Exercise 8.4.9

Question

\begin{enumerate}[label=(\alph*)] 
\item Show that the improper integral $\int_a^\infty f$ converges if and only if, for all $\epsilon > 0$ there exists $M > a$ such that whenever $d > c \geq M$ it follows that
$$\left|\int_c^d f\right| < \epsilon.$$
(In one direction it will be useful to consider the sequence $a_n =  \int_a^{a+n} f$.)
\item Show that if $0 \leq f \leq g$ and $\int_a^\infty g$ converges then $\int_a^\infty f$ converges.
\item Part (a) is a Cauchy criterion, and part (b) is a comparison test. State and prove an absolute convergence test for improper integrals.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item In the forward direction, assume that $\lim_{b \to \infty}\int_a^b f$ exists. Note that for any $c$, $$\int_c^\infty f = \lim_{b \to \infty} \int_a^c f + \int_c^b f - \int_a^c f = \int_c^a f + \int_a^\infty f $$ Now, choose $M$ large enough to ensure that for $m \geq M$, $$\left|\int_a^\infty f - \int_a^m f\right| = \left|\int_m^\infty f\right| < \frac{\epsilon}{2}$$ Now if $d > c \geq M$, $$\left|\int_c^d f\right| = \left|\int_c^\infty f - \int_d^\infty f\right| \leq \left|\int_c^\infty f\right| + \left|\int_d^\infty f\right| < 2 \frac{\epsilon}{2} = \epsilon$$ In the reverse direction, define $M_n$ large enough so that $d > c \geq M_n$ implies $\left|\int_c^d f\right| < 1/n $ and satisfying $M_n > a + n$. Define the sequence $x_n = \int_a^{M_n} f$, and note that $(x_n)$ is a Cauchy sequence and converges to some limit $x$. Thus for any $\epsilon > 0$, we can find $N_1$ so that for $n > N_1$, $$\left|x - \int_a^{M_n} f\right| < \frac{\epsilon}{2}$$ and set $N > \max\{2/\epsilon, N_1 \}$ so that for $b > c =M_{N} $, $$\left|\int_{M_N}^b f\right| = \left|\int_a^b f - \int_a^{M_N} f\right| < \frac{\epsilon}{2}$$ and so $b > M_N$ implies $$\left|\int_a^b - x\right| \leq \left|x - \int_a^{M_N} f\right| + \left|\int_a^{M_N} f - \int_a^b f\right| < \epsilon$$ showing that $\int_a^\infty f = x$. \item For any $\epsilon > 0$, from part (a) and since $\int_a^\infty g$ converges, $\exists M$ so that for $d > c \geq M$, $$\epsilon > \left|\int_c^d g\right| = \int_c^d g \geq \int_c^d f = \left|\int_c^d f\right|$$ implying $\int_a^\infty f$ converges, where we have taken advantage of the fact that $\int_c^d f \geq 0$ and $\int_c^d g \geq 0$. \item The test is that if $\int_a^\infty |f|$ converges, then so does $\int_a^\infty f$. To prove this, we use the same strategy as part (b): for any $\epsilon > 0$, $\exists M$ so that for $d > c \geq M$, $$\epsilon > \left|\int_c^d \left|f\right|\right| = \int_c^d \left|f\right| \geq \left|\int_c^d f\right|$$ implying $\int_a^\infty f$ converges. \end{enumerate}

Exercise 8.4.9

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Exercise 8.4.9

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