Opening the book…
Throughout this section, we assume that \(\mathbb{T}\) is unbounded above.
Suppose that the real-valued function \(f\) is defined on \([a,\infty)\) and is integrable from \(a\) to any point \(A\in\mathbb{T}\), \(A\geq a\). Assume that the integral
\[F(A)=\int_a^Af(t)\Delta t \]
approaches a finite limit as \(A\to\infty\). Then we call that limit the improper integral of the first kind from \(a\) to \(\infty\) and write
\begin{equation}\label{158}
\int_a^{\infty}f(t)\Delta t
=\lim_{A\to\infty}\left\{\int_a^Af(t)\Delta t\right\}.
\end{equation}
In such a case, we say that the improper integral
\begin{equation}\label{159}
\int_a^{\infty}f(t)\Delta t
\end{equation}
exists or that it is convergent. If the limit \eqref{158} does not exist, then the improper integral \eqref{159} is said to be not existent or divergent.
Let \(\mathbb{T}=\mathbb{Z}\). We consider
\[ I=\int_1^{\infty}\frac{3t^2+3t+1}{t^3(t+1)^3}\Delta t. \]
We have \(\sigma(t)=t+1\) and
\[\begin{aligned}
\left(\frac{1}{t^3}\right)^\Delta
=& -\frac{(t^3)^\Delta}{t^3(\sigma(t))^3}\\
=& -\frac{(\sigma(t))^2+t\sigma(t)+t^2}{t^3(t+1)^3}\\
=& -\frac{(t+1)^23+t(t+1)+t^2}{t^3(t+1)^3}\\
=& -\frac{t^2+2t+1+t^2+t+t^2}{t^3(t+1)^3}\\
=& -\frac{3t^2+3t+1}{t^3(t+1)^3}.
\end{aligned}\notag\]
Therefore,
\[\begin{aligned}
\int_1^{\infty}\frac{3t^2+3t+1}{t^3(t+1)^3}\Delta t
=& \lim_{A\to\infty}\int_1^A\frac{3t^2+3t+1}{t^3(t+1)^3}\Delta t\\
=& -\lim_{A\to\infty}\int_1^A\left(\frac{1}{t^3}\right)^\Delta\Delta t\\
=& -\lim_{A\to\infty}\frac{1}{t^3}\Bigl|_{t=1}^{t=A}\\
=& -\lim_{A\to\infty}\left(\frac{1}{A^3}-1\right)\\
=& 1.
\end{aligned}\notag\]
Let \(\mathbb{T}=2^{\mathbb{N}_0}\). We consider
\[ I=\int_1^{\infty}\frac{1}{t^3}\Delta t. \]
Here, \(\sigma(t)=2t\) and
\[\begin{aligned}
\left(\frac{1}{t^2}\right)^\Delta
=& -\frac{(t^2)^\Delta}{t^2(\sigma(t))^2}\\
=& -\frac{\sigma(t)+t}{4t^4}\\
=& -\frac{2t+t}{4t^4}\\
=& -\frac{3}{4t^3},
\end{aligned}\notag\]
whereupon
\[ \frac{1}{t^3}=-\frac{4}{3}\left(\frac{1}{t^2}\right)^\Delta. \]
Therefore
\[\begin{aligned}
I
=& \lim_{A\to\infty}\int_1^A\frac{1}{t^2}\Delta t\\
=& -\frac{4}{3}\lim_{A\to\infty}
\int_1^A\left(\frac{1}{t^2}\right)^\Delta\Delta t\\
=& -\frac{4}{3}\lim_{A\to\infty}\frac{1}{t^2}\Bigl|_{t=1}^{t=A}\\
=& -\frac{4}{3}\lim_{A\to\infty}\left(\frac{1}{A^2}-1\right)\\
=& \frac{4}{3}.
\end{aligned}\notag\]
Let \(\mathbb{T}=3^{\mathbb{N}_0}\). We consider
\[ I=\int_1^{\infty}\frac{1}{\sqrt{t}}\Delta t. \]
Here, \(\sigma(t)=3t\) and
\[\begin{aligned}
(\sqrt{t})^\Delta
=& \frac{\sqrt{\sigma(t)}-\sqrt{t}}{\sigma(t)-t}\\
=& \frac{\sqrt{\sigma(t)}-\sqrt{t}}
{(\sqrt{\sigma(t)}-\sqrt{t})(\sqrt{\sigma(t)}+\sqrt{t})}\\
=& \frac{1}{\sqrt{\sigma(t)}+\sqrt{t}}\\
=& \frac{1}{(1+\sqrt{3})\sqrt{t}},
\end{aligned}\notag\]
whereupon
\[ \frac{1}{\sqrt{t}}=(1+\sqrt{3})(\sqrt{t})^\Delta. \]
Therefore,
\[\begin{aligned}
I
=& \lim_{A\to\infty}\int_1^A \frac{1}{\sqrt{t}}\Delta t\\
=& (1+\sqrt{3})\lim_{A\to \infty}\int_1^A(\sqrt{t})^\Delta\Delta t\\
=& (1+\sqrt{3})\lim_{A\to\infty}\sqrt{t}\Bigl|_{t=1}^{t=A}\\
=& (1+\sqrt{3})\lim_{A\to\infty}(\sqrt{A}-1)\\
=& \infty.
\end{aligned}\notag\]
Consequently, this improper integral diverges to \(\infty\).
Let \(\mathbb{T}=2\mathbb{Z}\).
Investigate the following integrals for convergence and divergence.
1. \(\int_1^{\infty}\frac{2t+2}{(t^2+1)(t^2+4t+5)}\Delta t\),
2. \(\int_2^{\infty}\frac{1}{4t^2+12t+5}\Delta t\),
3. \(\int_0^{\infty}\frac{1}{t^2+10t+24}\Delta t\).
