- Use
[link] ,
[link] ,
[link] , and
properties of limits to prove the preceding theorem.
- Let
$f$ be defined and improperly-integrable on
$(a,b).$ Show that, given an
$\u03f5>0,$ there exists a
$\delta >0$ such that
for any
$a<{a}^{\text{'}}<a+\delta $ and any
$b-\delta <{b}^{\text{'}}<b$ we have
$|{\int}_{a}^{{a}^{\text{'}}}f|+|{\int}_{{b}^{\text{'}}}^{b}f|<\u03f5.$
- Let
$f$ be improperly-integrable on an open interval
$(a,b).$ Show that, given an
$\u03f5>0,$ there exists a
$\delta >0$ such that if
$(c,d)$ is any open subinterval of
$(a,b)$ for which
$d-c<\delta ,$ then
$|{\int}_{c}^{d}f|<\u03f5.$ HINT: Let
$\left\{{x}_{i}\right\}$ be a partition of
$[a,b]$ such that
$f$ is defined and improperly-integrable
on each subinterval
$({x}_{i-1},{x}_{i}).$ For each
$i,$ choose a
${\delta}_{i}$ using part (b).
Now
$f$ is bounded by
$M$ on all the intervals
$[{x}_{i-1}+{\delta}_{i},{x}_{i}-{\delta}_{i}],$ so
$\delta =\u03f5/M$ should work there.
- Suppose
$f$ is a continuous function on a closed bounded interval
$[a,b]$ and is continuously differentiable on the open interval
$(a,b).$ Prove that
${f}^{\text{'}}$ is improperly-integrable on
$(a,b),$ and evaluate
${\int}_{a}^{b}{f}^{\text{'}}.$ HINT: Fix a point
$c\in (a,b),$ and use the Fundamental Theorem of Calculus
to show that the two limits exist.
- (Integration by substitution again.)
Let
$g:[c,d]\to [a,b]$ be continuous on
$[c,d]$ and satisfy
$g\left(c\right)=a$ and
$g\left(d\right)=b.$ Suppose there exists a partition
$\{{x}_{0}<{x}_{1}<...<{x}_{n}\}$ of the interval
$[c,d]$ such that
$g$ is continuously differentiable on each subinterval
$({x}_{i-1},{x}_{i}).$ Prove that
${g}^{\text{'}}$ is improperly-integrable on the open interval
$(c,d).$ Show also that if
$f$ is continuous on
$[a,b],$ we have that
$${\int}_{a}^{b}f\left(t\right)\phantom{\rule{0.166667em}{0ex}}dt={\int}_{c}^{d}f\left(g\left(s\right)\right){g}^{\text{'}}\left(s\right)\phantom{\rule{0.166667em}{0ex}}ds.$$
HINT: Integrate over the subintervals
$({x}_{i-1},{x}_{i}),$ and use part (d).
REMARK Note that there are parts of
[link] and
[link] that are not asserted in
[link] .
The point is that these other properties do not hold forimproperly-integrable functions on open intervals.
See the following exercise.
- Define
$f$ to be the function on
$[1,\infty )$ given by
$f\left(x\right)={(-1)}^{n-1}/n$ if
$n-1\le x<n.$ Show that
$f$ is improperly-integrable on
$(1,\infty ),$ but that
$\left|f\right|$ is not
improperly-integrable on
$(1,\infty ).$ (Compare this with part (4) of
[link] .)
HINT: Verify that
${\int}_{1}^{N}f$ is a partial sum of a convergent infinite series,
and then verify that
${\int}_{1}^{N}\left|f\right|$ is a partial sum of a divergent infinite series.
- Define the function
$f$ on
$(1,\infty )$ by
$f\left(x\right)=1/x.$ For each positive integer
$n,$ define the function
${f}_{n}$ on
$(1,\infty )$ by
${f}_{n}\left(x\right)=1/x$ if
$1<x<n$ and
${f}_{n}\left(x\right)=0$ otherwise.
Show that each
${f}_{n}$ is improperly-integrable on
$(1,\infty ),$ that
$f$ is the uniform limit
of the sequence
$\left\{{f}_{n}\right\},$ but that
$f$ is not improperly-integrable on
$(1,\infty ).$ (Compare this with part (5) of Theorem 5.6.)
