# Suppose I know $\int h(t) dt = H(t)$, is there a way to find $\int h(t)^N dt$?

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....

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Hmm, I am not sure why all the downvotes, is this too easy somehow? – Bjørn Kjos-Hanssen Apr 15 '14 at 18:23

## 1 Answer

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.

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Bjørn, it works! Thank you! – Lei Yao Apr 15 '14 at 17:53