Answer
1. \(\frac{1}{2}\),
2. \(\frac{1}{10}\),
3. \(\frac{1}{4}\).
For the existence of the integral \eqref{158}, it is necessary and sufficient that for any given \(\varepsilon>0\), there exists \(A_0>a\) such that
\begin{equation}\label{160}
\left|\int_{A_1}^{A_2}f(t)\Delta t\right|<\varepsilon
\end{equation}
for any \(A_1,A_2\in\mathbb{T}\) satisfying the inequalities \(A_1>A_0\) and \(A_2>A_0\).
Proof
The convergence of the integral \eqref{158} is equivalent to the existence of the limit \(\lim_{A\to\infty}F(A)\). Using Cauchy's criterion for the existence of the limit of a function, it follows that the existence of the integral \eqref{158} is equivalent to the condition \eqref{160}.
Let \(\mathbb{T}=3\mathbb{Z}\). We prove that the integral
\[ \int_1^{\infty}\frac{1}{t^2+11t+28}\Delta t \]
is convergent. Note that \(\sigma(t)=t+3\) and
\[\begin{aligned}
\left(\frac{1}{t+4}\right)^\Delta
=& -\frac{(t+4)^\Delta}{(t+4)(\sigma(t)+4)}\\
=& -\frac{1}{(t+4)(t+7)}\\
=& -\frac{1}{t^2+11t+28}.
\end{aligned}\notag\]
Let \(\varepsilon>0\) be arbitrarily chosen. We take
\[ A>\max\left\{\frac{2-4\varepsilon}{\varepsilon},1\right\}. \]
Then, for any \(A_1,A_2>A\), we have
\[\begin{aligned}
\left|\int_{A_1}^{A_2}\frac{1}{t^2+11t+28}\Delta t\right|
=& \left|-\int_{A_1}^{A_2}\left(\frac{1}{t+4}\right)^\Delta\Delta t\right|\\
=& \left|\frac{1}{t+4}\Bigl|_{t=A_1}^{t=A_2}\right|\\
=& \left|\frac{1}{A_2+4}-\frac{1}{A_1+4}\right|\\
\leq& \frac{1}{A_2+4}+\frac{1}{A_1+4}\\
<& \frac{2}{A+4}\\
<& \varepsilon.
\end{aligned}\notag\]
Therefore, employing the Cauchy criterion, we conclude that the considered integral is convergent.
Let \(\mathbb{T}=2^{\mathbb{N}_0}\). We investigate the integral
\[ \int_1^{\infty}\frac{3t+7}{2(t+2)(t+3)(t+4)(2t+3)}\Delta t \]
for convergence. Here, \(\sigma(t)=2t\) and
\[\begin{aligned}
\left(\frac{1}{(t+3)(t+4)}\right)^\Delta
=& -\frac{((t+3)(t+4))^\Delta}{(t+3)(t+4)(\sigma(t)+3)(\sigma(t)+4)}\\
=& -\frac{(t+3)^\Delta(t+4)+(\sigma(t)+3)(t+4)^\Delta}
{2(t+2)(t+3)(t+4)(2t+3)}\\
=& -\frac{t+4+2t+3}{2(t+2)(t+3)(t+4)(2t+3)}\\
=& -\frac{3t+7}{2(t+2)(t+3)(t+4)(2t+3)}.
\end{aligned}\notag\]
Let \(\varepsilon>0\) be arbitrarily chosen. We take \(A>1\) such that
\[ A^2+7A+21-\frac{2}{\varepsilon}>0. \]
Then, for any \(A_1,A_2>A\), we have
\[\begin{aligned}
\left|\int_{A_1}^{A_2}
\frac{3t+7}{2(t+2)(t+3)(t+4)(2t+3)}\Delta t\right|
=& \left|-\int_{A_1}^{A_2}
\left(\frac{1}{(t+3)(t+4)}\right)^\Delta\Delta t\right|\\
=& \left|-\frac{1}{(t+3)(t+4)}\Bigl|_{t=A_1}^{t=A_2}\right|\\
=& \left|-\frac{1}{(A_2+3)(A_2+4)}+\frac{1}{(A_1+3)(A_1+4)}\right|\\
\leq& \frac{1}{(A_2+3)(A_2+4)}+\frac{1}{(A_1+3)(A_1+4)}\\
<& \frac{2}{(A+3)(A+4)}\\
<& \varepsilon.
\end{aligned}\notag\]
Thus, using the Cauchy criterion,it follows that the considered integral is convergent.
Let \(\mathbb{T}=2^{\mathbb{N}_0}\). We investigate the integral
\[ \int_1^{\infty}\frac{2\sin t-\sin(2t)}{2t^2}\Delta t \]
for convergence. Here, \(\sigma(t)=2t\) and
\[\begin{aligned}
\left(\frac{\sin t}{t}\right)^\Delta
=& \frac{t(\sin t)^\Delta-t^\Delta\sin t}{t\sigma(t)}\\
=& \frac{t\frac{\sin(2t)-\sin t}{2t-t}-\sin t}{2t^2}\\
=& \frac{\sin(2t)-2\sin t}{2t^2}.
\end{aligned}\notag\]
Let \(\varepsilon>0\) be arbitrarily chosen. We take \(A>\frac{2}{\varepsilon}\). Then, for any \(A_1,A_2>A\), we have
\[\begin{aligned}
\left|\int_{A_1}^{A_2}\frac{2\sin t-t\sin(2t)}{2t^2}\Delta t\right|
=& \left|-\int_{A_1}^{A_2}
\left(\frac{\sin t}{t}\right)^\Delta\Delta t\right|\\
=& \left|-\frac{\sin t}{t}\Bigl|_{t=A_1}^{t=A_2}\right|\\
=& \left|-\frac{\sin A_2}{A_2}+\frac{\sin A_1}{A_1}\right|\\
\leq& \frac{|\sin A_2|}{A_2}+\frac{|\sin A_1|}{A_1}\\
\leq& \frac{1}{A_2}+\frac{1}{A_1}\\
<& \frac{2}{A}\\
<& \varepsilon.
