11: The Delta Integral
- Page ID
- 204829
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Assume \(f: \mathbb{T} \rightarrow \mathbb{R}\) is a regulated function. Any function \(F\) as in Theorem 10.8 is called a preantiderivative of \(f\). We define the indefinite integral of a regulated function \(f\) by
\[\int f(t) \Delta t=F(t)+c\notag \]
where \(c\) is an arbitrary constant and \(F\) is a pre-antiderivative of \(f\). We define the Cauchy integral by
\[\int_s^t f(\tau) \Delta \tau=F(t)-F(s) \quad \text { for all } \quad s, t \in \mathbb{T}\notag \]
A function \(F: \mathbb{T} \rightarrow \mathbb{R}\) is called an antiderivative of \(f: \mathbb{T} \rightarrow \mathbb{R}\) provided
\[F^{\Delta}(t)=f(t) \quad \text { holds for all } \quad t \in \mathbb{T}^\kappa .\notag \]
Let \(\mathbb{T}=\mathbb{Z}\). Then \(\sigma(t)=t+1, t \in \mathbb{T}\). Assume \(f(t)=3 t^2+5 t+2, t \in \mathbb{T}\). Since \(g(t)=t^3+t^2\) satisfies
\[\begin{aligned}
g^{\Delta}(t) & =(\sigma(t))^2+t \sigma(t)+t^2+\sigma(t)+t \\
& =(t+1)^2+t(t+1)+t^2+t+1+t \\
& =t^2+2 t+1+t^2+t+t^2+2 t+1 \\
& =3 t^2+5 t+2
\end{aligned}\notag \]
we have
\[\int\left(3 t^2+5 t+2\right) \Delta t=t^3+t^2+c\notag \]
Let \(\mathbb{T}=2^{\mathbb{N}}\) and define \(f: \mathbb{T} \rightarrow \mathbb{R}\) by \(f(t)=\dfrac{2}{t} \sin \dfrac{t}{2} \cos \dfrac{3 t}{2}, t \in \mathbb{T}\). In this case, we have that \(\sigma(t)=2 t\). Since \(g(t)=\sin t\) satisfies
\[\begin{aligned}
g^{\Delta}(t) & =\dfrac{\sin \sigma(t)-\sin t}{\sigma(t)-t} \\
& =\dfrac{\sin (2 t)-\sin t}{t} \\
& =\dfrac{2}{t} \sin \dfrac{t}{2} \cos \dfrac{3 t}{2}
\end{aligned}\notag \]
we get
\[\int \dfrac{2}{t} \sin \dfrac{t}{2} \cos \dfrac{3 t}{2} \Delta t=\sin t+c\notag \]
Let \(\mathbb{T}=\mathbb{N}_0^2\) and define \(f: \mathbb{T} \rightarrow \mathbb{R}\) by \(f(t)=\dfrac{1}{1+2 \sqrt{t}} \log \dfrac{(1+\sqrt{t})^2}{t}, t \in \mathbb{T}\). Here, \(\sigma(t)=(1+\sqrt{t})^2\).
Since \(g(t)=\log t\) satisfies
\[\begin{aligned}
g^{\Delta}(t) & =\dfrac{\log \sigma(t)-\log t}{\sigma(t)-t} \\
& =\dfrac{\log (1+\sqrt{t})^2-\log t}{(1+\sqrt{t})^2-t} \\
& =\dfrac{1}{1+2 \sqrt{t}} \log \dfrac{(1+\sqrt{t})^2}{t}
\end{aligned}\notag \]
we get
\[\int \dfrac{1}{1+2 \sqrt{t}} \log \dfrac{(1+\sqrt{t})^2}{t} \Delta t=\log t+c\notag \]
Let \(\mathbb{T}=\left\{-\dfrac{1}{n}: n \in \mathbb{N}\right\} \cup \mathbb{N}_0\) and
\[f(t)=\left\{\begin{array}{lll}
\dfrac{6 t^2+9 t-7}{(1+7 t)(1+6 t)} & \text { if } & t \in\left\{-\dfrac{1}{n}: n \in \mathbb{N}\right\} \\
\dfrac{7 t^2+9 t-6}{(1+7 t)(8+7 t)} & \text { if } & t \in \mathbb{N}_0
\end{array}\right.\notag \]
Set
\[\dfrac{1+t^2}{1+7 t}, \quad t \in \mathbb{T}\notag \]
By Example 7.8, we get
\[F^{\Delta}(t)=f(t), \quad t \in \mathbb{T}\notag \]
Therefore
\[\int f(t) \Delta t=\dfrac{1+t^2}{1+7 t}+C, \quad t \in \mathbb{T}\notag \]where \(C\) is a constant.
Let \(\mathbb{T}=\left\{\left(\dfrac{1}{2}\right)^{2^n}: n \in \mathbb{N}_0\right\} \cup\{0,1\}\) and
\[f(t)=\left\{\begin{array}{l}
\dfrac{23}{(4+5 t)(4+5 \sqrt{t})} \quad \text { if } \quad t \in\left\{\left(\dfrac{1}{2}\right)^{2^n}: n \in \mathbb{N}\right\} \\
\dfrac{46}{117} \quad \text { if } \quad t=\dfrac{1}{2} \\
\dfrac{23}{16} \quad \text { if } \quad t=0 .
