$X$ and $Y$ are two random variables. $X$ with the parameter $\delta$ is a exponetial distributed random variable. And $ Y = \max(X, 1/\delta)$. Without going in detail I just want to know how do I get from
$\mathbf{E}({\max(X, 1/\delta)}) = \int^{+\infty}_{-\infty} \max(X, 1/ \delta) * f_x(x){\rm d}x$
to
$ =\int^{1/\delta}_{0} 1/\delta * f_x(x) {\rm d}x + \int^{+\infty}_{1/\delta} x * f_x(x){\rm d}x $
Since obviously: $$\max(X,1/\delta)=\begin{cases}(1/\delta)&:&X<1/\delta\\[1ex]X&:&X\geq 1/\delta\end{cases}$$
The integral can be separated into partitions, with the maximum taking on the relevant value for each partition.
$$\begin{align}\mathsf E(\max(X,1/\delta)) ~&=~ \mathsf E(\max(X,1/\delta)\mathbf 1_{X<1/\delta})+\mathsf E(\max(X,1/\delta)\mathbf 1_{X\geq 1/\delta})\\[1ex] &=~\mathsf E((1/\delta)\mathbf 1_{X<1/\delta})+\mathsf E(X\mathbf 1_{X\geq 1/\delta}) \\[2ex]\therefore\quad \int_0^\infty \max(x,1/\delta)\cdot f_X(x)\operatorname d x~ &=~\int_0^{1/\delta} (1/\delta)\cdot f_X(x)\operatorname d x+\int_{1/\delta}^\infty x\cdot f_X(s)\operatorname d x\end{align}$$