\end{aligned}\notag\]
Thus, using the Cauchy criterion, we conclude that the considered integral is convergent.
Let \(\mathbb{T}=3\mathbb{Z}\). Using Cauchy's criterion, prove that the integral
\[ \int_1^{\infty}\frac{1}{(2t+1)(2t+7)}\Delta t \]
is convergent.
An integral of type \eqref{159} is said to be absolutely convergent provided the integral
\begin{equation}\label{161}
\int_a^{\infty}|f(t)|\Delta t
\end{equation}
of the modulus of the function \(f\) is convergent. If an integral of type \eqref{159} is convergent but not absolutely convergent, then it is called conditionally convergent.
If the integral \eqref{159} is absolutely convergent, then it is convergent.
Proof
Let \eqref{159} be absolutely convergent. Then the integral \eqref{161} is convergent. Suppose that \(\varepsilon>0\) is arbitrarily chosen. Hence, employing the Cauchy criterion, it follows that there exists \(A>a\) such that for any \(A_1,A_2>A\), we have
\[ \left|\int_{A_1}^{A_2}|f(t)|\Delta t\right|<\varepsilon. \]
From here, for any \(A_1,A_2>A\), we have
\[ \left|\int_{A_1}^{A_2}f(t)\Delta t\right|
\leq\left|\int_{A_1}^{A_2}|f(t)|\Delta t\right|<\varepsilon, \]
which completes the proof.
An integral \eqref{159} with \(f(t)\geq 0\) for all \(t\geq a\) is convergent if and only if there exists a constant \(M>0\) such that
\[\int_a^A f(t)\Delta t\leq M \quad \mbox{whenever}\quad A\geq a. \]
Proof
1. Let \(F(A)=\int_a^Af(t)\Delta t\leq M\) whenever \(A\geq a\). Then
\[ \int_a^{\infty}f(t)\Delta t=\lim_{A\to\infty}F(A)\leq M. \]
Therefore, the integral \eqref{159} is convergent.
2. Let the integral \eqref{159} be convergent. Assume that the function \(F(A)\), \(A\geq a\), is unbounded. Then
\[ \int_a^{\infty}f(t)\Delta t=\lim_{A\to\infty}F(A)=\infty, \]
which is a contradiction.
Let the inequalities \(0\leq f(t)\leq g(t)\) be satisfied for all \(t\in[a,\infty)\). Then the convergence of the improper integral
\begin{equation}\label{162}
\int_a^{\infty}g(t)\Delta t
\end{equation}
implies the convergence of the improper integral \eqref{159}, while the divergence of the improper integral \eqref{159} implies the divergence of the improper integral \eqref{162}.
Proof
Since \(0\leq f(t)\leq g(t)\) for any \(t\in[a,\infty)\), we get
\[ 0\leq\int_a^Af(t)\Delta t\leq\int_a^Ag(t)\Delta t
\quad\mbox{for any}\quad A\in[a,\infty), \]
which completes the proof.
Let \(\mathbb{T}=\mathbb{Z}\). Consider the integral
\[ I=\int_1^{\infty}\log\frac{t+1}{t}(t^6+7t^3+100)\Delta t. \]
Here, \(\sigma(t)=t+1\) and
\[\log\frac{t+1}{t}(t^6+7t^3+100)\geq\log\frac{t+1}{t}
\quad \mbox{for any}\quad t\in[1,\infty) \]
and
\[ I\geq\int_1^{\infty}\log\frac{t+1}{t}\Delta t. \]
Note that
\[\begin{aligned}
(\log t)^\Delta
=& \frac{\log\sigma(t)-\log t}{\sigma(t)-t}\\
=& \log(t+1)-\log t\\
=& \log\frac{t+1}{t}.
\end{aligned}\notag\]
Therefore,
\[\begin{aligned}
\int_1^{\infty}\log\frac{t+1}{t}\Delta t
=& \lim_{A\to\infty}\int_1^A(\log t)^\Delta\Delta t\\
=& \lim_{A\to\infty}\log t\Bigl|_{t=1}^{t=A}\\
=& \infty.
\end{aligned}\notag\]
Hence, the improper integral \(I\) is divergent.
Let \(\mathbb{T}=2^{\mathbb{N}_0}\). Consider the integral
\[ I=\int_1^{\infty}\frac{1}{t^3(t^2+5)(t^2+7t+1)}\Delta t. \]
Here, \(\sigma(t)=2t\) and
\[ \frac{1}{t^3(t^2+5)(t^2+7t+1)}\leq\frac{1}{t^3}
\quad \mbox{for any}\quad t\in[1,\infty). \]
Also,
\[\begin{aligned}
\left(\frac{1}{t^2}\right)^\Delta
=& -\frac{(t^2)^\Delta}{t^2(\sigma(t))^2}\\
=& -\frac{\sigma(t)+t}{4t^4}\\
=& -\frac{3}{4t^3},
\end{aligned}\notag\]
whereupon
\[ \frac{1}{t^3}=-\frac{4}{3}\left(\frac{1}{t^2}\right)^\Delta. \]
Hence,
\[\begin{aligned}
\int_1^{\infty}\frac{1}{t^3}\Delta t
=& \lim_{A\to\infty}\int_1^A\frac{1}{t^3}\Delta t\\
=& -\frac{4}{3}\lim_{A\to\infty}
\int_1^A\left(\frac{1}{t^2}\right)^\Delta\Delta t\\
=& -\frac{4}{3}\lim_{A\to\infty}\frac{1}{t^2}\Bigl|_{t=1}^{A}\\
=& \frac{4}{3}.
