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Chapter 10: Problem 6

If \(f(x):=1 / x\) for \(x \in[1, \infty)\), show that \(f \notin \mathcal{R}^{*}[1, \infty)\).

Short Answer

Expert verified
The function \(f(x):=1 / x\) is not Riemann integrable over the interval \([1, \infty)\) as the improper integral over this interval diverges.

Step by step solution

01

Function Definition and Interval

Consider the function \(f(x):=1 / x\) for \(x \in[1, \infty)\). We wish to show that it is not Riemann integrable over the interval \([1, \infty)\).
02

Improper Integral Setup

To show that \(f(x)\) is not Riemann integrable over \([1, \infty)\), we set up an improper integral from 1 to infinity: \(\int_{1}^{\infty} f(x) \, dx = \int_{1}^{\infty} \frac{1}{x} \, dx\). We use the definition of improper integrals to rewrite this as a limit: \(\lim_{b \rightarrow \infty} \int_{1}^{b} \frac{1}{x} \, dx\).
03

Compute the Integral

The antiderivative of \(1/x\) is \(\ln|x|\), so we compute the integral from 1 to \(b\) as: \(\left[ \ln|b| - \ln|1|\right] = \ln|b|\).
04

Compute the Limit

Now, we compute the limit as \(b\) approaches infinity: \(\lim_{b \rightarrow \infty} \ln|b|\). Since the natural log function diverges as \(x\) approaches infinity, this limit does not exist or equivalent to infinity.
05

Conclude

Since the improper integral does not converge (the limit does not exist), \(f(x)\) is not Riemann integrable over the interval \([1, \infty)\). This means that \(f \notin \mathcal{R}^{*}[1, \infty)\).

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Most popular questions from this chapter

(a) If \(\left(f_{k}\right)\) is a bounded sequence in \(\mathcal{M}[a, b]\) and \(f_{k} \rightarrow f\) a.e., show that \(f \in \mathcal{M}[a, b]\). [Hint: Use the Dominated Convergence Theorem.] (b) If \(\left(g_{k}\right)\) is any sequence in \(\mathcal{M}[a, b]\) and if \(f_{k}:=\operatorname{Arctan} \circ g_{k}\), show that \(\left(f_{k}\right)\) is a bounded sequence in \(\mathcal{M}[a, b]\). (c) If \(\left(g_{k}\right)\) is a sequence in \(\mathcal{M}[a, b]\) and if \(g_{k} \rightarrow g\) a.e., show that \(g \in \mathcal{M}[a, b]\).

If \(f, g \in \mathcal{C}\\{a, b]\) and if \(g\) is bounded and monotone, show that \(f g \in \mathcal{L}[a, b]\). More exactly, if \(|g(x)| \leq B\), show that \(\|f g\| \leq B\|f\|\).

Let \(f \in \mathcal{R}^{*}[a, \gamma]\) for all \(\gamma \geq a\). Show that \(f \in R^{*}[a, \infty)\) if and only if for every \(\varepsilon>0\) there exists \(K(\varepsilon) \geq a\) such that if \(q>p \geq K(\varepsilon)\), then \(\left|\int_{\rho}^{q} f\right|<\varepsilon\).

Let \(g_{n}(x):=-1\) for \(x \in[-1,-1 / n)\). let \(g_{n}(x):=n x\) for \(x \in[-1 / n, 1 / n]\) and let \(g_{n}(x):=1\) for \(x \in(1 / n, 1]\). Show that \(\left\|g_{m}-g_{n}\right\| \rightarrow 0\) as \(m, n \rightarrow \infty\), so that the Completeness Theorem \(10.2 .12\) implies that there exists \(g \in \mathcal{L}[-1,1]\) such that \(\left(g_{n}\right)\) converges to \(g\) in \(\mathcal{L}[-1,1]\). Find such a function \(g\).

(a) If \(\dot{\mathcal{P}}\) is a tagged partition of \([a, b]\), show that cach tag can belong to at most two subintervals in \(\dot{\mathcal{P}}\) (b) Are there tagged partitions in which every tag belongs to exactly two subintervals?
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