Moment generating function of a distribution I want to find the moment generating function (mfg) and mean deviation of this distribution:
$$f(x,\epsilon,k,\theta) = k\theta^{(1+1/k+\epsilon/k)}x^{(k+\epsilon)}\exp{(-\theta x^k )}/(\Gamma(1+(1+\epsilon))/k)$$
where $\epsilon, k, \theta$ are the three parameters of this distribution. $x$ ranges from $0$ to infinity.
 A: This is a Generalized Gamma distribution: up to a scale factor, $x^k$ has a Gamma distribution. 
E. W. Stacy studied the Generalized Gamma distribution in a 1962 paper available on Project Euclid.  His parameterization is more natural for statistical applications and leads to simpler formulas.  The density function is determined by the density for a unit scale factor $a=1$ by
$$f(x;1,d,p)\mathrm{d}x =\frac{1}{C(p,d)} x^d \exp(-x^p) \frac{\mathrm{d}x}{x}$$
for $0 \lt x \lt \infty$ (see Stacy equation (1)).  The constant of proportionality is readily found via the substitution $x^p=u,$ yielding
$$C(p,d) = \int_0^\infty x^d \exp(-x^p)\frac{\mathrm{d}x}{x} =  \frac{\Gamma(d/p)}{p}.$$
To find the $r^\text{th}$ moment, integrate $x^r$ against $f\mathrm{d}x.$ No additional calculations are needed, because the preceding formula already gives the result upon replacing $d$ by $r+d$:
$$\mu_r(p,d) = \frac{1}{C(p,d)} \int_0^\infty x^{r+d} \exp(-x^p) \frac{\mathrm{d}x}{x} = \frac{C(p, r+d)}{C(p, d)} = \frac{\Gamma((r+d)/p)}{\Gamma(d/p)}.$$
By definition, these give the terms in the expansion of the moment generating function (mgf),
$$M_{p,d}(t) = \sum_{r=0}^\infty \frac{\mu_r(p,d)}{r!} t^r = \frac{1}{\Gamma(d/p)} \sum_{r=0}^\infty \frac{\Gamma((r+d)/p)}{r!} t^r.$$
(See Stacy equation (3).)  As usual, the scale factor $a \gt 0$ is accommodated by replacing $t$ by $a t$ in the mgf.
The parameterization given in the question identifies $k$ with $p,$ $1+k+\epsilon$ with $d$, and $1/\theta^{1/k}$ with $a.$
I presume the "mean deviation" refers to some function of these moments, but since I'm unsure which one, I will leave the algebra to the reader.