\end{aligned}\notag\]
Therefore, the integral \(I\) is convergent.
Let \(\mathbb{T}=3^{\mathbb{N}_0}\). Consider the integral
\[ I=\int_1^{\infty}\frac{\sin^3 t+e_1(t,1)+\cos^3 t+5}{t}\Delta t. \]
Here, \(\sigma(t)=3t\) and
\[ \frac{\sin^3t+e_1(t,1)+\cos^3t+5}{t}\geq\frac{1}{t}
\quad \mbox{for any}\quad t\in[1,\infty), \]
whereupon
\[ I\geq \int_1^{\infty}\frac{1}{t}\Delta t. \]
Also,
\[\begin{aligned}
(\log t)^\Delta
=& \frac{\log(\sigma(t))-\log t}{\sigma(t)-t}\\
=& \frac{\log(3t)-\log t}{2t}\\
=& \frac{\log 3}{2t},
\end{aligned}\notag\]
from where
\[ \frac{1}{t}=\frac{2}{\log 3}(\log t)^\Delta. \]
Hence,
\[\begin{aligned}
\int_1^{\infty}\frac{1}{t}\Delta t
=& \frac{2}{\log 3}\lim_{A\to\infty}\int_1^A(\log t)^\Delta\Delta t\\
=& \frac{2}{\log 3}\lim_{A\to\infty}\log t\Bigl|_{t=1}^{t=A}\\
=& \infty.
\end{aligned}\notag\]
Therefore, the considered integral is divergent.
Investigate the following integrals for convergence and divergence.
1. \(\int_1^{\infty}\log\frac{t+1}{t}(t^7+11t^6+12t^5+13t+4)\Delta t\), \(\mathbb{T}=\mathbb{Z}\),
2. \(\int_1^{\infty}\log\frac{t+3}{t}(t^4+11t^2+100)\Delta t\), \(\mathbb{T}=3\mathbb{Z}\),
3. \(\int_1^{\infty}\frac{1}{t^2}(t^4+5t^3+5t^2+5t+5)\Delta t\), \(\mathbb{T}=2^{\mathbb{N}_0}\),
4. \(\int_1^{\infty}\frac{1}{t^2(t^3+11t^2+12t+13)(t+1)}\Delta t\), \(\mathbb{T}=3^{\mathbb{N}_0}\),
5. \(\int_1^{\infty}\frac{1}{(t+1)^2(t+3)^3(t^4+10t+10)}\Delta t\), \(\mathbb{T}=\mathbb{Z}\),
6. \(\int_1^{\infty} t^2(t^6+5t+1)\Delta t\), \(\mathbb{T}=3^{\mathbb{N}_0}\).
Answer
1. Divergent,
2. divergent,
3. divergent,
4. convergent,
5. convergent,
6. divergent.
Let \(|f(t)|\leq g(t)\) for all \(t\in\mathbb{T}\) with \(t\geq a\). Then the convergence of the integral \(\int_a^{\infty}g(t)\Delta t\) implies the convergence of the integral \(\int_a^{\infty}f(t)\Delta t\).
Proof
Since \(\int_a^{\infty}g(t)\Delta t\) is convergent, we have that the integral \(\int_a^{\infty}|f(t)|\Delta t\) is convergent. Therefore, the integral \(\int_a^{\infty}f(t)\Delta t\) is absolutely convergent. From here, it follows that the integral \(\int_a^{\infty}f(t)\Delta t\) is convergent.
Let \(\int_a^{\infty}f(t)\Delta t\) and \(\int_a^{\infty}g(t)\Delta t\) be improper integrals of the first kind with positive integrands. Suppose that the limit
\begin{equation}\label{164}
\lim_{t\to\infty}\frac{f(t)}{g(t)}=L
\end{equation}
exists (finite) and is not zero. Then the integrals are simultaneously convergent or divergent.
Proof
Let \(\varepsilon\in(0,L)\) be arbitrarily chosen. From \eqref{164}, it follows that there exists \(A_0>a\) such that
\[ L-\varepsilon\leq\frac{f(t)}{g(t)}\leq L+\varepsilon \quad \mbox{for any}\quad t\geq A_0, \]
from where
\[ (L-\varepsilon)g(t)\leq f(t)\leq(L+\varepsilon)g(t) \quad \mbox{for any}\quad t\geq A_0. \]
Hence,
\begin{equation}\label{165}
(L-\varepsilon)\int_{A_0}^{\infty}g(t)\Delta t
\leq\int_{A_0}^{\infty}f(t)\Delta t
\leq(L+\varepsilon)\int_{A_0}^{\infty}g(t)\Delta t.
\end{equation}
1. Let \(\int_a^{\infty}g(t)\Delta t\) be convergent. Then \(\int_{A_0}^{\infty}g(t)\Delta t\) is convergent. Hence,
\[ (L+\varepsilon)\int_{A_0}^{\infty}g(t)\Delta t \]
is convergent. From here, using \eqref{165}, we obtain that \[\int_{A_0}^{\infty}f(t)\Delta t\] is convergent. Therefore, \[\int_a^{\infty}f(t)\Delta t\] is convergent.
2. Let \(\int_a^{\infty}f(t)\Delta t\) be convergent. Then \(\int_{A_0}^{\infty} f(t)\Delta t\) is convergent. From here, using \eqref{165}, we obtain that \(\int_{A_0}^{\infty}g(t)\Delta t\) is convergent. Therefore, \(\int_a^{\infty}g(t)\Delta t\) is convergent.
3. Let \(\int_a^{\infty}f(t)\Delta t\) be divergent. Hence, \(\int_{A_0}^{\infty}f(t)\Delta t\) is divergent. Thus, using \eqref{165}, it follows that \(\int_{A_0}^{\infty}g(t)\Delta t\) is divergent. Therefore, \(\int_a^{\infty}g(t)\Delta t\) is divergent.
