Find the moment generating function of the random variable

Question

$f_Y(y)=e^{-3y}+\frac{2}{3}e^{-y}$ if $y>0$ and $0$ else
a)find the MGF $M_Y(t)$ of Y (be careful to declare the domain)
The MGF is $E[e^{tY}]$ but is that $\int_{0}^{\infty} e^{ty}f(y) dy$ because when I tried that I got a messy integral so I feel like Im doing something wrong

Answer 1

$$\begin{align} M_{Y}(t) &= \int\limits_{0}^{\infty}e^{ty}(e^{-3y}+\dfrac{2}{3}e^{-y})\text{ d}y \\ &= \int\limits_{0}^{\infty}e^{ty}e^{-3y}\text{ d}y + \dfrac{2}{3}\int\limits_{0}^{\infty}e^{ty}e^{-y}\text{ d}y \\ &= \int\limits_{0}^{\infty}e^{ty-3y}\text{ d}y + \dfrac{2}{3}\int\limits_{0}^{\infty}e^{ty-y}\text{ d}y \\ &= \int\limits_{0}^{\infty}e^{y(t-3)}\text{ d}y + \dfrac{2}{3}\int_{0}^{\infty}e^{y(t-1)}\text{ d}y \\ &= \int\limits_{0}^{\infty}e^{-y(3-t)}\text{ d}y + \dfrac{2}{3}\int\limits_{0}^{\infty}e^{-y(1-t)}\text{ d}y \\ &= \dfrac{1}{3-t} + \dfrac{2}{3}\left(\dfrac{1}{1-t}\right)\text{ if } t < 1\text{.} \end{align}$$