Expected Value and Swapping integral and sum Consider $$E[e^{tX}]=\int_{-\infty}^{\infty} e^{tx} f(x) dx, t \in \mathbb{R}$$
f is an arbitrary probability density function
I can write $e^{tx} $ in its series. Then I want so swap integral and sum
I thought about Lebesgue's theorem for swapping integral and sum.
I need a integrable function that is an upper bound for the integrand. But how can I find such a function? 
I want to get to : $$E(e^{tx}) = 1+ E(X)t + \frac{E(X^2)t^2}{2!}+.....$$
 A: ANSWER FOR EDITED QUESTION
If there exist $t_{0}>0$ such that 
$$ \forall_{ t\in [-t_{0},t_{0}]} \ \ \mathbb{E}[e^{tX}]<+\infty \  \ $$ then for all $t\in (-t_{0}, t_{0})$ we have

$E(e^{tx}) = 1+ E(X)t + \frac{E(X^2)t^2}{2!}+.....$

because for all $n\in \mathbb{N}$
$$\Bigg|\sum_{k=0}^{n}\frac{(tX)^{k}}{k!}\Bigg| \le \sum_{k=0}^{n}\Bigg|\frac{(tX)^{k}}{k!}\Bigg| \le \sum_{k=0}^{n}\frac{(t_{0}|X|)^{k}}{k!} \le e^{t_{0}|X|} \le e^{t_{0}X}+e^{-t_{0}X}$$ 
and $e^{t_{0}X}+e^{-t_{0}X}$ is integrable by assumption. 
OLD ANSWER FOR OLD QUESTION
Hence $f$ is probability density function, let us denote corresponding probability measure as $\mathbb{P}$. We have
$$\mathbb{E}e^{tX}=\int_{\Omega}e^{tX}d\mathbb{P} = \int_{\Omega} \sum_{n=0}^{\infty}\frac{(tX)^{k}}{k!}d\mathbb{P}=\int_{\Omega}\lim_{k\to \infty}\sum_{k=0}^{n}\frac{(tX)^{k}}{k!}d\mathbb{P},$$
and we are wondering if
$$\int_{\Omega}\lim_{n\to \infty}\sum_{k=0}^{n}\frac{(tX)^{k}}{k!}d\mathbb{P}\stackrel{?}{=}\lim_{n\to \infty}\int_{\Omega}\sum_{k=0}^{n}\frac{(tX)^{k}}{k!}d\mathbb{P}=\sum_{k=0}^{\infty}\int_{\Omega}\frac{(tX)^{k}}{k!}d\mathbb{P}.$$
So this is question about swapping integral and limit and we were asked to use Lebesgue's  majorization theorem. If $X$ would be bounded, then convergence of $\ \sum_{k=0}^{n}\frac{(tX)^{k}}{k!} \ $ to $e^{tX}$ would be uniform on $\Omega$. As is mentioned in here -
Dominated convergence theorem and uniformly convergence - in such case we can easily find integrable upper bound.
Another way is to use monotone convergence theorem. It would require $\ t$ and  $X \ $ being of the same fixed sign, in order to get summands positive, i.e.
$$\forall k  \ \ \frac{(tX)^k}{k!}\ge 0,$$
so $t\ge 0$ and $X\ge 0$ or $t \le 0$ and $X\le 0$. 