4. Let \(\int_a^{\infty}g(t)\Delta t\) be divergent. Hence, \(\int_{A_0}^{\infty}g(t)\Delta t\) is divergent. Thus, using \eqref{165},
it follows that \(\int_{A_0}^{\infty}f(t)\Delta t\) is divergent. Therefore, \(\int_a^{\infty}f(t)\Delta t\) is divergent.
Let \(\mathbb{T}=\mathbb{Z}\). Consider the integral
\[ I=\int_1^{\infty}\frac{t^4}{(t^2+11t+30)(t^4+t^3+t^2+1)}\Delta t. \]
Define
\[ f(t)=\frac{t^4}{(t^2+11t+30)(t^4+t^3+t^2+1)},\quad
g(t)=\frac{1}{t^2+11t+30} \]
and
\[ J=\int_1^{\infty}\frac{1}{t^2+11t+30}\Delta t. \]
We have
\[\begin{aligned}
\lim_{t\to\infty}\frac{f(t)}{g(t)}
=& \lim_{t\to\infty}
\frac{\frac{t^4}{(t^2+11t+30)(t^4+t^3+t^2+1)}}{\frac{1}{t^2+11t+30}}\\
=& \lim_{t\to\infty}\frac{t^4}{t^4+t^3+t^2+1}\\
=& 1.
\end{aligned}\notag\]
Thus, it follows that the integrals \(I\) and \(J\) are simultaneously convergent or divergent. Note that
\[\begin{aligned}
\left(\frac{1}{t+5}\right)^\Delta
=& -\frac{(t+5)^\Delta}{(t+5)(\sigma(t)+5)}\\
=& -\frac{1}{(t+5)(t+6)}\\
=& -\frac{1}{t^2+11t+30}.
\end{aligned}\notag\]
Therefore,
\[\begin{aligned}
J
=& -\lim_{A\to\infty}\int_1^A\left(\frac{1}{t+5}\right)^\Delta\Delta t\\
=& -\lim_{A\to\infty}\frac{1}{t+5}\Bigl|_{t=1}^{t=A}\\
=& \frac{1}{6}.
\end{aligned}\notag\]
Consequently, the integral \(I\) is convergent.
Let \(\mathbb{T}=2^{\mathbb{N}_0}\). Consider the integral
\[ I=\int_1^{\infty}(t^2+2t+3)\Delta t. \]
Let
\[ f(t)=t^2+2t+3,\quad g(t)=t^2,\quad J=\int_1^{\infty}g(t)\Delta t. \]
We have that
\[\begin{aligned}
\lim_{t\to\infty}\frac{f(t)}{g(t)}
=& \lim_{t\to\infty}\frac{t^2+2t+3}{t^2}\\
=& 1.
\end{aligned}\notag\]
Thus, it follows that the integrals \(I\) and \(J\) are simultaneously convergent or divergent. Since
\[\begin{aligned}
J
=& \lim_{A\to\infty}\int_1^A t^2\Delta t\\
=& \frac{1}{7}\lim_{A\to\infty}\int_1^A(t^3)^\Delta\Delta t\\
=& \frac{1}{7}\lim_{A\to\infty}t^3\Bigl|_{t=1}^{t=A}\\
=& \infty,
\end{aligned}\notag\]
we conclude that \(I\) is divergent.
Let \(\mathbb{T}=2^{\mathbb{N}_0}\). Consider the integral
\[ I=\int_1^{\infty}\frac{e^{-t}-e^{-2t}}{t^2}(t+1)\Delta t. \]
We set
\[ f(t)=\frac{e^{-t}-e^{-2t}}{t^2}(t+1),\quad
g(t)=\frac{e^{-t}-e^{-2t}}{t} \]
and
\[ J=\int_1^{\infty}\frac{e^{-t}-e^{-2t}}{t}\Delta t. \]
Then
\[\begin{aligned}
\lim_{t\to\infty}\frac{f(t)}{g(t)}
=& \lim_{t\to\infty}
\frac{\frac{e^{-t}-e^{-2t}}{t^2}(t+1)}{\frac{e^{-t}-e^{-2t}}{t}}\\
=& \lim_{t\to\infty}\frac{t+1}{t}\\
=& 1.
\end{aligned}\notag\]
We conclude that the integrals \(I\) and \(J\) are simultaneously convergent or divergent. Note that
\[\begin{aligned}
(e^{-t})^\Delta
=& \frac{e^{-\sigma(t)}-e^{-t}}{\sigma(t)-t}\\
=& \frac{e^{-2t}-e^{-t}}{t}.
\end{aligned}\notag\]
Hence,
\[\begin{aligned}
J
=& -\lim_{A\to\infty}\int_1^A(e^{-t})^\Delta\Delta t\\
=& -\lim_{A\to\infty}e^{-t}\Bigl|_{t=1}^{t=A}\\
=& \frac{1}{e}.
\end{aligned}\notag\]
Consequently, the integral \(I\) is convergent.
Let \(\mathbb{T}=2^{\mathbb{N}_0}\). Investigate the integral
\[ \int_2^{\infty}\frac{1}{t+1}{\mbox{arctanh}}\frac{t}{1-2t^2}\Delta t \]
for convergence and divergence.
Answer
Convergent.
Hint
Use
\[ f(t)=\frac{1}{t+1}{\mbox{arctanh}}\frac{t}{1-2t^2},\quad
g(t)=\frac{1}{t}{\mbox{arctanh}}\frac{t}{1-2t^2} \]
and
\[ ({\mbox{arctanh}}t)^\Delta=\frac{1}{t}{\mbox{arctanh}}\frac{t}{1-2t^2}. \]
Let \(f\) be integrable from \(a\) to any point \(A\in\mathbb{T}\), \(A>a\). Suppose that the integral \( F(A)=\int_a^Af(t)\Delta t \) is bounded for any \(A\geq a\). Suppose that \(g\) is monotone on \([a,\infty)\) and \(\lim_{t\to\infty}g(t)=0\). Then the improper integral of the first kind of the form
\begin{equation}\label{168}
\int_a^{\infty}f(t)g(t)\Delta t
\end{equation}
is convergent.
