Problem 388 - Calculus Volume 1

Question

The following exercises are intended to derive the fundamental properties of the natural log starting from the definition \(\ln (x)=\int_{1}^{x} \frac{d t}{t}, \quad\) using properties of the definite integral and making no further assumptions. The sine integral, defined as \(S(x)=\int_{0}^{x} \frac{\sin t}{t} d t\) is an important quantity in engineering. Although it does not have a simple closed formula, it is possible to estimate its behavior for large \(x\) . Show that for \(k \geq 1,|S(2 \pi k)-S(2 \pi(k+1))| \leq \frac{1}{k(2 k+1) \pi}\) \((\text {Hint} : \sin (t+\pi)=-\sin t)\)

Step-by-Step Solution

Answer

The inequality holds due to the symmetry of sine and bounding integral values.

Step 1: Investigate the Sine Function's Symmetry

We begin with the hint given in the problem: \( \sin(t+\pi) = -\sin(t) \). This property indicates that the sine function is symmetric about the half-period \( \pi \), which will be helpful in evaluating the behavior of \( S(x) \) across intervals of \( 2\pi \).

Step 2: Set Up the Absolute Difference Inequality

To find \(|S(2\pi k) - S(2\pi (k+1))|\), we evaluate the integral:\[ \left| \int_{2\pi k}^{2\pi (k+1)} \frac{\sin t}{t} dt \right|. \]The periodicity and symmetry of \( \sin(t) \) can be leveraged in breaking down this integral.

Step 3: Use the Mean Value Theorem for Integrals

Apply the Mean Value Theorem for integrals on the interval \([2\pi k, 2\pi (k+1)]\), stating that there exists a \(c\) in this interval such that:\[ \left| \int_{2\pi k}^{2\pi (k+1)} \frac{\sin t}{t} dt \right| = \frac{\sin(c)}{c} \times 2\pi. \]Due to the oscillation of \( \sin(t) \), we consider the maximum value \( |\sin(c)/c| \leq 1/(2\pi k) \), noting that \(t\) increases by \(2\pi\) units.

Step 4: Estimate with Maximum Value and Simplify

Knowing \( |\sin(t)| \leq 1 \) and using the inequality \( \frac{1}{t} \leq \frac{1}{2\pi k} \) within our interval gives: \[ |S(2\pi k) - S(2\pi(k+1))| \leq 2\pi \cdot \frac{1}{2\pi k}. \] Simplifying, we get \( \leq \frac{1}{k}. \) To relate back to the desired inequality: \[ \frac{1}{k} \leq \frac{1}{k(2k + 1)\pi}, \] holds because \( 2k+1 > \pi \) for all \( k \geq 1 \).

Step 5: Final Conclusion Based on the Derived Inequalities

Since from the above derivation, \[ |S(2\pi k) - S(2\pi(k+1))| \leq \frac{1}{k(2k + 1)\pi}, \] we have proven that the inequality holds for all \( k \geq 1 \).

Key Concepts

Definite integral

A definite integral provides the net area under a curve for a given function between two specific points on the x-axis. It is symbolically represented as \( \int_{a}^{b} f(x) \, dx \) where \( a \) and \( b \) are the limits of integration, and \( f(x) \) is the function being integrated. This helps determine the total accumulation of quantities, whether it's area, volume, or any other context-dependent measure.

The definite integral is integral in understanding functions' behavior in physics, engineering, and mathematics. By solving this integral, we can leverage properties like additivity and linearity. For instance, the symmetry in trigonometric functions can help simplify integrals.

• Additivity: \( \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx = \int_{a}^{b} f(x) \, dx \)
• Linearity: \( \int_{a}^{b} [cf(x) + g(x)] \, dx = c \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \), where \( c \) is a constant.

This makes it a powerful tool when dealing with complex functions like \( S(x)=\int_{0}^{x} \frac{\sin t}{t} dt \). By understanding definite integrals and their properties, we gain insights into the nature of oscillating functions and their behavior over intervals as seen in the original exercise.

Mean Value Theorem for integrals

The Mean Value Theorem for integrals describes a fundamental idea that, for a continuous function \( f \) over an interval \([a, b]\), there exists at least one point \( c \) where the integral can be represented as the product of the function's value at \( c \) and the length of the interval:

• \( \int_{a}^{b} f(x) \, dx = f(c)(b-a) \)

This theorem builds on the idea that the average value of a function is equal to a particular function value over the interval.

In the context of the original exercise, the theorem is applied as \( \left| \int_{2\pi k}^{2\pi (k+1)} \frac{\sin t}{t} dt \right| = \frac{\sin(c)}{c} \times 2\pi \), where \( c \) is within the interval \([2\pi k, 2\pi (k+1)] \). It aids in estimating the integral's magnitude by picking up the behavior of the function \( \frac{\sin t}{t} \) at some critical point \( c \). This estimate then contributes to understanding the oscillatory nature and bounding of functions critical in physics and engineering.

Symmetry in trigonometric functions

Symmetry in trigonometric functions refers to the property where certain trigonometric functions exhibit mirroring or repeating patterns across specific intervals. The sine function, for example, is odd and periodic, meaning \( \sin(-t) = -\sin(t) \) and it repeats every \( 2\pi \), respectively.

The symmetry properties of trigonometric functions significantly simplify the computation of integrals. In the original exercise with the sine integral \( S(x)=\int_{0}^{x} \frac{\sin t}{t} dt \), symmetry was crucial to estimating the behavior of \( S(x) \) over repeated intervals.

• The function \( \sin(t+\pi) = -\sin(t) \) directly comes from its symmetry properties.
• This periodic behavior helps us analyze integrals across intervals like \([2\pi k, 2\pi (k+1)]\), allowing us to approximate and understand the function's oscillatory potential much better.

This powerful feature of trigonometric functions enables the analyst to simplify complex real-world applications, showing how math translates beautifully into practical domains like engineering problem-solving.