Why is $\int_0^t ds \int_0^{t'} ds' \delta(s-s')= \min(t,t')$

Question

I don't understand this equation $\int_0^t ds \int_0^{t'} ds' \delta(s-s')= \min(t,t')$. I tried to work with the property of the dirac delta function that $\int_a^b \delta(x-c)dx = 1$ if $c \in [a,b]$, but I can't see how I can obtain the minimum. Can someone help me?

Thank you in advance!

Answer 1

let $\chi_t(x)$ be the indicator function for the interval $[0,t]$. then, as you point out: $$ \int_0^{t'}ds'\delta(s-s') = \chi_{t'}(s) $$ but now: $$ \int_0^t\chi_{t'}(s) ds = \int_0^{\infty} \chi_{t}(s)\chi_{t'}(s) ds = \int_0^{\infty} \chi_{\min(t,t')}(s) ds = \min(t,t') $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{\int_{b}^{c}\delta\pars{x - a}\dd x} = \ds{\bracks{b < a < c} - \bracks{c < a < b}}$ where $\ds{\bracks{\cdots}}$ is an Iverson Bracket.

\begin{align} &\int_{0}^{t}\dd s\int_{0}^{t'}\dd s'\,\delta\pars{s - s'} = \int_{0}^{t}\int_{0}^{t'}\delta\pars{s' - s}\dd s'\,\dd s \\[5mm] = &\ \int_{0}^{t}\bracks{0 < s < t'}\dd s - \int_{0}^{t}\bracks{t' < s < 0}\dd s = \int_{0}^{t}\bracks{0 < s < t'}\dd s + \int_{t}^{0}\bracks{t' < s < 0}\dd s \\[1cm] = &\ \bracks{t > 0}\bracks{t' > 0}\braces{\vphantom{\Large A}% \bracks{t' < t}t' + \bracks{t' > t}t} \\[5mm] + &\ \bracks{t < 0}\bracks{t' < 0}\braces{\vphantom{\Large A}% \bracks{t' < t}\pars{-t} + \bracks{t' > t}\pars{-t'}} \\[1cm] = &\ \bracks{t > 0}\bracks{t' > 0}\min\braces{t,t'} - \bracks{t < 0}\bracks{t' < 0}\max\braces{t,t'} \\[5mm] = &\ \bracks{t > 0}\bracks{t' > 0}\min\braces{t,t'} + \bracks{t < 0}\bracks{t' < 0}\min\braces{-t,-t'} \\[5mm] = &\ \braces{\vphantom{\Large A}% \bracks{t > 0}\bracks{t' > 0} + \bracks{t < 0}\bracks{t' < 0}} \min\braces{\verts{t},\verts{t'}} \\[5mm] = &\ \bbox[#ffe,20px,border:2px dotted navy]{% \ds{\bracks{tt' > 0}\min\braces{\verts{t},\verts{t'}}}} \end{align}

Answer 3

Using the notation for positive and negative parts, we calculate

$$I(t,t^{\prime})~:=~ \int_0^t \!\mathrm{d}s \int_0^{t^{\prime}} \! \mathrm{d}s^{\prime} ~\delta(s\!-\!s^{\prime}) ~=~{\rm sgn}(t)~{\rm sgn}(t^{\prime}) ~J(t,t^{\prime}), \tag{1}$$ where $$J(t,t^{\prime})~:=~ \iint_{\mathbb{R^2}}\!\mathrm{d}s~\mathrm{d}s^{\prime} ~1_{[-t^{-},t^{+}]}(s)~1_{[-t^{\prime -},t^{\prime +}]}(s^{\prime})~\delta(s\!-\!s^{\prime})~=~ \int_{\mathbb{R}}\!\mathrm{d}s ~1_{[-t^{-},t^{+}]}(s)~1_{[-t^{\prime -},t^{\prime +}]}(s)$$ $$ ~=~ \int_{\mathbb{R}}\!\mathrm{d}s ~1_{[-\min(t^{-},t^{\prime -}),\min(t^{+},t^{\prime +})]}(s) ~=~\min(t^{-},t^{\prime -})+\min(t^{+},t^{\prime +}).\tag{2}$$ Combining eqs. (1) & (2), we find that OP's double integral reads $$I(t,t^{\prime})~=~\theta(tt^{\prime})\min(|t|,|t^{\prime}|),\tag{3}$$ where $\theta$ denotes the Heaviside step function.