Chapter 14: Improper Integrals of the Second Kind
- Page ID
- 210055
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In the following situation, the ordinary Riemann integral of \(f\) on \([a,b]\) cannot exist since a Riemann integrable function from \(a\) to \(b\) must be bounded on \([a,b)\).
Let \(\mathbb{T}\) be a time scale, \(a,b\in\mathbb{T}\) with \(a<b\). Suppose that \(b\) is left-dense. Assume that the function \(f\) is defined in the interval \([a,b)\). Suppose that \(f\) is integrable on any interval \([a,c]\) with \(c<b\)and is unbounded on \([a,b)\). The formal expression
\begin{equation}\label{175}
\int_a^b f(t)\Delta t
\end{equation}
is called the improper integral of the second kind. We say that the integral \eqref{175} is improper at \(t=b\). We also say that \(f\) has a singularity at \(t=b\). If the left-sided limit
\begin{equation}\label{176}
\lim_{c\to b-}\int_a^c f(t)\Delta t
\end{equation}
exists as a finite number, then the improper integral \eqref{175} is said to exist or be convergent. In such a case, we call this limit the value of the improper integral \eqref{175} and write
\[ \int_a^b f(t)\Delta t=\lim_{c\to b-}\int_a^c f(t)\Delta t. \]
If the limit \eqref{176} does not exist, then the integral \eqref{175} is said to be not existent or divergent.
Let \(\mathbb{T}=[0,1]\cup 2^{\mathbb{N}}\), where \([0,1]\) is the real-valued interval. Define
\[ f(t)=\begin{cases}
\sqrt{1-t^2} & \quad\mbox{for}\quad t\in[0,1]\\
t^4 & \quad\mbox{for}\quad t\in 2^{\mathbb{N}}.
\end{cases} \]
Consider the integral
\[ I=\int_0^8\frac{1}{f(t)}\Delta t. \]
We have
\[\begin{aligned}
I
=& \int_0^1\frac{1}{f(t)}\Delta t+\int_2^8\frac{1}{f(t)}\Delta t\\
=& \int_0^1\frac{1}{\sqrt{1-t^2}}dt+\int_2^8\frac{1}{t^4}\Delta t\\
=& \lim_{c\to 1-}\int_0^c\frac{1}{\sqrt{1-t^2}}dt
+\frac{1}{t^4}\mu(t)\Bigl|_{t=2}+\frac{1}{t^4}\mu(t)\Bigl|_{t=4}\\
=& \lim_{c\to 1-}\arcsin t\Bigl|_{t=0}^{t=c}+\frac{2}{16}+\frac{4}{256}\\
=& \frac{\pi}{2}+\frac{9}{64}.
\end{aligned}\notag\]
Therefore, the considered integral is convergent.
Let \(\mathbb{T}=\{-4,-2\}\cup[0,1]\), where \([0,1]\) is the real-valued interval. Consider the integral
\[ I=\int_{-4}^1\frac{\Delta t}{\sqrt{1-t}}. \]
We have
\[\begin{aligned}
I
=& \int_{-4}^{-2}\frac{\Delta t}{\sqrt{1-t}}
+\int_0^1\frac{dt}{\sqrt{1-t}}\\
=& \frac{1}{\sqrt{1-t}}\mu(t)\Bigl|_{t=-4}
+\lim_{c\to 1-}\int_0^c\frac{dt}{\sqrt{1-t}}\\
=& \frac{2}{\sqrt{5}}-2\lim_{c\to 1-}\sqrt{1-t}\Bigl|_{t=0}^{t=c}\\
=& \frac{2}{\sqrt{5}}+2.
\end{aligned}\notag\]
Therefore, the considered integral is convergent.
Let \(\mathbb{T}=\{-1,0\}\cup[1,2]\), where \([1,2]\) is the real-valued interval. Consider the integral
\[ I=\int_{-1}^2\frac{t^3}{\sqrt{4-t^2}}\Delta t. \]
We have
\[\begin{aligned}
I
=& \int_{-1}^0\frac{t^3}{\sqrt{4-t^2}}\Delta t
+\int_1^2 \frac{t^3}{\sqrt{4-t^2}}dt\\
=& \frac{t^3\mu(t)}{\sqrt{4-t^2}}\Bigl|_{t=-1}
-\lim_{c\to 2-}\int_1^c t^2d\sqrt{4-t^2}\\
=& -\frac{1}{\sqrt{3}}-\lim_{c\to 2-}t^2\sqrt{4-t^3}\Bigl|_{t=1}^{t=c}
+2\lim_{c\to 2-}\int_1^ct\sqrt{4-t^2}dt\\
=& -\frac{1}{\sqrt{3}}+\sqrt{3}
-\lim_{c\to 2-}\int_1^c\sqrt{4-t^2}{\mathbb{R}m d}(4-t^2)\\
=& \frac{2\sqrt{3}}{3}-\lim_{c\to 2-}
\frac{(4-t^2)^{\frac{3}{2}}}{\frac{3}{2}}\Bigl|_{t=1}^{t=c}\\
=& \frac{2\sqrt{3}}{3}+2\sqrt{3}\\
=& \frac{8\sqrt{3}}{3}.
