I am trying to find the -1 moments of sum of N geometric random variable, i.e. $E[\frac{1}{\sum\_{i=1}^N X\_i}]$


Suppose the probability mass function is $f\_X(x) = (1 - p)^{x - 1} p$


The moment generating function of $\sum\_{i = 1}^N X\_i$ will be


$E[e^{t \sum X\_i}] = ( \frac{pe^t}{1 - (1-p)e^t})^N$


Let $h(t) = \frac{pe^t}{1 - (1-p)e^t}$, the -1 moments of $\sum\_{i = 1}^N X\_i$, will just be $\int h(t)^N dt |\_{t = 0}$. This is where I got stuck....



Suppose I know $\int h(t) dt = H(t)$, is there a way to find $\int h(t)^N dt$?
If you set $u=e^t$, we get
$$
\int \left(\frac{pu}{1-(1-p)u}\right)^N\frac{du}{u}
$$
According to Wolfram Alpha, with $p=1/2$,
$$
\int \left(\frac{e^t}{2 (1-e^t/2)}\right)^N dt = \left(\frac{e^t}{2-e^t}\right)^N (1-e^t/2)^n \frac{{}\_2F\_1(n, n, n+1, e^t/2))}{n}+\text{constant}
$$
where ${}\_2F\_1$ is the [hypergeometric function](http://en.wikipedia.org/wiki/Hypergeometric_function).


