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Chapter 10: Problem 14

Show that the integral \(\int_{0}^{\infty}(1 / \sqrt{x}) \sin x d x\) converges. [Hint: Integrate by Parts.]

Short Answer

Expert verified
The integral \(\int_{0}^{\infty}(1 / \sqrt{x}) \sin x d x\) converges.

Step by step solution

01

Applying Integration by Parts

We start by identifying parts as \(u\) and \(dv\). We let \(u = \frac{1}{\sqrt{x}}\), which implies \(du = -\frac{1}{2x\sqrt{x}}dx\), and let \(dv = \sin x dx\), which implies \(v=-\cos x\). Now we have everything we need to apply the integration by parts.
02

Running Integration by Parts

\(\int udv=uv-\int vdu\) gives us \(\int_0^\infty {(1/ \sqrt{x}) \sin x dx = [-\cos x/\sqrt{x}]_0^\infty + \int_0^\infty {\frac{\cos x}{2x\sqrt{x}} dx}}\)
03

Calculation of the limits

Calculating the limits of \([- \cos x/\sqrt{x}]_0^\infty\), first \((- \cos x/\sqrt{x})\) as \(x \rightarrow \infty\) goes to 0. And \((- \cos x/\sqrt{x})\) as \(x \rightarrow 0\) is undefined but the limit exists and is 0. So the first part of the integral results in 0.
04

Remaining Integral Calculation

The integral of the remaining function is less than or equal to \(\int_0^\infty {\frac{dx}{2x\sqrt{x}}}\) due to the fact that absolute value of cosine does not exceed 1. This integral is easily computed and found to be finite by direct computation or by the p-test for integrals of power functions.
05

Convergence Declaration

Since the two parts obtained as a result of integration by parts both approached a finite value, the integral \(\int_{0}^{\infty}(1 / \sqrt{x}) \sin x d x\) converges.

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Most popular questions from this chapter

(a) Show that \(\lim _{k \rightarrow \infty} \int_{0}^{1} \frac{x^{k} d x}{(1+x)^{2}}=0\). (b) Show that \(\lim _{k \rightarrow \infty} \int_{0}^{1} \frac{k x^{k} d x}{1+x}=\frac{1}{2}\).

(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 \(g \in \mathcal{M}[a, b]\).

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

Let \(\Phi(x):=x|\cos (\pi / x)|\) for \(x \in(0,1)\) and let \(\Phi(0):=0 .\) Then \(\Phi\) is continuous on \([0,1]\) and \(\Phi^{\prime}(x)\) exists for \(x \notin E:=\\{0\\} \cup\left\\{a_{k}: k \in \mathbb{N}\right\\}\), where \(a_{k}:=2 /(2 k+1) .\) Let \(\varphi(x):=\Phi^{\prime}(x)\) for \(x \notin E\) and \(\varphi(x):=0\) for \(x \in E\). Show that \(\varphi\) is not bounded on \([0,1]\). Using the Fundamental Theorem \(10.1 .9\) with \(E\) countable, conclude that \(\varphi \in \mathcal{R}^{*}[0,1]\) and that \(\int_{a}^{b} \varphi=\Phi(b)-\Phi(a)\) for \(a, b \in[0,1] .\) As in Exercise 19, show that \(|\varphi| \notin \mathcal{R}^{*}[0,1]\).

Establish the convergence of Fresnel's integral \(\int_{0}^{\infty} \sin \left(x^{2}\right) d x .\) [Hint: Use the Substitution Theorem \(10.1 .12 .\) ]
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