How to solve an expression with inverse of exponential distribution

[user_01]

I have an Energy harvesting expression something like the following
$R = \tau B \log\Big(1 + \frac{E h^2}{\tau r^\alpha\sigma^2} \Big)$
$E = \tau(2^{R/\tau B}-1 )\frac{r^\alpha\sigma^2}{h^2}$

Let all constant terms as $a$ to simplify the expression into : $E = a\frac{1}{h^2}$
$E$ is a random variable because of $h^2\sim\exp(1)$ i.e. $h$ is exponentially distributed random variable with unit parameter.

Consider $x= h^2 \sim \exp(1)$,
$\mathbb{E}[E] = E_x= a \mathbb{E}\Big[\frac{1}{x}\Big] = a\int_0^\infty \frac{e^{-x}}{x} \to \infty\ (\text{Divergent})$

I have to solve this problem because my circuit cannot harvest an infinite amount of energy (this is impractical).
How can I solve this problem? Any way around to find the approximate solution?

I will really appreciate any suggestions.

 

[Jhero]

One possible solution is to use a truncated exponential distribution. This means that you can bound the range of possible values for h and use the truncated distribution to compute the expected value of E.For example, if you set a hard upper bound of h at h_max, then you can compute the expected value of E as: $E_{x|h<h_{max}} = a \mathbb{E}\Big[\frac{1}{x}\Big]_{x<h_{max}^2} = a\int_0^{h_{max}^2} \frac{e^{-x}}{x}$This should give you a finite expected value of E that is much better than the original divergent result.