Tail probability integral I'm trying to show that the following integral is finite:
\begin{equation}
\displaystyle\int_0^\infty{e^xP\left(\sum_{i=1}^k{\ln(X_i)}>x\right)}dx
\end{equation}where I actually know the density of the iid random variables $X_i$. It is proportional to $x^{-\alpha},\hspace{3pt}2<\alpha<3$, for $x\geq M>0$
I can show the finiteness of the integral for $k=1$, but for higher values of $k$ I'm trying to use 
$$ P\left(\sum_{i=1}^k{\ln(X_i)}>x\right)\leq kP\left(\sum_{i=1}^k{\ln(X_i)}>\frac{x}{k}\right)$$
but it does not work. 
I believe this should hold true, so can a smarter person suggest a more clever inequality to use when dealing with $k>1$?
 A: For any real valued random variable $Z$,
$$
\int_0^{+\infty}\mathrm e^x\,\mathbf 1_{Z\geqslant x}\,\mathrm dx=\mathbf 1_{Z\geqslant0}\int_0^Z\mathrm e^x\,\mathrm dx=(\mathrm e^Z-1)^+.
$$
Using this for the random variable $Z=\sum\limits_{i=1}^k\log Y_i$ and integrating it with respect to $\mathbb P$, one sees that the integral to compute is 
$$
\mathbb E\left(\left(\prod_{i=1}^kY_i-1\right)^+\right).
$$
This expectation is finite if and only if $\prod\limits_{i=1}^kY_i$ is integrable at $+\infty$. Since the support of each $Y_i$ is bounded below and the random variables $Y_i$ are independent, this happens if and only if every $Y_i$ is integrable. If the density of each $Y_i$ is of order $x^{-\alpha}$ at $+\infty$ with $\alpha\gt2$, this is so.
The result holds for every density such that every $Y_i$ is integrable, for example a density of order $x^{-2}(\log x)^{-\beta}$ at $+\infty$, with $\beta\gt1$.
A: $\sum_1^k\log X_i>x\Leftrightarrow\prod_{i=1}^kX_i^p>e^{px}$, hence by independence,
$$P\left(\sum_1^k\log X_i>x\right)\leqslant e^{-px}\prod_{i=1}^kE(X_i^p).$$
The assumption about density gives that $E(X_i^p)$ is finite for some $p>1$ and the conclusion follows.