Proof
Let \(A_1,A_2\in\mathbb{T}\), \(A_2>A_1\geq a\). By the mean value theorem, it follows that there exists \(\Lambda\) between \(\inf_{A\in[A_1,A_2]}F(A)\) and \(\sup_{A\in[A_1,A_2]}F(A)\) such that
\begin{equation}\label{166}
\int_{A_1}^{A_2}f(t)g(t)\Delta t
=(g(A_1)-g(A_2))\Lambda+g(A_2)\int_{A_1}^{A_2}f(t)\Delta t.
\end{equation}
Let \(M>0\) be a constant such that
\[ |F(A)|\leq M \quad \mbox{on}\quad [a,\infty). \]
From \eqref{166}, we get
\[ \int_{A_1}^{A_2}f(t)g(t)\Delta t
=(g(A_1)-g(A_2))\Lambda+g(A_2)(F(A_2)-F(A_1)) \]
and
\begin{equation}\label{167}
\begin{aligned}
\left|\int_{A_1}^{A_2}f(t)g(t)\Delta t\right|
&= |(g(A_1)-g(A_2))\Lambda+g(A_2)(F(A_2)-F(A_1))|\\
&\leq |g(A_1)||\Lambda|+|g(A_2)||\Lambda|+|g(A_2)|(|F(A_2)|+|F(A_1)|)\\
&\leq M|g(A_1)|+3M|g(A_2)|\\
&= M(|g(A_1)|+3|g(A_2)|).
\end{aligned}
\end{equation}
Let \(\varepsilon>0\) be arbitrarily chosen. Since \(g(t)\to 0\) as \(t\to\infty\), there exists \(A_3>a\) such that
\[ |g(A)|<\frac{\varepsilon}{4M} \quad \mbox{for any}\quad A>A_3. \]
Hence, using \eqref{167}, for \(A_1,A_2>A_3\), we get
\[\begin{aligned}
\left|\int_{A_1}^{A_2}f(t)g(t)\Delta t\right|
<& M\left(\frac{\varepsilon}{4M}+\frac{3\varepsilon}{4M}\right)\\
=& \varepsilon.
\end{aligned}\notag\]
From here and from Cauchy's criterion, it follows that the integral \eqref{168} is convergent.
Let \(\mathbb{T}=\mathbb{Z}\). Consider the integral
\[ I=\int_1^{\infty}\frac{t\sin t}{(t^2+3t+2)(t^2+1)}\Delta t. \]
Let
\[ f(t)=\frac{\sin t}{t^2+3t+2},\quad g(t)=\frac{t}{t^2+1}. \]
We have
\[\begin{aligned}
g^\Delta(t)
=& \frac{t^\Delta(t^2+1)-t(t^2+1)^\Delta}{(t^2+1)((\sigma(t))^2+1)}\\
=& \frac{t^2+1-t(\sigma(t)+t)}{(t^2+1)((t+1)^2+1)}\\
=& \frac{t^2+1-t(2t+1)}{(t^2+1)(t^2+2t+2)}\\
=& \frac{t^2+1-2t^2-t}{(t^2+1)(t^2+2t+2)}\\
=& \frac{1-t-t^2}{(t^2+1)(t^2+2t+2)}.
\end{aligned}\notag\]
Therefore, the function \(g\) is monotone on \([1,\infty)\). Moreover,
\[ \lim_{t\to\infty}g(t)=\lim_{t\to\infty}\frac{t}{t^2+1}=0 \]
and
\[\begin{aligned}
\left|\int_1^{\infty}f(t)\Delta t\right|
\leq& \int_1^{\infty}|f(t)|\Delta t\\
=& \int_1^{\infty}\frac{|\sin t|}{(t+1)(t+2)}\Delta t\\
\leq& \int_1^{\infty}\frac{1}{(t+1)(t+2)}\Delta t\\
=& -\int_1^{\infty}\left(\frac{1}{t+1}\right)^\Delta\Delta t\\
=& -\frac{1}{t+1}\Bigl|_{t=1}^{t=\infty}\\
=& \frac{1}{2}.
\end{aligned}\notag\]
Thus, it follows that the integral \(I\) is convergent.
Let \(\mathbb{T}=2^{\mathbb{N}_0}\). Consider the integral
\[ I=\int_1^{\infty}\frac{\sin^2t+\cos t+3}{t^2(t^2+1)}\Delta t. \]
Let
\[ f(t)=\frac{\sin^2t +\cos t+3}{t^2},\quad g(t)=\frac{1}{t^2+1}. \]
Then
\[\begin{aligned}
\left(\frac{1}{t^2+1}\right)^\Delta
=& -\frac{(t^2+1)^\Delta}{(t^2+1)((\sigma(t))^2+1)}\\
=& -\frac{\sigma(t)+t}{(t^2+1)(4t^2+1)}\\
=& -\frac{3t}{(t^2+1)(4t^2+1)}.
\end{aligned}\notag\]
Therefore, the function \(g\) is monotone on \([1,\infty)\). Moreover,
\[ \lim_{t\to\infty}g(t)=\lim_{t\to\infty}\frac{1}{t^2+1}=0 \]
and
\[\begin{aligned}
\left|\int_1^{\infty}f(t)\Delta t\right|
\leq& \int_1^{\infty}|f(t)|\Delta t\\
=& \int_1^{\infty}\frac{|\sin^2 t+\cos t+3|}{t^2}\Delta t\\
\leq& \int_1^{\infty}\frac{\sin^2t+|\cos t|+3}{t^2}\Delta t\\
\leq& 5\int_1^{\infty}\frac{1}{t^2}\Delta t\\
=& -10\int_1^{\infty}\left(\frac{1}{t}\right)^\Delta\Delta t\\
=& -10\frac{1}{t}\Bigl|_{t=1}^{t=\infty}\\
=& 10.
