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Chapter 1: Problem 173

Find \(\lim _{n \rightarrow \infty} \int_0^{\infty} \frac{n \cos \left(\sqrt[4]{x / n^2}\right)}{1+n^2 x^2} d x\)

Short Answer

Expert verified
\( \lim_{n \rightarrow \infty} \int_0^{\infty} \frac{n \cos \left(\sqrt[4]{\frac{x}{n^2}} \right)}{1+n^2 x^2} dx = 0 \)

Step by step solution

01

Analyze the integrand

First, examine the given integrand: \[ I(n) = \frac{n \cos \left( \sqrt[4]{\frac{x}{n^2}} \right) }{1 + n^2 x^2} \] As \( n \rightarrow \infty \), the behavior of the integrand will depend on the term in the denominator, \( 1 + n^2 x^2 \).
02

Change of variables

Use the substitution method to simplify the integrand. Let \(t = n^2 x\), then \(x = \frac{t}{n^2}\) and \(dx = \frac{dt}{n^2}\). The integral becomes: \[ I(n) = \frac{1}{n} \int_0^{\infty} \frac{ \cos \left(\frac{t^{1/4}}{n^{1/2}} \right)}{1+t^2} dt \]
03

Simplify the cosine term

As \( n \rightarrow \infty \), \( \frac{t^{1/4}}{n^{1/2}} \rightarrow 0 \). Using the approximation for cosine for small values, \( \cos(x) \approx 1 - \frac{x^2}{2} \), we get: \[ \cos \left( \frac{t^{1/4}}{n^{1/2}} \right) \approx 1 - \frac{1}{2} \left( \frac{t^{1/4}}{n^{1/2}} \right)^2 = 1 - \frac{t^{1/2}}{2 n} \]
04

Simplify the integral

Using the approximation: \[ I(n) \approx \frac{1}{n} \int_0^{\infty} \frac{ 1 - \frac{t^{1/2}}{2n} }{1 + t^2} dt \] Split this into two integrals: \[ I(n) \approx \frac{1}{n} \int_0^{\infty} \frac{1}{1+t^2} dt - \frac{1}{2n^2} \int_0^{\infty} \frac{t^{1/2}}{1+t^2} dt \]
05

Evaluate the integrals

The first integral is a standard result: \[ \int_0^{\infty} \frac{1}{1+t^2} dt = \frac{\pi}{2} \] The second integral involves a more complex function, but it can be shown that it is finite and not dependent on \(n\): \[ \frac{1}{2n^2} \int_0^{\infty} \frac{t^{1/2}}{1+t^2} dt \]
06

Take the limit as n approaches infinity

As \(n \rightarrow \infty\), the second term will go to 0 because of the \( \frac{1}{n^2} \) term: \[ \lim_{n \rightarrow \infty} I(n) = \frac{1}{n} \cdot \frac{\pi}{2} = 0 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

improper integrals
When working on integrals with infinite limits or integrands with unbounded integrability at some points, we encounter improper integrals. These require special techniques to evaluate. In the given exercise, the integral from 0 to infinity is improper because of the upper limit extending to infinity.

Evaluating an improper integral often involves considering the limit of a proper integral as it approaches infinity. For example, to solve \(\lim_{n \to \infty} \, \int_0^{\infty} \, \frac{1}{x+1} \, dx\), you first solve the integral from 0 to some finite value and then take the limit as it goes to infinity.
change of variables in integrals
A common technique in calculus for simplifying integrals is the change of variables strategy. This method often transforms a complex integral into an easier one. In the provided exercise, the change of variables simplified the given integrand. By setting \(t = n^2 x\), we transformed the original integral into a form that is easier to evaluate.

After the substitution and simplifications, the new integral became: \(\frac{1}{n} \int_0^{\infty} \frac{\cos \left(\frac{t^{1/4}}{n^{1/2}} \right)}{1+t^2} dt\). By simplifying further, complex parts of the integrand reduced to more manageable terms.
asymptotic analysis
Asymptotic analysis helps in understanding the behavior of functions as the argument tends towards a limit, often infinity. In the exercise, we analyzed the terms in the integrand as \(n\) approaches infinity. This technique helped in approximating the behavior of the cosine term in the integrand.

Using the small-angle approximation for cosine (\(\cos(x) \approx 1 \frac{\x^{\2}}{2}\)), we evaluated how the terms change and simplify in the limiting process. This analysis showed that certain terms become negligible as \(n\) becomes very large, simplifying the integral further and aiding in finding the final limit.

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Most popular questions from this chapter

The MAA Student Chapter at Wohascum College is about to organize an icosahedron-building party. Each participant will be provided twenty congruent equilateral triangles cut from old ceiling tiles. The edges of the triangles are to be beveled so they will fit together at the correct angle to form a regular icosahedron. What is this angle (between adjacent faces of the icosahedron)?

Let \(a\) and \(d\) be relatively prime positive integers, and consider the sequence \(a, a+d, a+4 d, a+9 d, \ldots, a+n^2 d, \ldots\) Given a positive integer \(b\), can one always find an integer in the sequence which is relatively prime to \(b\) ?

Let \(E\) be an ellipse in the plane. Describe the set \(S\) of all points \(P\) outside the ellipse such that the two tangent lines to the ellipse that pass through \(P\) form a right angle.

Let \(\mathbb{Z} / n \mathbb{Z}\) be the set \(\\{0,1, \ldots, n-1\\}\) with addition modulo \(n\). Consider subsets \(S_n\) of \(\mathbb{Z} / n \mathbb{Z}\) such that \(\left(S_n+k\right) \cap S_n\) is nonempty for every \(k\) in \(\mathbb{Z} / n \mathbb{Z}\). Let \(f(n)\) denote the minimal number of elements in such a subset. Find $$ \lim _{n \rightarrow \infty} \frac{\ln f(n)}{\ln n} $$ or show that this limit does not exist.

Consider the sequence \(4,1 / 3,4 / 3,4 / 9,16 / 27,64 / 81, \ldots\), in which each term (after the first two) is the product of the two previous ones. Note that for this particular sequence, the first and third terms are greater than 1 while the second and fourth terms are less than 1 . However, after that the "alternating" pattern fails: the fifth and all subsequent terms are less than 1. Do there exist sequences of positive real numbers in which each term is the product of the two previous terms and for which all odd-numbered terms are greater than 1, while all even-numbered terms are less than 1? If so, find all such sequences. If not, prove that no such sequence is possible.
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