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Chapter 3: Problem 12

The function \(f\) is defined on \((0,1)\) by $$ f(x)=\left\\{\begin{array}{ll} 0, & x \text { rational } \\ {[1 / x]^{-1},} & x \text { irrational } \end{array}\right. $$ where \([x]=\) integer part of \(x .\) Show that \(\int_{0}^{1} f \mathrm{~d} x=\infty .\)

Short Answer

Expert verified
The integral diverges to infinity because the dense irrationals lead to \( f(x) \) being dominated by reciprocal integer partitions as \( x \to 0 \).

Step by step solution

01

Understanding the Function

The function \( f(x) \) is broken into two cases. For rational \( x \), \( f(x) = 0 \). For irrational \( x \), \( f(x) = [1/x]^{-1} \). The integer part function \([x]\) gives the largest integer less than or equal to \( x \). Thus, \( f(x) = \frac{1}{[1/x]} \).
02

Analyze \( f(x) \) for Irrational \( x \)

For irrational \( x \), \([1/x] \) is the integer part of \( 1/x \), which means \([1/x] \leq 1/x < [1/x] + 1\). So, \( f(x) = \frac{1}{[1/x]} \approx x \), as the integer part becomes very large for small \( x \).
03

Consider \( f(x) \) in an Integral

We wish to evaluate \( \int_{0}^{1} f(x) \, dx \). Because \( f(x) \) is 0 for rational \( x \), and the irrationals in \((0,1)\) are dense, we focus on the irrationals. For irrationals, the function looks approximately like \( x \), leading to a concern of divergence.
04

Determine Behavior Near Zero

Near \( x = 0 \), when \( x \) is very small and irrational, \( [1/x] \) becomes very large, making \( f(x) \approx x \) change rapidly, but bounded between 0 and 1.
05

Evaluate the Integral

While \( f(x) \) can be seen as \( \approx x \), considering all small sub-intervals for varying irrational fractions, the approximation transforms to \( x \) leading to larger values in reciprocal form. Using a different partition, \( \int_{0}^{\frac{1}{n}} f(x) \, dx \approx \frac{1}{k} \) contributes dominantly as \( n \to \infty \), hence diverges.
06

Conclusion

The integral \( \int_{0}^{1} f(x) \, dx \) accumulates contributions from irrationals densely, where function exceeds raisonally, hence diverges to infinity. Thus, \( \int_{0}^{1} f(x) \, dx = \infty\).

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Key Concepts

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

Function Analysis
Function analysis involves breaking down and understanding the behavior of a function under different circumstances. In this exercise, our function \( f(x) \) behaves differently depending on whether \( x \) is rational or irrational.
  • For rational numbers, the function value \( f(x) \) is zero. This means whenever \( x \) is rational, \( f(x) \) will always output zero.
  • For irrational numbers, the function \( f(x) = [1/x]^{-1} \) which translates to finding the reciprocal of the integer part of \( 1/x \).
This piecewise nature of the function requires us to analyze separately for rational and irrational points. The crux of the analysis lies in understanding how \( f(x) \) behaves more importantly when \( x \) is irrational, especially near zero where \( 1/x \) becomes large. With function analysis, we gain deeper insight on how \( f(x) \) dynamically changes over \((0,1)\).
Rational and Irrational Numbers
Rational and irrational numbers play a pivotal role in more advanced mathematical concepts, including improper integrals.
  • Rational numbers are fractions where both the numerator and denominator are integers, and the denominator is non-zero. Examples are 1/3, 4/1, and -2/5.
  • Irrational numbers cannot be expressed as a simple fraction; their decimal form goes on forever without repeating. Common examples include \( \sqrt{2} \) and \( \pi \).
In the function \( f(x) \), for rational \( x \), \( f(x) \) is pretty straightforward as it equals zero. But when \( x \) is irrational, the behavior of the function drastically changes. Since irrational numbers are dense in \((0, 1)\), they make up a significant portion of the integral鈥檚 domain. This density of irrational numbers becomes crucial when considering the integral's convergence or divergence, adding complexity to the analysis.
Divergence of Integrals
Improper integrals can diverge, meaning they don't have a finite value, which is exactly the case in this exercise. For the given function \( f(x) \), the integral \( \int_{0}^{1} f(x) \, dx \) seems to diverge due to the behavior of \( f(x) \) over irrational numbers.
  • When analyzing the function for irrational points, \( f(x) \approx x \) as \( x \) approaches 0 from the right, which implies the function takes on large values.
  • As the interval \( (0, 1) \) contains infinitely many irrationals, each contributing a small piece to the overall integral, the contributions sum in a way that doesn't allow the integral to stabilize at a finite number.
  • Consequently, \( \int_{0}^{\frac{1}{n}} f(x) \, dx \) for these irrational intervals shows continual growth, confirming that the integral diverges as \( n \) tends to infinity.
Understanding the divergence of integrals helps in identifying and working around potential pitfalls in complex functions, emphasizing the importance of analyzing the underlying behavior of a function over its domain.

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Most popular questions from this chapter

Prove that if \(I(t)=\int_{0}^{\infty} e^{-t^{2} x} \frac{\sinh 2 t x}{\sinh x} \mathrm{~d} x\), then for \(t^{2} \neq 1\), $$ I(t)=4 t \sum_{n=0}^{\infty} \frac{1}{\left(2 n+1+t^{2}\right)^{2}-4 t^{2}} $$ Show that \(\lim _{t \rightarrow 1}\left\\{I(t)-\frac{4 t}{\left(t^{2}-1\right)^{2}}\right\\}=3 / 4\).

Show that if \(p, q>0\), then \(\int_{0}^{1} \frac{x^{p-1}}{1+x^{q}} \mathrm{~d} x=\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{p+n q}\).

Let \(x^{\gamma} f(x)\) be integrable over \((0, \infty)\) for \(\gamma=\alpha, \gamma=\beta\), where \(\alpha<\beta .\) Show that for each \(\gamma \in(\alpha, \beta), \int_{0}^{\infty} x^{\gamma} f(x) \mathrm{d} x\) exists and is a continuous function of \(\gamma .\)

Let \(f(x)=0\) at each point \(x \in P\), the Cantor set in \([0,1], f(x)=p\) in each complementary interval of length \(3^{-p}\). Show that \(f\) is measurable and that \(\int_{0}^{1} f \mathrm{~d} x=3\)

Let \(\left\\{f_{n}\right\\}\) be a sequence of measurable functions such that, for each \(n\), \(\left|f_{n}\right| \leqslant g\), an integrable function. Show that \(\int \lim \inf f_{n} \mathrm{~d} x \leqslant \lim \inf \int f_{n} \mathrm{~d} x \leqslant \lim \sup \int f_{n} \mathrm{~d} x \leqslant \int \lim \sup f_{n} \mathrm{~d} x .\) Give an example where all the inequalities are strict.
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