Expectation Random Variables Say $X$ to be uniformly distributed from $[0,1]$. Say $k_1$ and $k_2$ to be two non negative constants (that is, they take values from $[0,+inft]$.
I want to compute the expectation of the following max involving the two constants:
$$E[\max(X+k_1,k_2)]$$
(that is, the expectation of the max between the sum "random variable plus a constant k1" and "a constant k2").  
As someone suggested:
let $Y=X+k_1$ be the uniform variable on $[k_1,k_1+1]$. Then your variable is $\max \{ Y,k_2 \}$. Cases:


*

*If $k_2 < k_1$, then $\max \{ Y,k_2 \} = Y$, so you have $E(Y)=\frac{2k_1+1}{2}$.

*If $k_2 > k_1+1$, then $\max \{ Y,k_2 \}=k_2$, so you have just $k_2$. 

*If $k_1 \leq k_2 \leq k_1+1$, then $\mathbb{P}(Y \leq k_2) = k_2-k_1$.


how to handle the last case?
 A: Let $Y=X+k_1$ be the uniform variable on $[k_1,k_1+1]$. Then your variable is $\max \{ Y,k_2 \}$. Cases:


*

*If $k_2 < k_1$, then $\max \{ Y,k_2 \} = Y$, so you have $E(Y)=\frac{2k_1+1}{2}$.

*If $k_2 > k_1+1$, then $\max \{ Y,k_2 \}=k_2$, so you have just $k_2$. 

*If $k_1 \leq k_2 \leq k_1+1$, then $\mathbb{P}(Y \leq k_2) = k_2-k_1$.


Can you handle the last case?
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

Note that $\ds{\max\pars{a,b} = {a + b + \verts{a - b} \over 2}}$. Then,

\begin{align}&\color{#66f}{\large{\mathbb E}\bracks{\max\pars{X + k_{1},k_{2}}}}
={\mathbb E}\bracks{X + k_{1} + k_{2} + \verts{X + k_{1} - k_{2}} \over 2}
\\[5mm]&={1 \over 4} + {k_{1} + k_{2} \over 2}
+\half\,\color{#c00000}{{\mathbb E}\pars{\verts{X + k_{1} - k_{2}}}}
\end{align}

\begin{align}&\color{#c00000}{{\mathbb E}\bracks{\verts{X + k_{1} - k_{2}}}}
=\int_{0}^{1}\verts{X + k_{1} - k_{2}}\,\dd X
\\[5mm]&=\verts{1 + k_{1} - k_{2}}
-\int_{0}^{1}X\sgn\pars{X + k_{1} - k_{2}}\,\dd X
\\[5mm]&=\verts{1 + k_{1} - k_{2}} - \half\,\sgn\pars{1 + k_{1} - k_{2}}
+ \int_{0}^{1}X^{2}\,\delta\pars{X + k_{1} - k_{2}}\,\dd X 
\\[5mm]&=\verts{1 + k_{1} - k_{2}} - \half\,\sgn\pars{1 + k_{1} - k_{2}}
+ \pars{k_{2} - k_{1}}^{2}
\int_{0}^{1}\delta\pars{X - \bracks{k_{2} - k_{1}}}\,\dd X 
\\[5mm]&=\color{#c00000}{%
\verts{1 + k_{1} - k_{2}} - \half\,\sgn\pars{1 + k_{1} - k_{2}}
+ \pars{k_{2} - k_{1}}^{2}\Theta\pars{k_{2} - k_{1}}\Theta\pars{1 - k_{2} + k_{1}}}
\end{align}

\begin{align}
&\color{#66f}{\large{\mathbb E}\bracks{\max\pars{X + k_{1},k_{2}}}}
\\[5mm]&=\color{#66f}{\large{1 \over 4} + {k_{1} + k_{2} \over 2}
+\half\,\verts{1 + k_{1} - k_{2}} - {1 \over 4}\,\sgn\pars{1 + k_{1} - k_{2}}}
\\[5mm]&+ \color{#66f}{\large\half\,\pars{k_{2} - k_{1}}^{2}
\Theta\pars{k_{2} - k_{1}}\Theta\pars{1 - k_{2} + k_{1}}}
\end{align}

$\ds{\delta\pars{x}}$ is the
  Dirac Delta Function and $\ds{\Theta\pars{x}}$ is the
  Heaviside Step Function .

