Trouble integrating tail bound I have the following bound on the deviation of a non-negative random variable:
$$\mathbb P (X \geq t) \leq ce^{-t^\alpha}$$
for some $\alpha \geq 1, c>1$.
I want to show that:
$$\mathbb E X \leq (\log c)^{1/\alpha}(1+1/(\alpha \log c))$$
My thought is to apply the equality:
$$\mathbb E X = \int_0^\infty \mathbb P(X \geq t)dt$$
and use the naive bound: $\min\{1, ce^{-t^\alpha}\}$.
But after splitting the integral into the two regions, I do not know how to handle the second term which comes out as:
$$\int_a^\infty ce^{-t^\alpha}dt$$
For $\alpha = 2$ I could use a mills ratio bound, but for other values of $\alpha \neq 1$ I do not know how to bound this integral.
 A: update: it seems that I had made a mistake in my substitution. After fixing it the result follows easily. See below.
Based on the form of the expectation bound, it makes sense to split the integral into two pieces before applying the tail bound. Also from the form of the expectation bound it makes sense to break the integral at the point $(\ln c)^{1/\alpha}$.
We have
\begin{align*}
\mathbb{E} X 
&= \int_0^{\infty} \mathbb{P}(X \ge t) dt\\ 
&= \int_0^{(\ln c)^{1/\alpha}} \mathbb{P}(X \ge t)dt + 
\int_{(\ln c)^{1/\alpha}}^{\infty} \mathbb{P}(X \ge t)dt\\
&\le (\ln c)^{1/\alpha} + \int_{(\ln c)^{1/\alpha}}^{\infty} c\exp({-t^\alpha})dt.
\end{align*}
The first term follows since $\mathbb{P}(X\ge t) \le 1$, and the second term follows from the tail bound. So we have to bound this second integral. Substituting $u=t^\alpha$ gives
\begin{align*}
\int_{(\ln c)^{1/\alpha}}^{\infty} c\exp({-t^\alpha})dt
&= \int_{\ln c}^{\infty} \frac{c}{\alpha} u^{\frac{1}{\alpha}-1} e^{-u}du \\
&\le  (\ln c)^{\frac{1}{\alpha}-1} \frac{c}{\alpha}  
\int_{\ln c}^{\infty}e^{-u}du \\
&=  (\ln c)^{\frac{1}{\alpha}-1} \frac{c}{\alpha}  
\frac{1}{c}\\
&= \frac{(\ln c)^{1/\alpha}}{\alpha\ln c}.
\end{align*}
Therefore,
$$
\mathbb{E} X \le (\ln c)^{1/\alpha} \left ( 1+\frac{1}{\alpha\ln c} \right ).
$$