\end{array}\right.\notag \]
Set
\[F(t)=\dfrac{1+7 t}{4+5 t}, \quad t \in \mathbb{T}\notag \]
By Example 7.9, we get
\[F^{\Delta}(t)=f(t), \quad t \in \mathbb{T}\notag \]
Therefore
\[\int f(t) \Delta t=\dfrac{1+7 t}{4+5 t}+C, \quad t \in \mathbb{T}\notag \]
where \(C\) is a constant.
Let \(U=\left\{\dfrac{1}{2^n}: n \in \mathbb{N}\right\}\) and
\[\mathbb{T}=U \cup(1-U) \cup(1+U) \cup(2-U) \cup(2+U) \cup\{0,1,2\},\notag \]
and
\[f(t)=\left\{\begin{array}{l}
-\dfrac{151}{64} \quad \text { if } \quad t=\dfrac{1}{2} \\
-\dfrac{1195}{64} \quad \text { if } \quad t=\dfrac{3}{2} \\
-15 t^3-3 t \quad \text { if } \quad t \in U \backslash\left\{\dfrac{1}{2}\right\} \\
-\dfrac{15 t^3+11 t^2+17 t+5}{8} \quad \text { if } \quad t \in(1-U) \backslash\left\{\dfrac{1}{2}\right\} \\
-15 t^3+17 t^2-10 t+2 \quad \text { if } \quad t \in(1+U) \backslash\left\{\dfrac{3}{2}\right\} \\
-\dfrac{15 t^3+22 t^2+32 t+8}{8} \quad \text { if } \quad t \in(2-U) \backslash\left\{\dfrac{3}{2}\right\} \\
-15 t^3+34 t^2-31 t+10 \quad \text { if } \quad t \in(2+U) \backslash\left\{\dfrac{5}{2}\right\} \\
0 \quad \text { if } \quad t=0 \\
-6 \text { if } \quad t=1 \\
-36 \text { if } \quad t=2 .
\end{array}\right.\notag \]
Set
\[F(t)=\left(2+t^2\right)\left(1-t^2\right), \quad t \in \mathbb{T} .\notag \]
By Example 7.10, we get
\[F^{\Delta}(t)=f(t), \quad t \in \mathbb{T}^\kappa\notag \]
Therefore
\[\int f(t) \Delta t=\left(2+t^2\right)\left(1-t^2\right)+C, \quad t \in \mathbb{T}^\kappa\notag \]
where \(C\) is a constant.
Let \(\mathbb{T}=\mathbb{N}_0^3\). Prove that
\[
\int\left(2 t+3 \sqrt[3]{t^2}+3 \sqrt[3]{t}+2\right) \Delta t=t^2+t+c
\notag \]
- Answer
-
Add texts here. Do not delete this text first.
Every rd-continuous function \(f: \mathbb{T} \rightarrow \mathbb{R}\) has an antiderivative. In particular, if \(t_0 \in \mathbb{T}\), then \(f\) defined by
\[F(t)=\int_{t_0}^t f(\tau) \Delta \tau \quad \text { for } \quad t \in \mathbb{T},\notag \]
is an antiderivative of \(f\).
Proof. Since \(f\) is rd-continuous, it is regulated. Let \(F\) be a function guaranteed to exist by Theorem 0.13, together with \(D\), satisfying
\[F^{\Delta}(t)=f(t) \quad \text { for } \quad t \in D\notag \]
We have that \(F\) is pre-differentiable with \(D\). Let \(t \in \mathbb{T}^\kappa \backslash D\). Then \(t\) is right-dense. Since \(f\) is rd-continuous, \(f\) is continuous at \(t\). Let \(\varepsilon>0\) be arbitrarily chosen. Then there exists a neighbourhood \(U\) of \(t\) such that
\[|f(s)-f(t)| \leq \varepsilon \quad \text { for all } \quad s \in \notag \]
We define
\[h(\tau)=F(\tau)-f(t)\left(\tau-t_0\right) \quad \text { for } \quad \tau \in \mathbb{T} .\notag \]
Then \(h\) is pre-differentiable with \(D\) and
\[\begin{aligned}
h^{\Delta}(\tau) & =F^{\Delta}(\tau)-f(t) \\
& =f(\tau)-f(t) \quad \text { for all } \quad \tau \in D
\end{aligned}\notag \]
Hence,
\[\begin{aligned}
\left|h^{\Delta}(s)\right| & =|f(s)-f(t)| \\
& \leq \varepsilon \quad \text { for all } \quad s \in D \cap U
\end{aligned}\notag \]
Therefore,
\[\sup _{s \in D \cap U}\left|h^{\Delta}(s)\right| \leq \varepsilon,\notag \]
whereupon
\[\begin{aligned}
|F(t)-F(r)-f(t)(t-r)|= & \mid h(t)+f(t)\left(t-t_0\right)-(h(r) \\
& \left.+f(t)\left(r-t_0\right)\right)-f(t)(t-r) \mid \\
= & |h(t)-h(r)| \\
\leq & \left\{\sup _{s \in D \cap U}\left|h^{\Delta}(s)\right|\right\}|t-r| \\
\leq & \varepsilon|t-r|
\end{aligned}\notag \]
which shows that \(f \) is differentiable at \(t\) and \(F^{\Delta}(t)=f(t)\).