Let $Y$ denote a random variable with probability density function given by $$f(y) = \frac{1}{2}e^{-|{y}|}, \quad -\infty < y < \infty$$
Find the moment generating function of $Y$.
\begin{align}M_y(t) = E(e^{ty})&=\frac{1}{2}\int^\infty_{0}e^{-y}e^{ty}dy +\frac12 \int^0_{-\infty}e^{y}e^{ty}dy\\&=\frac{1}{2}\int^\infty_{0}e^{y(t-1)}dy + \frac12\int^0_{-\infty}e^{y(t+1)}dy \\&=\frac{1}{2}\left(\frac{e^{\infty(t-1)}}{t-1}-\frac{1}{t-1}+\frac{1}{t+1}-\frac{e^{-\infty(t+1)}}{t+1}\right)\end{align}
I am currently stuck here. The results will differ according to the value of $t$. I've heard from another source that we can assume $t<1$. Is the claim valid? Am I missing another step?
If $(t-1)>0$, then $e^{\infty(t-1)}=+\infty$ and if $t+1<0$ then $e^{-\infty(t+1)}=+\infty$ and everything goes wrong (i.e. the integrals do not converge). You need to exclude this cases for the integrals to converge. So you must (not can) assume that $t-1<0 \iff t<1$ and $t+1>0 \iff t>-1$. If you combine these two you get that the moment generating function is defined for $|t|<1$.