\end{aligned}\notag\]
Therefore, it follows that the integral \(I\) is convergent.
Let \(\mathbb{T}=3^{\mathbb{N}_0}\). Consider the integral
\[ I=\int_1^{\infty}\frac{1}{t(t^{10}+t^{11}+t^{12}+1)}\Delta t. \]
We set
\[ f(t)=\frac{1}{t^{10}+t^{11}+t^{12}+1},\quad g(t)=\frac{1}{t}. \]
Then
\[ g^\Delta(t)=-\frac{t^\Delta}{t\sigma(t)}=-\frac{1}{3t^2}. \]
Therefore, the function \(g\) is a monotonic function on \([1,\infty)\). Also,
\[\lim_{t\to\infty}g(t)=\lim_{t\to\infty}\frac{1}{t}=0 \]
and
\[\begin{aligned}
\left|\int_1^{\infty}f(t)\Delta t\right|
\leq& \int_1^{\infty}|f(t)|\Delta t\\
\leq& \int_1^{\infty}\frac{1}{t^{10}+t^{11}+t^{12}+1}\Delta t\\
\leq& \int_1^{\infty}\frac{1}{t^2}\Delta t\\
=& -3\int_1^{\infty}\left(\frac{1}{t}\right)^\Delta\Delta t\\
=& -3\frac{1}{t}\Bigl|_{t=1}^{t=\infty}\\
=& 3.
\end{aligned}\notag\]
Hence, it follows that the integral \(I\) is convergent.
Let \(\mathbb{T}=\mathbb{Z}\). Prove that the integral
\[\int_1^{\infty}\frac{\sin t-2\cos t+10}{(t^2+1)(t^2-3t+5)}\Delta t \]
is convergent.
Let \(\mathbb{T}\) be a time scale of the form
\begin{equation}\label{169}
\mathbb{T}=\{t_k:\;k\in\mathbb{N}_0\} \quad \mbox{with}\quad 0<t_0<t_1<\ldots
\quad \mbox{and}\quad \lim_{k\to\infty}t_k=\infty.
\end{equation}
Suppose that \(f:[t_0,\infty)\to\mathbb{R}\) is nonincreasing with \(\int_{t_0}^{\infty}f(t)\Delta t<\infty\).Assume that \(g:\mathbb{T}\to\mathbb{R}_+\) satisfies
\[ g(t_k)\leq Kf(t_{k+1}) \quad \mbox{for all}\quad k\in\mathbb{N}_0, \]
where \(K>0\) is a constant.
Prove that \(\int_{t_0}^{\infty}g(t)\Delta t<\infty\).
Solution
We have
\[\begin{aligned}
\int_{t_0}^{\infty}g(t)\Delta t
=& \sum_{k=0}^{\infty}g(t_k)\mu(t_k)\\
\leq& K\sum_{k=0}^{\infty}f(t_{k+1})\mu(t_k)\\
\leq& K\sum_{k=0}^{\infty}f(t_k)\mu(t_k)\\
=& K\int_{t_0}^{\infty}f(t)\Delta t\\
<& \infty.
\end{aligned}\notag\]
Let \(\mathbb{T}=2^{\mathbb{N}_0}\). Prove that the integral
\[ \int_1^{\infty}\frac{1}{t^p}\Delta t \]
is divergent for \(p\in[0,1]\) and convergent for \(p>1\).
Solution
Let \(t_k=2^k\), \(k\in\mathbb{N}_0\). Then
\[ t_{k+1}=2^{k+1},\quad \mu(t_k)=t_{k+1}-t_k=2^{k+1}-2^k \]
and
\[\begin{aligned}
\int_1^{\infty}\frac{1}{t^p}\Delta t
=& \sum_{k=0}^{\infty}\frac{1}{t_k^p}\mu(t_k)\\
=& \sum_{k=0}^{\infty}\frac{1}{2^{kp}}(2^{k+1}-2^k)\\
=& 2\sum_{k=0}^{\infty}\frac{1}{2^{k(p-1)}}
-\sum_{k=0}^{\infty}\frac{1}{2^{k(p-1)}}\\
=& \sum_{k=0}^{\infty}\frac{1}{2^{k(p-1)}}
\begin{cases}
=\infty & \quad \mbox{if}\quad p\in[0,1]\\
<\infty & \quad \mbox{if}\quad p>1.
\end{cases}
\end{aligned}\notag\]
Let \(\mathbb{T}\) be a time scale that satisfies \eqref{169}. If \(f:[t_0,\infty)\to\mathbb{R}\) is nonincreasing, then
\[ \int_{t_0}^{\infty}f(t)\nabla t
\leq\int_{t_0}^{\infty}f(t)dt
\leq\int_{t_0}^{\infty}f(t)\Delta t, \]
where the first and last integrals are taken over \(\mathbb{T}\), while the middle integral is taken over the interval \([t_0,\infty)\) of \(\mathbb{R}\).