\end{aligned}\notag\]
Therefore, the considered integral is convergent.
Investigate the following integrals for convergence and divergence.
1. \(\int_{-3}^1\frac{\Delta t}{(1-t)(2t-1)}\), \(\mathbb{T}=\{-3,-2,-1,0\}\cup\left[\frac{1}{2},1\right]\), where \(\left[\frac{1}{2},1\right]\) is the real-valued interval.
2. \(\int_{-3}^{10}\frac{2t}{(t^2-1)^2}\Delta t\), \(\mathbb{T}=[-3,3]\cup\{4,7,10\}\), where \([-3,3]\) is the real-valued interval.
3. \(\int_{-3}^2\frac{\Delta t}{t\sqrt{3t^2-2t-1}}\), \(\mathbb{T}=\{-3,-2,-1,0\}\cup[1,2]\), where \([1,2]\) is the real-valued interval.
4. \(\int_{-3}^3\frac{\Delta t}{(t-7)\sqrt{t^2-3}}\), \(\mathbb{T}=\{-3,-2,-1\}\cup[\sqrt{3},3]\), where \([\sqrt{3},3]\) is the real-valued interval.
5. \(\int_{-7}^1\frac{\Delta t}{\sqrt[3]{t(1-t)}}\), \(\mathbb{T}=\{-7,-4,-1\}\cup[0,1]\), where \([0,1]\) is the real-valued interval.
6. \(\int_{-\frac{1}{2}}^1\frac{\Delta t}{(10-t)\sqrt{1-t^2}}\), \(\mathbb{T}=\left\{-\frac{1}{2},-\frac{1}{4},0\right\}\cup\left[\frac{1}{2},1\right]\), where \(\left[\frac{1}{2},1\right]\) is the real-valued interval.
Answer
1. Divergent,
2. divergent,
3. convergent,
4. convergent,
5. convergent,
6. convergent.
All theorems for improper integrals of the first kind have exact analogues for improper integrals of the second kind.
1. For the existence of the integral \eqref{175}, it is necessary and sufficient that for any given \(\varepsilon>0\), there exists \(b_0<b\) such that
\[ \left|\int_{c_1}^{c_2}f(t)\Delta t\right|<\varepsilon \]
for any \(c_1,c_2\in\mathbb{T}\) satisfying the inequalities
\(b_0<c_1<b\) and \(b_0<c_2<b\).
2. Suppose that \(f(t)\geq 0\). Then, for any \(c\in[a,b]\),
\[ F(c)=\int_a^c f(t)\Delta t \]
does not decrease as \(c\) increases, and the integral \eqref{175} is convergent if and only if \(f\) is bounded, in which case the value of the integral is \(\lim_{c\to b-}F(c)\).
3. Let the limit
\[ \lim_{t\to b-}\frac{f(t)}{g(t)}=L \]
exist (finite) and suppose it is not zero.
Then the integrals \(\int_a^b f(t)\Delta t\) and \(\int_a^b g(t)\Delta t\) are simultaneously convergent or divergent.
Similar definitions are made and entirely similar results are obtained for integrals of the second kind that are improper at the lower limit of integration.
Let \(\mathbb{T}\) be a time scale satisfying \eqref{177}. Investigate the integral
\[ \int_a^b(t-a)^{\alpha}(b-t)^{\beta}\Delta t \]
for convergence and divergence.
Answer
Convergent for \(\alpha>-1\) and \(\beta>-1\), divergent for \(\alpha\leq -1\) or \(\beta\leq -1\).
Let \(\mathbb{T}\) be an arbitrary time scale, \(a,b\in\mathbb{T}\) with \(a<b\), and suppose that \(b\) is left-dense. Let \(p\geq 1\). We prove that the integral
\begin{equation}\label{178}
\int_a^b\frac{\Delta t}{(b-t)^p}
\end{equation}
is divergent.
1. Let \(p=1\). Let us choose points \(t_n\in\mathbb{T}\) for \(n\in\mathbb{N}_0\) such that
\begin{equation}\label{177}
a=t_0<t_1<\ldots<b \quad\mbox{and}\quad \lim_{n\to\infty}t_n=b.
