/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Show that the integral \(\int_{0... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 19

Show that the integral \(\int_{0}^{\infty} \sqrt{x} \cdot \sin \left(x^{2}\right) d x\) is convergent, even though the integrand is not bounded as \(x \rightarrow \infty\). [Hint: Make a substitution.]

Short Answer

Expert verified
The integral \( \int_{0}^{\infty} \sqrt{x} \cdot \sin(x^{2}) dx \) is convergent.

Step by step solution

01

Identify the Right Substitution

The key hint given is to make a substitution which, with the presence of the \( x^{2} \) term and the square-root of \( x \) in the integrand, gives the idea to use the substitution \( t = x^{2} \). Consequently, \( dt = 2x dx \) or in terms of differential of x, \( dx = dt/2x \). The given integral will change its form after this substitution.
02

Apply the Substitution

Perform the variable substitution to modify the integral into a more convenient form. Our substitution gives us \( \int_{0}^{\infty} \sqrt{x} \cdot \sin(x^{2}) dx \ = \int_{0}^{\infty} \sqrt{t} \cdot \sin(t) (dt/2\sqrt{t}) \). Here, \( \sqrt{t} \) from the original integrand and \( \sqrt{t} \) from the denominator will cancel out, resulting in \( \int_{0}^{\infty} \sin(t) dt/2 \).
03

Apply the Comparison Test

The comparison test for convergence can be used on the new integral expression \( \int_{0}^{\infty} \sin(t) dt/2 \). Since the absolute value \( |\sin(t)/2| \leq 1/2 \) for all \( t \geq 0 \), and the integral \( \int_{0}^{\infty} 1/2 dt \) is clearly convergent, by comparison test we can assess that our original integral converges.
04

Rounding off the Conclusion

By applying the substitution and subsequently using the comparison test, we have shown the given integral \( \int_{0}^{\infty} \sqrt{x} \cdot \sin(x^{2}) dx \) is indeed convergent, in spite of its integrand not being bounded as \( x \rightarrow \infty \).

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Most popular questions from this chapter

If \(E \in \mathbb{M}[a, b]\), we define the (Lebesgue) measure of \(E\) to be the number \(m(E):=\int_{a}^{b} \mathbf{1}_{\varepsilon}\). In this exercise, we develop a number of properties of the measure function \(m: \mathbb{M}[a, b] \rightarrow \mathbb{R}\). (a) Show that \(m(\emptyset)=0\) and \(0 \leq m(E) \leq b-a\). (b) Show that \(m([c, d])=m([c, d))=m((c, d])=m((c, d))=d-c\). (c) Show that \(m\left(E^{\prime}\right)=(b-a)-m(E)\). (d) Show that \(m(E \cup F)+m(E \cap F)=m(E)+m(F)\). (e) If \(E \cap F=\emptyset\), show that \(m(E \cup F)=m(E)+m(F)\). (This is the additivity property of the measure function.) (f) If \(\left(E_{k}\right)\) is an increasing sequence in \(\mathbb{M}[a, b]\), show that \(m\left(\bigcup_{k=1}^{\infty} E_{k}\right)=\lim _{k} m\left(E_{k}\right)\). [Hint: Use the Monotone Convergence Theorem.] (g) If \(\left(C_{k}\right)\) is a sequence in \(\mathbb{M}[a, b]\) that is pairwise disjoint (in the sense that \(C_{j} \cap C_{k}=\emptyset\) whenever \(j \neq k\) ), show that (18) $$m\left(\bigcup_{k=1}^{\infty} C_{k}\right)=\sum_{k=1}^{\infty} m\left(C_{k}\right) .$$ (This is the countable additivity property of the measure function.)

(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?

If \(f \in \mathcal{L}[a, b]\), show that \(f^{2}\) is not necessarily in \(\mathcal{L}[a, b]\).

If \(f_{k}(x):=k\) for \(x \in[1 / k, 2 / k]\) and \(f_{k}(x):=0\) elsewhere on \([0,2]\), show that \(f_{k}(x) \rightarrow 0\) but that \(\int_{0}^{2} f_{k}=1\)

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 \(g\).
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