Proof
Since \(\mathbb{T}\) is a time scale that satisfies \eqref{169}, we have
\[ \int_{t_0}^{\infty}f(t)\nabla t
=\sum_{k=0}^{\infty}f(t_{k+1})(t_{k+1}-t_k) \quad \mbox{and}\quad
\int_{t_0}^{\infty}f(t)\Delta t=\sum_{k=0}^{\infty}f(t_k)(t_{k+1}-t_k). \]
Because the function \(f\) is nonincreasing on \([t_0,\infty)\), we get
\[ f(t_{k+1})(t_{k+1}-t_k)
\leq\int_{t_k}^{t_{k+1}}f(t)dt
\leq f(t_k)(t_{k+1}-t_k) \]
for all \(k\in\mathbb{N}_0\). Hence,
\[ \sum_{k=0}^{\infty}f(t_{k+1})(t_{k+1}-t_k)
\leq\sum_{k=0}^{\infty}\int_{t_k}^{t_{k+1}} f(t)dt
\leq\sum_{k=0}^{\infty}f(t_k)(t_{k+1}-t_k), \]
which completes the proof.
Let \(\mathbb{T}\) be a time scale that satisfies \eqref{169} and assume that \(f:[t_0,\infty)\to\mathbb{R}_+\) is nonincreasing.
1. If \(\int_{t_0}^{\infty}f(t)dt=\infty\), then \(\int_{t_0}^{\infty}f(t)\Delta t=\infty\).
2. If \(\int_{t_0}^{\infty}f(t)dt<\infty\), then \(\int_{t_0}^{\infty}f(t)\nabla t<\infty\).
Let \(\mathbb{T}\) be a time scale that satisfies \eqref{169}. Then
\begin{equation}\label{170}
\int_{t_0}^{\infty}\frac{1}{t^p}\Delta t=\infty \quad \mbox{if}\quad 0\leq p\leq 1.
\end{equation}
Proof
Let \(f(t)=\frac{1}{t^p}\). Then \(f\) is nonincreasing on \([t_0,\infty)\) and
\[ \int_{t_0}^{\infty}\frac{1}{t^p}dt=\infty. \]
Thus, we get \eqref{170}.
Let \(\mathbb{T}\) be a time scale satisfying \eqref{169}. Then
\begin{equation}\label{171}
\int_{t_0}^{\infty}\frac{\nabla t}{t^p}<\infty \quad\mbox{if}\quad p>1.
\end{equation}
Proof
Let \(f(t)=\frac{1}{t^p}\). Then \(f\) is nonincreasing on \([t_0,\infty)\). Since
\[ \int_{t_0}^{\infty}\frac{1}{t^p}dt<\infty \quad\mbox{for}\quad p>1, \]
we get \eqref{171}.
Let \(\mathbb{T}\) be a time scale satisfying \eqref{169}. Then
\begin{equation}\label{172}
\int_{t_0}^{\infty}\frac{\nabla t}{t^p}=\infty \quad\mbox{if}\quad 0\leq p\leq 1.
\end{equation}
Proof
1. Firstly, we prove that
\begin{equation}\label{173}
\int_{t_0}^{\infty}\frac{\nabla t}{t}=\infty.
\end{equation}
We have
\[ \int_{t_0}^{\infty}\frac{\nabla t}{t}
=\sum_{k=0}^{\infty}\frac{t_{k+1}-t_k}{t_{k+1}}. \]
Assume that
\[ \sum_{k=0}^{\infty}\frac{t_{k+1}-t_k}{t_{k+1}}<\infty. \]
Then
\[ \lim_{k\to\infty}\frac{t_{k+1}-t_k}{t_{k+1}}=0, \]
whereupon
\[ \lim_{k\to\infty}\frac{t_{k+1}}{t_k}=1. \]
Since \(t_k<t_{k+1}\) for any \(k\in\mathbb{N}_0\), there exists \(N\in\mathbb{N}\) such that
\[ \frac{t_{k+1}}{t_k}<2 \quad\mbox{for any}\quad k>N. \]
Then we have
\[ \sum_{k=0}^{\infty}\frac{t_{k+1}-t_k}{t_k}=\infty. \]
On the other hand, we have
\[\begin{aligned}
\sum_{k=0}^{\infty}\frac{t_{k+1}-t_k}{t_k}
=& \sum_{k=0}^{N-1}\frac{t_{k+1}-t_k}{t_k}
+\sum_{k=N}^{\infty}\frac{t_{k+1}-t_k}{t_k}\\
=& \sum_{k=0}^{N-1}\frac{t_{k+1}-t_k}{t_k}
+\sum_{k=N}^{\infty}\frac{t_{k+1}-t_k}{t_{k+1}}\frac{t_{k+1}}{t_k}\\
<& \sum_{k=0}^{N-1}\frac{t_{k+1}-t_k}{t_k}
+2\sum_{k=N}^{\infty}\frac{t_{k+1}-t_k}{t_{k+1}}\\
<& \infty,
\end{aligned}\notag\]
which is a contradiction. Therefore
\[ \sum_{k=0}^{\infty}\frac{t_{k+1}-t_k}{t_{k+1}}=\infty, \]
from where \eqref{173} follows.
2. Let \(p\in[0,1)\). Since \(\lim_{k\to\infty}t_k=\infty\), nthere exists \(t_l\in\mathbb{T}\) so that \(t_l>1\). Hence,
\begin{equation}\label{174}
\int_{t_0}^{\infty}\frac{1}{t^p}\nabla t
=\int_{t_0}^{t_l}\frac{1}{t^p}\nabla t
+\int_{t_l}^{\infty}\frac{1}{t^p}\nabla t.
\end{equation}
Because \(t_l>1\) and \(p\in[0,1)\), we have
\[ \int_{t_l}^{\infty}\frac{1}{t}\nabla t=\infty \quad\mbox{and}\quad
\int_{t_l}^{\infty}\frac{1}{t^p}\nabla t
\geq\int_{t_l}^{\infty}\frac{1}{t}\nabla t. \]
Thus, using \eqref{174}, we get the desired result \eqref{172}.This completes the proof.
This page titled Chapter 13: Improper Integrals of the First Kind was last modified on Wed, 03 Dec 2025 07:46:20 GMT and is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Svetlin G. Georgiev.
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