Let $X$ be a random variable s.t. for $v, b > 0$ and $C \geq 1$:


$$ P(X \geq t) \leq C\exp\left(-\frac{t^2}{2(v^2 + bt)} \right) $$


I am trying to show that $\mathbb{E}X \leq 2v(\sqrt{\pi} + \sqrt{\log C}) + 4b(1 + \log C)$. 


In terms of progress I have made so far, I set $X = \int\_0^\infty P(X \geq t)dt$, and have also made cases for when $v^2 > bt$ and otherwise (i.e. simplifying the exponentiated value to be $-t^2/4v^2$ in one case and $-t^2/4bt$ in the other). 


This obtainis all the non $C$ values in the final inequality; However, I can't figure out how to squash the $C$ into the $\log C$ and $\sqrt{\log C}$ terms.



Bounding integral arising from expectation of a random variable satisfying Bernstein's inequality
For $t>0$, we have $v^2+bt\le\max(2bt,2v^2)$, whence 
\begin{equation}
 P(X\ge t)\le1\wedge\max[Q\_1(t),Q\_2(t)]\le[1\wedge Q\_1(t)]+[1\wedge Q\_2(t)], 
\end{equation}
where $x\wedge y:=\min(x,y)$, 
\begin{equation}
Q\_1(t):=C\exp\Big(-\frac{t^2}{4bt}\Big),\quad 
 Q\_2(t):=C\exp\Big(-\frac{t^2}{4v^2}\Big).
\end{equation} So, 
\begin{equation}
 EX=\int\_0^\infty P(X\ge t)\,dt\le I\_1+I\_2,
\end{equation}
where 
\begin{multline}
 I\_1:=\int\_0^\infty[1\wedge Q\_1(t)]\,dt
 =4b\ln C+C\int\_{4b\ln C}^\infty \exp\Big(-\frac{t}{4b}\Big)\,dt=4b(1+\ln C)
\end{multline}
and 
\begin{multline}
 I\_2:=\int\_0^\infty[1\wedge Q\_2(t)]\,dt
 =v\sqrt 2\int\_0^\infty\min(1,Ce^{-s^2/2})\,ds
\\ 
 =v\sqrt 2\Big(\sqrt{2\ln C}+C\int\_{\sqrt{2\ln C}}^\infty e^{-s^2/2}\,ds\Big)
 \le 2v\big(\sqrt{\ln C}+\sqrt\pi/2\big); 
\end{multline} 
the latter inequality follows because $\int\_t^\infty e^{-s^2/2}\,ds\le\sqrt{\pi/2}\,e^{-t^2/2}$ for $t\ge0$. 


Thus, 
\begin{equation}
 EX\le 4b(1+\ln C)+2v\big(\sqrt{\ln C}+\sqrt\pi/2\big),
\end{equation}
which is a bit better than the bound requested in the OP, where we see $\sqrt\pi$ in place of $\sqrt\pi/2$.


