Derivation of PMF of Poisson Distribution from its Characteristic Function I came across a question which asked to obtain the probability function of $X$ (a discrete random variable) with its characteristic function given as follows:
$${\phi _X}(t) = {e^{\lambda ({e^{it}} - 1)}}$$
I know that this is the characteristic function of a Poisson Distribution. So,
$$X \sim Poi(\lambda )$$
However, I was unable to show this mathematically. I started to answer this question as follows:
$$P(X = x) = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi  {{e^{ - itx}}{\phi _X}(t)dt} $$
$$ = \frac{1}{{2\pi }}{e^{ - \lambda }}\int\limits_{ - \pi }^\pi  {{e^{ - itx}}{e^{\lambda {e^{it}}}}dt} $$
But after this step I am unable to figure how will be evaluate the integral.
Can someone please suggest how should go ahead after this step?
Thanks in advance!
 A: The formula I used (see exercise $26.12$, Probability and Measure by Patrick Billingsley), similar to the celebrated inversion formula, is (the formula that you gave can be derived from it):
$$P[X = a] = \lim_{T \to \infty} \frac{1}{2T}\int_{-T}^T e^{-ita}\phi_X(t) dt.$$
Notice that $\phi_X(t) = e^{-\lambda}\sum_{k = 0}^\infty\frac{\lambda^k e^{itk}}{k!}$. Evaluate the integral by Fubini's theorem
\begin{align}
& \int_{-T}^T e^{-ita}\phi_X(t) dt \\
= & \int_{-T}^T e^{-ita}e^{-\lambda}\sum_{k = 0}^\infty\frac{\lambda^k e^{itk}}{k!} dt \\
= & e^{-\lambda}\sum_{k = 0}^\infty\frac{\lambda^k}{k!}\int_{-T}^Te^{it(k - a)}dt \\
= & 2Te^{-\lambda}\frac{\lambda^a}{a!} + 2e^{-\lambda}\sum_{k \neq a}\frac{\lambda^k\sin[(k - a)T]}{k!(k - a)}
\end{align}
where we used that if $k = a$, then $\int_{-T}^T e^{it(k - a)} dt = 2T$, and if $k \neq a$, 
$$\int_{-T}^T e^{it(k - a)} dt = 2\int_0^T \cos[(k - a)t] dt = \frac{2}{k - a}\sin[(k - a)T].$$
Notice by the dominated convergence theorem, 
$$\lim_{T \to \infty}\frac{1}{2T}\sum_{k \neq a}\frac{\lambda^k\sin[(k - a)T]}{k!(k - a)} = \sum_{k \neq a}\lim_{T \to \infty}\frac{\lambda^k\sin[(k - a)T]}{2k!(k - a)T} = 0.$$
Therefore, $P[X = a] = e^{-\lambda}\frac{\lambda^a}{a!}$, for $a = 0, 1, 2, \ldots$, the proof is complete.
A: 
Given cf ${\phi _X}(t) = {e^{\lambda ({e^{it}} - 1)}}$, derive the pmf $P(X=x)$ ... 

There is a simpler way. For a discrete random variable $X$ taking non-negative integer values, the cf  $E[e^{i t X}]$ is related to the probability generating function pgf $E[s^{X}]$ via $s = e^{i t}$, so the pgf has form:
$$\Pi(s) = E\left[s^X\right] = e^{\lambda  (s-1)}$$
The pgf generates probabilities via the relation:
$$P(X=x) \quad = \quad \frac{1}{x!}\frac{d^x\Pi(s)}{ds^x}|_{s=0} \quad \text{for} \quad x\in \{0,1,2,\ldots \}$$
The $n^{th}$ derivative of  $e^{\lambda  (s-1)}$ wrt $s$ has form $\lambda ^n e^{\lambda  (s-1)}$, and when $s = 0$, the latter has form:  $e^{-\lambda } \lambda ^n$. Replacing $n$ with $x$, the pmf is thus:
$$P(X=x) = \frac{e^{-\lambda } \lambda ^x}{x!} \quad \text{for} \quad  x\in \{0,1,2,\ldots \}$$