\end{equation}
We set
\begin{equation}\label{179}
\tau_n=\frac{1}{b-t_n} \quad\mbox{for any}\quad n\in\mathbb{N}_0.
\end{equation}
Then \(\lim_{n\to\infty}\tau_n=\infty\), \(t_n=b-\frac{1}{\tau_n}\), and
\[\begin{aligned}
t_{n+1}-t_n
=& \frac{1}{\tau_n}-\frac{1}{\tau_{n+1}}\\
=& \frac{\tau_{n+1}-\tau_n}{\tau_n\tau_{n+1}} \quad\mbox{for all}\quad n\in\mathbb{N}_0.
\end{aligned}\notag\]
Hence,
\[\begin{aligned}
\int_a^b\frac{\Delta t}{b-t}
=& \sum_{n=0}^{\infty}\int_{t_n}^{t_{n+1}}\frac{\Delta t}{b-t}\\
&\geq & \sum_{n=0}^{\infty}\frac{1}{b-t_n}\int_{t_n}^{t_{n+1}}\Delta t\\
=& \sum_{n=0}^{\infty}\frac{t_{n+1}-t_n}{b-t_n}\\
=& \sum_{n=0}^{\infty}\tau_n\frac{\tau_{n+1}-\tau_n}{\tau_n\tau_{n+1}}\\
=& \sum_{n=0}^{\infty}\frac{\tau_{n+1}-\tau_n}{\tau_{n+1}}\\
=& \infty.
\end{aligned}\notag\]
2. Let \(p>1\). There exists \(d\in[a,b)\) such that
\[ 0<b-t<1 \quad\mbox{for}\quad t\in[d,b). \]
Then
\[ (b-t)^p<b-t \quad\mbox{for}\quad t\in[d,b). \]
Hence,
\[\begin{aligned}
\int_a^b\frac{\Delta t}{(b-t)^p}
=& \int_a^d\frac{\Delta t}{(b-t)^p}+\int_d^b\frac{\Delta t}{(b-t)^p}\\
&>& \int_a^d\frac{\Delta t}{(b-t)^p}+\int_d^b\frac{\Delta t}{b-t}\\
=& \infty.
\end{aligned}\notag\]
Let \(\mathbb{T}\) be a time scale satisfying \eqref{177}. We consider the integral
\[ I=\int_a^b\frac{\Delta t}{(t^4+t^2+1)(b-t)^{\frac{1}{2}}}. \]
We have
\[I\leq\int_a^b\frac{\Delta t}{(b-t)^{\frac{1}{2}}}<\infty. \]
Let \(\mathbb{T}\) be a time scale satisfying \eqref{177}. Assume \(a=0,b=2,\frac{1}{2},1\in\mathbb{T}\). We consider the integral
\[ I=\int_0^2\frac{t^{\alpha-1}}{|1-t|}\Delta t. \]
We have
\[ I=\int_0^{\frac{1}{2}}\frac{t^{\alpha-1}}{1-t}\Delta t
+\int_{\frac{1}{2}}^1\frac{t^{\alpha-1}}{1-t}\Delta t
+\int_1^2\frac{t^{\alpha-1}}{t-1}\Delta t. \]
Since \[\int_{\frac{1}{2}}^1\frac{t^{\alpha-1}}{1-t}\Delta t\]
is divergent for all \(\alpha\in\mathbb{R}\), we conclude that the integral \(I\) is divergent.
Let \(\mathbb{T}\) be a time scale satisfying \eqref{177}. Assume \(a=0,b=1,\frac{1}{2}\in\mathbb{T}\). Consider the integral
\[ I=\int_0^1t^{\alpha-1}(1-t)^{\beta-1}\Delta t. \]
We have
\[\begin{aligned}
I
=& \int_0^{\frac{1}{2}}t^{\alpha-1}(1-t)^{\beta-1}\Delta t
+\int_{\frac{1}{2}}^1t^{\alpha-1}(1-t)^{\beta-1}\Delta t\\
=& I_1+I_2.
\end{aligned}\notag\]
Note that \(I_1\) is convergent for \(\alpha>0\) and divergent for \(\alpha\leq 0\). Also, \(I_2\) is convergent for \(\beta>0\) and divergent for \(\beta\leq 0\). Therefore, \(I\) is convergent for \(\alpha>0\) and \(\beta>0\), and \(I\) is divergent for \(\alpha\leq 0\) or \(\beta\leq 0\).
Let \(\mathbb{T}\) be a time scale satisfying \eqref{177}. Investigate the integral
\[\int_a^b(t-a)^{\alpha}(b-t)^{\beta}\Delta t \]
for convergence and divergence.
Answer
Convergent for \(\alpha>-1\) and \(\beta>-1\), divergent for \(\alpha\leq -1\) or \(\beta\leq -1\).