What is the value of this integral?

Question

Suppose that $f(z)=1/(b-a) , a≤z≤b$, and $f(z)=0$ , otherwise.

Now, I want to find the following integral: $$g(x,y)=\int_{-\infty} ^\infty f(x-z)f(y-z)dz$$

I would appreciate if you could help me. Thanks.

Answer 1

Note that $$\int_\mathbb R\mathbf 1_{x-b\lt z\lt x-a}\mathbf 1_{y-b\lt z\lt y-a}\mathrm dz=\int_\mathbb R\mathbf 1_{\max(x,y)-b\lt z\lt\min(x,y)-a}\mathrm dz $$ hence $$ g(x,y)=\frac{(b-a-|x-y|)^+}{(b-a)^2}. $$

Answer 2

Hints:

$$f(x-z)f(y-z)\neq 0\iff a\le x-z\,,\,y-z\le b\iff \begin{cases}x-b\le z\le x-a&,\;\;\text{and also}\\{}\\y-b\le z\le y-a\end{cases}$$

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} {\rm g}\pars{x,y} &= \int_{-\infty}^{\infty}\fermi\pars{x - z}\fermi\pars{y - z}\,\dd z = \int_{-\infty}^{\infty}\fermi\pars{x - y - z}\fermi\pars{-z}\,\dd z = \int_{-\infty}^{\infty}\fermi\pars{-\bracks{y - x} + z}\fermi\pars{z}\,\dd z \\[3mm]&=\int_{-\infty}^{\infty}\fermi\pars{z - \mu}\fermi\pars{z}\,\dd z ={1 \over b - a}\int_{a}^{b}\fermi\pars{z - \mu}\,\dd z =\left.{1 \over \pars{b - a}^{2}}\int_{a}^{b}\,\dd z \right\vert_{\mu + a\ <\ z\ <\ \mu + b}\quad \\&\mbox{where}\quad\mu \equiv y - x \end{align}

1. $\mu + b < a\quad\imp\quad{\rm g}\pars{x,y} = 0$
2. $a < \mu + b < b\quad\imp\quad {\rm g}\pars{x,y} = \ds{\mu + b - a \over \pars{b - a}^{2}}$
3. $a < \mu + a < b\quad\imp\quad {\rm g}\pars{x,y} = \ds{b - \mu - a \over \pars{b - a}^{2}}$
4. $\mu + a > b\quad\imp\quad{\rm g}\pars{x,y} = 0$

$$ {\rm g}\pars{x,y} =\left\lbrace% \begin{array}{lcrcccl} {b - a - \pars{x - y} \over \pars{b - a}^{2}} & \mbox{if} & a - b & < & y - x & \leq & 0 \\[2mm] {b - a - \pars{y - x} \over \pars{b - a}^{2}} & \mbox{if} & 0 & < & y - x & \leq & b - a \\[2mm] 0 & \mbox{otherwise} &&&&& \end{array}\right. $$

$$\color{#00f}{\large% {\rm g}\pars{x,y} = {b - a - \verts{x - y} \over \pars{b -a}^{2}} \quad\mbox{if}\quad\verts{x - y} \leq b - a\,,\qquad \color{#000}{\Large 0}\quad\mbox{otherwise}} $$