# 5.6 Extending the definition of integrability  (Page 2/2)

1. Use [link] , [link] , [link] , and properties of limits to prove the preceding theorem.
2. Let $f$ be defined and improperly-integrable on $\left(a,b\right).$ Show that, given an $ϵ>0,$ there exists a $\delta >0$ such that for any $a<{a}^{\text{'}} and any $b-\delta <{b}^{\text{'}} we have $|{\int }_{a}^{{a}^{\text{'}}}f|+|{\int }_{{b}^{\text{'}}}^{b}f|<ϵ.$
3. Let $f$ be improperly-integrable on an open interval $\left(a,b\right).$ Show that, given an $ϵ>0,$ there exists a $\delta >0$ such that if $\left(c,d\right)$ is any open subinterval of $\left(a,b\right)$ for which $d-c<\delta ,$ then $|{\int }_{c}^{d}f|<ϵ.$ HINT: Let $\left\{{x}_{i}\right\}$ be a partition of $\left[a,b\right]$ such that $f$ is defined and improperly-integrable on each subinterval $\left({x}_{i-1},{x}_{i}\right).$ For each $i,$ choose a ${\delta }_{i}$ using part (b). Now $f$ is bounded by $M$ on all the intervals $\left[{x}_{i-1}+{\delta }_{i},{x}_{i}-{\delta }_{i}\right],$ so $\delta =ϵ/M$ should work there.
4. Suppose $f$ is a continuous function on a closed bounded interval $\left[a,b\right]$ and is continuously differentiable on the open interval $\left(a,b\right).$ Prove that ${f}^{\text{'}}$ is improperly-integrable on $\left(a,b\right),$ and evaluate ${\int }_{a}^{b}{f}^{\text{'}}.$ HINT: Fix a point $c\in \left(a,b\right),$ and use the Fundamental Theorem of Calculus to show that the two limits exist.
5. (Integration by substitution again.) Let $g:\left[c,d\right]\to \left[a,b\right]$ be continuous on $\left[c,d\right]$ and satisfy $g\left(c\right)=a$ and $g\left(d\right)=b.$ Suppose there exists a partition $\left\{{x}_{0}<{x}_{1}<...<{x}_{n}\right\}$ of the interval $\left[c,d\right]$ such that $g$ is continuously differentiable on each subinterval $\left({x}_{i-1},{x}_{i}\right).$ Prove that ${g}^{\text{'}}$ is improperly-integrable on the open interval $\left(c,d\right).$ Show also that if $f$ is continuous on $\left[a,b\right],$ 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 $\left({x}_{i-1},{x}_{i}\right),$ 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.

1. Define $f$ to be the function on $\left[1,\infty \right)$ given by $f\left(x\right)={\left(-1\right)}^{n-1}/n$ if $n-1\le x Show that $f$ is improperly-integrable on $\left(1,\infty \right),$ but that $|f|$ is not improperly-integrable on $\left(1,\infty \right).$ (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 ${\int }_{1}^{N}|f|$ is a partial sum of a divergent infinite series.
2. Define the function $f$ on $\left(1,\infty \right)$ by $f\left(x\right)=1/x.$ For each positive integer $n,$ define the function ${f}_{n}$ on $\left(1,\infty \right)$ by ${f}_{n}\left(x\right)=1/x$ if $1 and ${f}_{n}\left(x\right)=0$ otherwise. Show that each ${f}_{n}$ is improperly-integrable on $\left(1,\infty \right),$ that $f$ is the uniform limit of the sequence $\left\{{f}_{n}\right\},$ but that $f$ is not improperly-integrable on $\left(1,\infty \right).$ (Compare this with part (5) of Theorem 5.6.)
3. Suppose $f$ is a nonnegative real-valued function on the half-open interval $\left(a,\infty \right)$ that is integrable on every closed bounded subinterval $\left[a,{b}^{\text{'}}\right].$ 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 $\left[a,\infty \right)$ 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 $\left[1,\infty \right),$ assume that $f$ is integrable on every closed bounded interval $\left[1,b\right],$ and suppose that $f$ is nonincreasing; i.e., if $x 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:

1. For each $N,$ we have
${S}_{N}-{a}_{1}\le {\int }_{1}^{N}f\left(x\right)\phantom{\rule{0.166667em}{0ex}}dx\le {S}_{N-1}.$
2. 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 $\left\{{S}_{N-1}-{\int }_{1}^{N}f\right\}$ is bounded above.
3. The sequence $\left\{{S}_{N-1}-{\int }_{1}^{N}f\right\}$ is nondecreasing.
4.  (Integral Test) The infinite series $\sum {a}_{i}$ converges if and only if the function $f$ is improperly-integrable on $\left(1,\infty \right).$

For each positive integer $N,$ define a step function ${k}_{N}$ on the interval $\left[1,N\right]$ as follows. Let $P=\left\{{x}_{0}<{x}_{1}<...<{x}_{N-1}\right\}$ be the partition of $\left[1,N\right]$ given by the points $\left\{1<2<3<... i.e., ${x}_{i}=i+1.$ Define ${k}_{N}\left(x\right)$ to be the constant ${c}_{i}=f\left(i+1\right)$ on the interval $\left[{x}_{i-1},{x}_{i}\right)=\left[i,i+1\right).$ 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 $x\in \left[1,N\right].$ Also,

$\begin{array}{ccc}\hfill {\int }_{1}^{N}{k}_{N}& =& \sum _{i=1}^{N-1}{c}_{i}\left({x}_{i}-{x}_{i-1}\right)\hfill \\ & =& \sum _{i=1}^{N-1}f\left(i+1\right)\hfill \\ & =& \sum _{i=2}^{N}f\left(i\right)\hfill \\ & =& \sum _{i=2}^{N}{a}_{i}\hfill \\ & =& {S}_{N}-{a}_{1},\hfill \end{array}$

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 for $1\le i 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 $x\in \left[1,N\right].$ 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 $\left\{{S}_{N-1}-{\int }_{1}^{N}f\right\}$ 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 $\left[1,\infty \right),$ 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 $\left[1,\infty \right).$

We may now resolve a question first raised in [link] . That is, for $1 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.

1. 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 $s>1.$
2. 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 $\left[1,\infty \right)$ defined by $f\left(x\right)=1/x.$

1. Use [link] to prove that the sequence $\left\{{\sum }_{i=1}^{N}\frac{1}{i}-lnN\right\}$ 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.
2. Prove that
$\sum _{i=1}^{\infty }\frac{{\left(-1\right)}^{i+1}}{i}=ln2.$
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).

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