Problem 388 - Calculus Volume 1

Question

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 \(k \geq 1,|S(2 \pi k)-S(2 \pi(k+1))| \leq \frac{1}{k(2 k+1) \pi}\) \((\) Hint \(: \sin (t+\pi)=-\sin t)\)

Step-by-Step Solution

Answer

The difference is bounded by symmetry and cancellation properties of sine, leading to the inequality.

Step 1: Understand the Problem

We need to find an upper bound for the difference \(|S(2\pi k) - S(2\pi(k+1))|\) for integers \(k \geq 1\). This integral of \(\frac{\sin t}{t}\) has to be analyzed over the intervals \([2\pi k, 2\pi(k+1)]\).

Step 2: Investigate the Integral Behavior

The hint suggests using the periodicity and symmetry property \(\sin(t + \pi) = -\sin(t)\). This property implies that the behavior of \(rac{\sin t}{t}\) over one full period \([0, 2\pi]\) will repeat in an altered form, possibly helping in bounding the integral.

Step 3: Develop the Integral Difference

Consider the integral difference: \( |S(2\pi k) - S(2\pi (k+1))| = \left| \int_{2\pi k}^{2\pi(k+1)} \frac{\sin t}{t} dt \right| \). Because \(\sin(t+\pi) = -\sin(t)\), integrals over [\( 2\pi k, \pi(2k+1) \)] and [\(\pi(2k+1),2\pi(k+1)\)] have opposite contributions.

Step 4: Symmetry and Positive Contribution

The integral \(\int_{2\pi k}^{\pi(2k+1)} \frac{\sin t}{t} dt + \int_{\pi(2k+1)}^{2\pi(k+1)} \frac{\sin t}{t} dt\) evaluates largely to zero due to symmetry. Thus, any remaining positive contribution should be very small.

Step 5: Bound the Remaining Contribution

We need to find bounds on \(\int_{2\pi k}^{2\pi (k+1)} \left| \frac{\sin t}{t} \right| dt\). For large \(k\), \(\frac{1}{t} \approx \frac{1}{2\pi k} \), so the bound becomes: \(\int_{2\pi k}^{2\pi(k+1)} \left| \frac{\sin t}{t} \right| dt \leq \frac{1}{(2k\pi)} \int_{2\pi k}^{2\pi(k+1)} dt = \frac{1}{k(2k+1)\pi}\).

Step 6: Conclude the Bound

The symmetry and alternating contributions imply that the integral over each period largely cancels out, and a small error remains, leading to the inequality:\(|S(2\pi k) - S(2\pi(k+1))| \leq \frac{1}{k(2k+1) \pi}\).

Key Concepts

Integral Calculus

Integral Calculus is a vital branch of mathematics that helps in determining the total accumulation of quantities, among other applications. It is concerned with the concepts of integrating functions to find areas, volumes, central points, and many useful things. In this specific problem, we examine the sine integral shown as \(S(x) = \int_{0}^{x} \frac{\sin t}{t} dt\). This function doesn't simplify to a basic formula but is used in real-world applications nonetheless. Integral Calculus allows us to compute an approximation or estimation of the sine integral over defined intervals.

When dealing with this particular integral, the focus is on estimating its behavior and finding bounds for the differences between two sine integrals over specific intervals. We are interested in how the integral accumulates over one period from \(2\pi k\) to \(2\pi(k+1)\), especially since it involves functions like \(\sin t\) that have periodic properties.

This exercise shows that even though an exact formula for \(S(x)\) is not available, techniques from Integral Calculus, such as boundary estimation and symmetry, can be used to solve practical problems and understand the behavior of functions.

Periodic Functions

Periodic Functions are those that repeat their values at regular intervals. A classic example of a periodic function is the sine function, which repeats every \(2\pi\). This aspect of periodicity is crucial in this problem as it simplifies the integral and helps in boundary estimation.

The hint provided: \(\sin(t + \pi) = -\sin(t)\) aids in revealing a pattern. This essentially means that the behavior of the sine function over one complete cycle can help deduce results about subsequent cycles. When these functions are part of sinusoidal integrals, understanding periodicity helps in predicting how integrals will behave over certain intervals.

In the context of this exercise, the periodic nature of \(\sin t\) allows us to use its repetitive values to form bounds or limits on the integral, contributing to an easier calculation and estimation of \(|S(2\pi k) - S(2\pi(k+1))|\). Its periodic nature also leads to parts of the integral canceling out or simplifying due to symmetry, as discussed in upcoming sections.

Symmetry in Integration

Integrals benefit greatly from symmetry, as symmetrical properties can help simplify complex calculations. In this exercise, using the knowledge that \(\sin(t + \pi) = -\sin(t)\), indicates a symmetrical property useful for evaluating integrals over certain intervals.

Consider integrating \( \int_{2\pi k}^{2\pi(k+1)} \frac{\sin t}{t} dt\). The function can be divided from \([2\pi k, \pi(2k+1)]\) and \([\pi(2k+1), 2\pi(k+1)]\). Due to symmetry, these split integrals mostly cancel each other out since they have opposite contributions. This effectively reduces the complexity and amount of calculation needed.

As a result, much of the integral will evaluate to near zero due to the alternating signs of the sine function over these split intervals. Thereby, remaining positive contributions become negligible. This achievement is primarily through leveraging the symmetric aspects of these intervals, critical in finding bound estimates for the sine integral differences.

Bound Estimation

Bound Estimation is a technique in calculus to determine and limit the possible values an integral can take. Estimating bounds is particularly important when direct solutions are complex or unavailable, as is the case with non-elementary integrals such as the sine integral.

In this context, we estimate the bounds of \(|S(2\pi k) - S(2\pi(k+1))|\) by examining the integral of \(\frac{\sin t}{t}\) and using the periodic and symmetric properties already discussed. It is essential to approximate the contributions of \(\frac{\sin t}{t}\) over large intervals such that it is manageable and understandable.

For large values of \(k\), we approximate \(\frac{1}{t} \approx \frac{1}{2\pi k}\), leading to the bound \(\int_{2\pi k}^{2\pi(k+1)} \left| \frac{\sin t}{t} \right| dt \leq \frac{1}{k(2k+1)\pi}\). This concludes with observing that the calculated bound satisfies the initial estimation requirement. Thus, even with the complex nature of the sine integral, bound estimation provides a useful way to understand and approximate its behavior across intervals, enhancing the entire analysis.