Let $X$ be a Poisson variable with
$$P(X=n) = \frac{\lambda ^{n}}{n!}e^{-\lambda }, \lambda >0 , n = 0,1,2,...$$
Obtain a closed-form expression for $E[{e^{iuX}}]$ for $u$ real, and $i=\sqrt{-1}$.
I'm not really sure how to approach this problem. :( Any help would be appreciated. Thanks in advance!
This is known as the characteristic function of a distribution. It is essentially the Fourier transform of the distribution. Since the Fourier transform is a 1-1 map, the characteristic function is an equivalent representation of the distribution, and is convenient for some operations.
In this case, for a given $u$, you evaluate (1) $$ \sum_{n=0}^{\infty} e^{i u n} \frac{\lambda^n}{n!} e^{-\lambda} = e^{-\lambda} \big[\sum_{n=0}^{\infty} \frac{{(e^{i u}\lambda)}^n}{n!} \big]$$
Remember that $e^{-\lambda} {\lambda}^n/n!$ is a distribution, and sums to 1, That is, $$ \sum_{n=0}^{\infty}e^{-\lambda} \frac{{\lambda}^n}{n!} = 1, \quad \forall \lambda \textrm{ such that the sum converges}.$$ In particular, replacing $\lambda$ by $e^{iu}\lambda$ leads us to $$e^{-(e^{i u}\lambda)} \sum_{n=0}^{\infty} \frac{{(e^{i u}\lambda)}^n}{n!} = 1,$$ $$\sum_{n=0}^{\infty} \frac{{(e^{i u}\lambda)}^n}{n!} = e^{(e^{i u}\lambda)}.$$ Substitute this in (1) leads us to famous expression that the characteristic polynomial of the Poisson is:
$$ e^{\lambda (e^{i u} -1)}$$