- Suppose
$f$ is a nonnegative real-valued function on the
half-open interval
$(a,\infty )$ that is integrable on every closed bounded subinterval
$[a,{b}^{\text{'}}].$ For each positive integer
$n\ge a,$ define
${y}_{n}={\int}_{a}^{n}f\left(x\right)\phantom{\rule{0.166667em}{0ex}}dx.$ Prove that
$f$ is improperly-integrable on
$[a,\infty )$ if and only if the sequence
$\left\{{y}_{n}\right\}$ is convergent. In that case, show that
${\int}_{a}^{\infty}f=lim{y}_{n}.$
We are now able to prove an important result relating
integrals over infinite intervals and convergence of infinite series.
Let
$f$ be a positive function on
$[1,\infty ),$ assume that
$f$ is
integrable on every closed bounded interval
$[1,b],$ and suppose that
$f$ is nonincreasing; i.e., if
$x<y$ then
$f\left(x\right)\ge f\left(y\right).$ For each positive integer
$i,$ set
${a}_{i}=f\left(i\right),$ and let
${S}_{N}$ denote the
$N$ th partial sum of the infinite series
$\sum {a}_{i}:$
${S}_{N}={\sum}_{i=1}^{N}{a}_{i}.$ Then:
- For each
$N,$ we have
$${S}_{N}-{a}_{1}\le {\int}_{1}^{N}f\left(x\right)\phantom{\rule{0.166667em}{0ex}}dx\le {S}_{N-1}.$$
- For each
$N,$ we have that
$${S}_{N-1}-{\int}_{1}^{N}f\left(x\right)\phantom{\rule{0.166667em}{0ex}}dx\le {a}_{1}-{a}_{N}\le {a}_{1};$$
i.e., the sequence
$\{{S}_{N-1}-{\int}_{1}^{N}f\}$ is bounded above.
- The sequence
$\{{S}_{N-1}-{\int}_{1}^{N}f\}$ is nondecreasing.
- (Integral Test)
The infinite series
$\sum {a}_{i}$ converges if and only if
the function
$f$ is improperly-integrable on
$(1,\infty ).$
For each positive integer
$N,$ define
a step function
${k}_{N}$ on the interval
$[1,N]$ as follows.
Let
$P=\{{x}_{0}<{x}_{1}<...<{x}_{N-1}\}$ be the partition of
$[1,N]$ given by the points
$\{1<2<3<...<N\},$ i.e.,
${x}_{i}=i+1.$ Define
${k}_{N}\left(x\right)$ to be the constant
${c}_{i}=f(i+1)$ on the interval
$[{x}_{i-1},{x}_{i})=[i,i+1).$ Complete the definition of
${k}_{N}$ by setting
${k}_{N}\left(N\right)=f\left(N\right).$ Then, because
$f$ is nonincreasing, we have that
${k}_{N}\left(x\right)\le f\left(x\right)$ for all
$x\in [1,N].$ Also,
$$\begin{array}{ccc}\hfill {\int}_{1}^{N}{k}_{N}& =& \sum _{i=1}^{N-1}{c}_{i}({x}_{i}-{x}_{i-1})\hfill \\ & =& \sum _{i=1}^{N-1}f(i+1)\hfill \\ & =& \sum _{i=2}^{N}f\left(i\right)\hfill \\ & =& \sum _{i=2}^{N}{a}_{i}\hfill \\ & =& {S}_{N}-{a}_{1},\hfill \end{array}$$
which then implies that
$${S}_{N}-{a}_{1}={\int}_{1}^{N}{k}_{N}\left(x\right)\phantom{\rule{0.166667em}{0ex}}dx\le {\int}_{1}^{N}f\left(x\right)\phantom{\rule{0.166667em}{0ex}}dx.$$
This proves half of part (1).
