Expectation of a distribution involving exponential $ f(y) = \begin{cases} e^y, & \text{if $y<0$} \\\ 0, & \text{if elsewhere } \end{cases}$ Find $E[e^{\frac{3Y}{2}}]$. I wrote the integration $\int^{\infty}_{-\infty} e^{\frac{5y}{2}} dy$ Then how do I proceed. Kind request: Please give an elaborated answer because I am asked the moment generating function in the next part.

Recall that for a measurable function $g$, and a random variable $Y$ with density $f_Y$ we have $$\mathbb E[g(Y)] = \int_{\mathbb R}g(y)f_Y(y)\,\mathsf dy, $$ provided that $\mathbb E[|g(Y)|]<\infty$. In this case, $f_Y(y) = e^y\cdot\mathsf 1_{(-\infty,0)}(y)$ and $g(y) = e^{ty}$, where $t\in\mathbb R$. We compute \begin{align} \mathbb E\left [e^{tY}\right] &= \int_{\mathbb R} e^{ty}e^y\cdot\mathsf 1_{(-\infty,0)}(y)\,\mathsf dy\\\ &= \int_{-\infty}^0 e^{y(1+t)}\,\mathsf dy\\\ &= \frac1{1+t}, \quad t>-1. \end{align} When $t=\frac32$, we have $$\mathbb E\left[e^{\frac32Y} \right] = \frac1{1+\frac32} = \frac25. $$