Proof of finiteness of the following integral Assume a discrete positive random variable $\pmb{r}$ with probability mass function (p.m.f.) containing countably infinite number of mass points such that in any closed interval we have finite number of mass points. Denote the masses by $r_i,~i=1,...,\infty$ and corresponding probabilities by $p_i,~i=1,...,\infty$.
For a given number $a$, the random variable $\pmb{r}$ satisfies the following 
$E[\pmb{r}^2]\leq a$.
Denote $I_0(\alpha)$ as the modified Bessel function of first kind of order zero, i.e., 
$I_0(\alpha)=\frac{1}{\pi}\int\limits_{0}^{\pi}e^{\alpha\cos(\theta)}d\theta$.
For a given positive number $r$, consider the following integral 
$\int\limits_0^{\infty}Re^{-\frac{1}{2}(R^2+r^2)}I_0(rR) \ln\left\{\sum\limits_{i=1}^{\infty}e^{-\frac{r_i^2}{2}}p_iI_0(r_iR)\right\}dR\triangleq F(r)$.
I am trying to prove that $F(r)< c+o(r)$, where $c$ is a constant and $o(r)$ goes to infinity with $r$. I would really appreciate if anyone can help me about this. 
 A: Thanks for those who spent time on this. Finally, I have been able to solve this problem. The solution is as follows. Please feel free to correct me if you see something wrong.
Since $\ln x < x$ for $x>0$, we have
\begin{align}
&\int\limits_0^{\infty}Re^{-\frac{1}{2}(R^2+r^2)}I_0(rR) \ln\left\{\sum\limits_{i=1}^{\infty}e^{-\frac{r_i^2}{2}}p_iI_0(r_iR)\right\}dR \\
&< \int\limits_0^{\infty}Re^{-\frac{1}{2}(R^2+r^2)}I_0(rR) \cdot \left\{\sum\limits_{i=1}^{\infty}e^{-\frac{r_i^2}{2}}p_iI_0(r_iR)\right\}dR\\
&=e^{-\frac{r^2}{2}}\sum\limits_{i=1}^{\infty}p_ie^{-\frac{r_i^2}{2}}\int\limits_0^{\infty}Re^{-\frac{1}{2}R^2}I_0(rR) \cdot I_0(r_iR)dR.
\end{align}
Due to $I_0(x)<e^x$ we have 
\begin{align}
&e^{-\frac{r^2}{2}}\sum\limits_{i=1}^{\infty}p_ie^{-\frac{r_i^2}{2}}\int\limits_0^{\infty}Re^{-\frac{1}{2}R^2}I_0(rR) \cdot I_0(r_iR)dR\\
&<e^{-\frac{r^2}{2}}\sum\limits_{i=1}^{\infty}p_ie^{-\frac{r_i^2}{2}}\int\limits_0^{\infty}Re^{-\frac{1}{2}R^2+(r+r_i)R}dR\\
&=e^{-\frac{r^2}{2}}\sum\limits_{i=1}^{\infty}p_ie^{-\frac{r_i^2}{2}}\left\{1+e^{\frac{(r+r_i)^2}{2}}\sqrt{\frac{\pi}{2}}(r+r_i)\left(1+erf\left(\frac{r+r_i}{2}\right)\right)\right\}\\
&\leq \sum\limits_{i=1}^{\infty}p_i\left\{1+2\sqrt{\frac{\pi}{2}}(r+r_i)\right\}\\
&=1+r+2\sqrt{\frac{\pi}{2}}E[\pmb{r}].
\end{align}
Since the random variable $\pmb{r}$ has finite second moment, therefore we have $E[\pmb{r}]\leq \sqrt{E[\pmb{r}^2]}=\sqrt{a}$. Therefore,
\begin{align}
1+r+2\sqrt{\frac{\pi}{2}}E[\pmb{r}]\leq1+\sqrt{a}+2\sqrt{\frac{\pi}{2}}r.
\end{align}
The result is therefore,
$F(r)<1+\sqrt{a}+2\sqrt{\frac{\pi}{2}}r $.
Note that the inequality is strict. 