For each positive integer
$N>1$ define another step function
${l}_{N},$ using the same partition
$P$ as above,
by setting
${l}_{N}\left(x\right)=f\left(i\right)$ if
$i\le x<i+1$ for
$1\le i<N,$ and complete the definition of
${l}_{N}$ by setting
${l}_{N}\left(N\right)=f\left(N\right).$ Again, because
$f$ is nonincreasing, we have that
$f\left(x\right)\le {l}_{N}\left(x\right)$ for all
$x\in [1,N].$ Also
$$\begin{array}{ccc}\hfill {\int}_{1}^{N}{l}_{N}& =& \sum _{i=1}^{N-1}f\left(i\right)\hfill \\ & =& \sum _{i=1}^{N-1}{a}_{i}\hfill \\ & =& {S}_{N-1},\hfill \end{array}$$
which then implies that
$${\int}_{1}^{N}f\left(x\right)\phantom{\rule{0.166667em}{0ex}}dx\le {\int}_{1}^{N}{l}_{N}\left(x\right)\phantom{\rule{0.166667em}{0ex}}dx={S}_{N-1},$$
and this proves the other half of part (1).
It follows from part (1) that
$${S}_{N-1}-{\int}_{1}^{N}f\left(x\right)\phantom{\rule{0.166667em}{0ex}}dx\le {S}_{N-1}-{S}_{N}+{a}_{1}={a}_{1}-{a}_{N},$$
and this proves part (2).
We see that the sequence
$\{{S}_{N-1}-{\int}_{1}^{N}f\}$ is nondecreasing
by observing that
$$\begin{array}{ccc}\hfill {S}_{N}-{\int}_{1}^{N+1}f-{S}_{N-1}+{\int}_{1}^{N}f& =& {a}_{N}-{\int}_{N}^{N+1}f\hfill \\ & =& f\left(N\right)-{\int}_{N}^{N+1}f\hfill \\ & \ge & 0,\hfill \end{array}$$
because
$f$ is nonincreasing.
Finally, to prove part (4), note that both of the
sequences
$\left\{{S}_{N}\right\}$ and
$\left\{{\int}_{1}^{N}f\right\}$ are nondecreasing.
If
$f$ is improperly-integrable on
$[1,\infty ),$ then
${lim}_{N}{\int}_{1}^{N}f$ exists, and
${S}_{N}\le {a}_{1}+{\int}_{1}^{\infty}f\left(x\right)\phantom{\rule{0.166667em}{0ex}}dx$ for all
$N,$ which implies that
$\sum {a}_{i}$ converges by
[link] .
Conversely, if
$\sum {a}_{i}$ converges, then
$lim{S}_{N}$ exists.
Since
${\int}_{1}^{N}f\left(x\right)\phantom{\rule{0.166667em}{0ex}}dx\le {S}_{N-1},$ it then follows, again from
[link] , that
${lim}_{N}{\int}_{1}^{N}f\left(x\right)\phantom{\rule{0.166667em}{0ex}}dx$ exists.
So, by the preceding exercise,
$f$ is improperly-integrable on
$[1,\infty ).$
We may now resolve a question first raised in
[link] .
That is, for
$1<s<2,$ is the infinite series
$\sum 1/{n}^{s}$ convergent or divergent?
We saw in that exercise that this series is convergent if
$s$ is a rational number.
- Let
$s$ be a real number.
Use the Integral Test toprove that the infinite series
$\sum 1/{n}^{s}$ is convergent if and only if
$s>1.$
- Let
$s$ be a complex number
$s=a+bi.$ Prove that the infinite series
$\sum 1/{n}^{s}$ is absolutely convergent
if and only if
$a>1.$
Let
$f$ be the function on
$[1,\infty )$ defined by
$f\left(x\right)=1/x.$
- Use
[link] to prove that the
sequence
$\{{\sum}_{i=1}^{N}\frac{1}{i}-lnN\}$ converges to a positive number
$\gamma \le 1.$ (This number
$\gamma $ is called Euler's constant.)
HINT: Show that this sequence is bounded above and nondecreasing.
- Prove that
$$\sum _{i=1}^{\infty}\frac{{(-1)}^{i+1}}{i}=ln2.$$
HINT: Write
${S}_{2N}$ for the
$2N$ th partial sum of the series.
Use the fact that
$${S}_{2N}=\sum _{i=1}^{2N}\frac{1}{i}-2\sum _{i=1}^{N}\frac{1}{2i}.$$
Now add and subtract
$ln\left(2N\right)$ and use part (a).