Symmetric fat-tailed distributions where $\mathbb{E} e^X < \infty$ What "fat-tailed distributions" $p(x)$, symmetric about zero, have the property
$$\newcommand{\e}{\mathbb{E}}\newcommand{\rd}{\mathrm{d}}
\e e^X = \int_{-\infty}^{\infty} e^x p(x) \rd x < \infty \> ?
$$
Context
I'm attempting to price financial options for $X$ without using the Black–Scholes formula. It is usually easier to work with the log-price $Y = \log(X)$ and often it is assumed that $Y$ is normally distributed.
Empirical observations (eg, the "volatility smile") suggest that $Y$ isn't normal; the normal distribution decreases too rapidly away from 0. Thus, we need a fat-tailed distribution. 
The value of a call option increases exponentially as $Y$ increases linearly. Therefore, $p(x)$ must decline fast enough that $\e e^X < \infty$. In other words, the distribution must decline slower than the normal distribution, but still fast enough that $\e e^X$ converges. 
I tried the Cauchy and Student's $t$ distributions, but $\e e^X$ diverges for both, regardless of parameters. 
I also realize I can create arbitrary distributions meeting my conditions (though I'm not exactly sure how), but I'm looking for a well-known (parametrized family of) distribution. 
Even more details (for the masochist): 
https://github.com/barrycarter/bcapps/blob/master/bc-imp-vol.m 
 A: The definition of fat-tail in wikipedia is that 
$$p(x)\sim x^{-(\alpha+1)}$$
as $x\to\infty$ for some $\alpha>0$. Now 
$$\frac{e^x}{x^{\alpha+1}}\to\infty,$$
as $x\to\infty$, so the $Ee^X$ cannot exist for such type of distributions. So you need to precise what do you have in mind by saying fat-tailed.
A: The variance gamma process is a useful way to go.  It is an extension of the standard brownian motion process, $Z(t)$
$$Y(t;\mu,\sigma,\nu)=\mu \Gamma(t;1,\nu) + \sigma Z(\Gamma[t;1,\nu])$$
Where $\Gamma(t;1,\nu)$ is a gamma process, with independent gamma distributed increments.  So $\Gamma(t+s;1,\nu)-\Gamma(t;1,\nu)$ has a gamma distribution with mean $s$ and variance $s\nu$.  (you could replace the $1$ by another parameter $\mu$ if desired, to have mean $s\mu$, as is done here).  The intuition is that "time" is measured in some sort of trading volume rather than calendar time.  So it incorporates a sense of the market "getting hot" and "getting cold".
This process has the "fat tail" property that you seek, and is commonly used as an option pricing model.  It can be used by averaging the Black–Scholes option price with respect to gamma distribution for the time.
The increments of this distribution $Y(t)-Y(s)$ follow a Laplace distribution.
More generally one speaks of what are called Levy process, which have characteristic function $\varphi_{y}(\theta)=E[\exp(i\theta Y_{t})]$:
$$\varphi_{t}(\theta)=\exp\left(it\mu\theta - \frac{t}{2}\sigma^{2}\theta^{2}+t\int_{|x|<\epsilon}[e^{i\theta x}-1-i\theta x]d\Pi(x)+t\int_{|x|>\epsilon}[e^{i\theta x}-1]d\Pi(x)\right)$$
Where $d\Pi(x)$ is called the Levy measure, and governs discontinuities in the process (or the "jumps").  The measure must satisfy:
$$\int_{|x|<\epsilon}x^{2}d\Pi(x)<\infty$$
$$\int_{|x|>\epsilon}d\Pi(x)<\infty$$
Thus it can have a singularity at zero as long it is not "too big".  Can show for the variance gamma process the Levy measure is given by:
$$d\Pi(x)=\frac{dx}{\nu|x|}\exp(-\frac{1}{\nu}|x|)$$
UPDATE
In regards to Cardinal's well directed comments (I have a tendency to "wander-off" when I'm doing some sort of maths) you can see that the only criterion for your "fat tailed" distribution to satisfy your criterion is that the moment generating function $m_{X}(t)=E[\exp(tX)]$ exists when evaluated at $t=1$.  Equivalently we require the characteristic function to exist when evaluated at $\theta=-i$.  Because the Levy-process is defined by its CF then it is always finite when the two integral equations above are satisfied.  The expectation is given by the above formula
