
\mathbb EZ\leq 2v\left(\sqrt{\pi}+\sqrt{\log C}\right)+4b(1+\log C).
$$
