1.2: Preparation

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We begin with two useful convergence criteria for improper integrals that do not involve a parameter. Consistent with the definition on p. 152, we say that \(f\) is locally integrable on an interval \(I\) if it is integrable on every finite closed subinterval of \(I\).

[theorem:2] Suppose \(g\) is locally integrable on \([a,b)\) and denote

\[G(r)=\int_{a}^{r}g(x)\,dx,\quad a\le r<b.\]

Then the improper integral \(\int_{a}^{b}g(x)\,dx\) converges if and only if\(,\) for each \(\epsilon >0,\) there is an \(r_{0}\in[a,b)\) such that

\[\label{eq:9} |G(r)-G(r_{1})|<\epsilon,\quad r_{0}\le r,r_{1}<b.\]

For necessity, suppose \(\int_{a}^{b}g(x)\,dx=L\). By definition, this means that for each \(\epsilon>0\) there is an \(r_{0}\in [a,b)\) such that

\[|G(r)-L|<\frac{\epsilon}{2} \text{\quad and\quad} |G(r_{1})-L|<\frac{\epsilon}{2},\quad r_{0}\le r,r_{1}<b.\]

Therefore

\[\begin{aligned} |G(r)-G(r_{1})|&=&|(G(r)-L)-(G(r_{1})-L)|\\ &\le& |G(r)-L|+|G(r_{1})-L|< \epsilon,\quad r_{0}\le r,r_{1}<b.\end{aligned}\]

For sufficiency, [eq:9] implies that

\[|G(r)|= |G(r_{1})+(G(r)-G(r_{1}))|< |G(r_{1})|+|G(r)-G(r_{1})|\le |G(r_{1})|+\epsilon,\]

\(r_{0}\le r\le r_{1}<b\). Since \(G\) is also bounded on the compact set \([a,r_{0}]\) (Theorem 5.2.11, p. 313), \(G\) is bounded on \([a,b)\). Therefore the monotonic functions

\[\overline{G}(r)=\sup\left\{G(r_{1})\, \big|\, r\le r_{1}<b\right\} \text{\quad and\quad} \underline{G}(r)=\inf\left\{G(r_{1})\, \big|\, r\le r_{1}<b\right\}\]

are well defined on \([a,b)\), and

\[\lim_{r\to b-}\overline{G}(r)=\overline{L} \text{\quad and\quad} \lim_{r\to b-}\underline{G}(r)=\underline{L}\]

both exist and are finite (Theorem 2.1.11, p. 47). From [eq:9],

\[\begin{aligned} |G(r)-G(r_{1})|&=&|(G(r)-G(r_{0}))-(G(r_{1})-G(r_{0}))|\\ &\le &|G(r)-G(r_{0})|+|G(r_{1})-G(r_{0})|< 2\epsilon,\end{aligned}\]

\[\overline{G}(r)-\underline{G}(r)\le 2\epsilon, \quad r_{0}\le r, r_{1}<b.\]

Since \(\epsilon\) is an arbitrary positive number, this implies that

\[\lim_{r\to b-}(\overline{G}(r)-\underline{G}(r))=0,\]

so \(\overline{L}=\underline{L}\). Let \(L=\overline{L}=\underline{L}\). Since

\[\underline{G}(r)\le G(r)\le \overline{G}(r),\]

it follows that \(\lim_{r\to b-} G(r)=L\).

We leave the proof of the following theorem to you (Exercise [exer:2]).

[theorem:3] Suppose \(g\) is locally integrable on \((a,b]\) and denote

\[G(r)=\int_{r}^{b}g(x)\,dx,\quad a\le r<b.\]

Then the improper integral \(\int_{a}^{b}g(x)\,dx\) converges if and only if\(,\) for each \(\epsilon >0,\) there is an \(r_{0}\in(a,b]\) such that

\[|G(r)-G(r_{1})|<\epsilon,\quad a<r,r_{1}\le r_{0}.\]

To see why we associate Theorems [theorem:2] and [theorem:3] with Cauchy, compare them with Theorem 4.3.5 (p. 204)