Question

Let \(P=\left\\{x_{0}, x_{1}, \ldots, x_{n}\right\\}\) be a partition of \([a, b], c_{0}=x_{0}=a, c_{n+1}=x_{n}=b\) and \(x_{j-1} \leq c_{j} \leq x_{j}, j=1,2, \ldots, n\). Verify that $$ \sum_{j=1}^{n} g\left(c_{j}\right)\left[f\left(x_{j}\right)-f\left(x_{j-1}\right)\right]=g(b) f(b)-g(a) f(a)-\sum_{j=0}^{n} f\left(x_{j}\right)\left[g\left(c_{j+1}\right)-g\left(c_{j}\right)\right] $$ Use this to prove that if \(\int_{a}^{b} f(x) d g(x)\) exists, then so does \(\int_{a}^{b} g(x) d f(x),\) and $$ \int_{a}^{b} g(x) d f(x)=f(b) g(b)-f(a) g(a)-\int_{a}^{b} f(x) d g(x) $$

Short answer

Question: Verify the given formula and use it to prove the existence of the integral \(\int_{a}^{b} g(x) df(x)\) given that the integral \(\int_{a}^{b} f(x) dg(x)\) exists. Answer: The given formula has been verified as true through algebraic manipulations. Using this formula, we have proven that the integral \(\int_{a}^{b} g(x) df(x)\) exists because the limit of the left side of the verified formula exists. The value of the integral is given by the expression: $$ \int_{a}^{b} g(x) df(x) = g(b) f(b) - g(a) f(a) - \int_{a}^{b} f(x) dg(x) $$

Step-by-step solution

Verify the given formula

We are given the formula: $$ \sum_{j=1}^{n} g\left(c_{j}\right)\left[f\left(x_{j}\right)-f\left(x_{j-1}\right)\right] = g(b) f(b) - g(a) f(a) - \sum_{j=0}^{n} f\left(x_{j}\right)\left[g\left(c_{j+1}\right) - g\left(c_{j}\right)\right] $$ We start by expanding the summations on both sides of the equation. On the left side, expand the sum: $$ \sum_{j=1}^{n} g\left(c_{j}\right)\left[f\left(x_{j}\right)-f\left(x_{j-1}\right)\right] = g\left(c_{1}\right)(f\left(x_{1}\right) - f\left(x_{0}\right)) + g\left(c_{2}\right)(f\left(x_{2}\right) - f\left(x_{1}\right)) + \dots + g\left(c_{n}\right)(f\left(x_{n}\right) - f\left(x_{n-1}\right)) $$ On the right side, expand the sum: $$ g(b) f(b) - g(a) f(a) - \sum_{j=0}^{n} f\left(x_{j}\right)\left[g\left(c_{j+1}\right) - g\left(c_{j}\right)\right] = g(b) f(b) - g(a) f(a) - \left[f\left(x_{0}\right)(g\left(c_{1}\right) - g\left(c_{0}\right)) + f\left(x_{1}\right)(g\left(c_{2}\right) - g\left(c_{1}\right)) + \dots + f\left(x_{n}\right)(g\left(c_{n+1}\right) - g\left(c_{n}\right))\right] $$ Now, compare the left and right sides: $$ g\left(c_{1}\right)(f\left(x_{1}\right) - f\left(x_{0}\right)) + g\left(c_{2}\right)(f\left(x_{2}\right) - f\left(x_{1}\right)) + \dots + g\left(c_{n}\right)(f\left(x_{n}\right) - f\left(x_{n-1}\right)) \\ = g(b) f(b) - g(a) f(a) - \left[f\left(x_{0}\right)(g\left(c_{1}\right) - g\left(c_{0}\right)) + f\left(x_{1}\right)(g\left(c_{2}\right) - g\left(c_{1}\right)) + \dots + f\left(x_{n}\right)(g\left(c_{n+1}\right) - g\left(c_{n}\right))\right] $$ The summation terms on both sides telescope, yielding: $$ g\left(c_{1}\right)f\left(x_{0}\right) + g\left(c_{2}\right)f\left(x_{1}\right) + \dots + g\left(c_{n}\right)f\left(x_{n-1}\right) = g(b) f(b) - g(a) f(a) - \left[f\left(x_{0}\right)g\left(c_{0}\right) + f\left(x_{1}\right)g\left(c_{1}\right) + \dots + f\left(x_{n}\right)g\left(c_{n}\right)\right] $$ The formula is true, as the terms on both sides are equal.

Prove the existence of \(\int_{a}^{b} g(x) df(x)\) and its value

Given that the integral \(\int_{a}^{b} f(x) dg(x)\) exists, we can use the verified formula to prove the existence of the integral \(\int_{a}^{b} g(x) df(x)\) and its value. By the verified formula, we have: $$ \sum_{j=1}^{n} g\left(c_{j}\right)\left[f\left(x_{j}\right)-f\left(x_{j-1}\right)\right] = g(b) f(b) - g(a) f(a) - \sum_{j=0}^{n} f\left(x_{j}\right)\left[g\left(c_{j+1}\right) - g\left(c_{j}\right)\right] $$ As the integral \(\int_{a}^{b} f(x) dg(x)\) exists, the Riemann-Stieltjes sums approach this integral as the partition gets finer. In other words, $$ \lim_{n \to \infty} \sum_{j=0}^{n} f\left(x_{j}\right)\left[g\left(c_{j+1}\right) - g\left(c_{j}\right)\right] = \int_{a}^{b} f(x) dg(x) $$ Now, for the existence of the integral \(\int_{a}^{b} g(x) df(x)\), we need to show that the Riemann-Stieltjes sums for \(g(x)\) approach a certain value as the partitions get finer, i.e., we need to show that the limit as \(n\) approaches infinity exists for the left side of the verified formula. Notice that, by taking limit on both sides of the verified formula we get: $$ \lim_{n \to \infty} \sum_{j=1}^{n} g\left(c_{j}\right)\left[f\left(x_{j}\right)-f\left(x_{j-1}\right)\right] = g(b) f(b) - g(a) f(a) - \lim_{n \to \infty} \sum_{j=0}^{n} f\left(x_{j}\right)\left[g\left(c_{j+1}\right) - g\left(c_{j}\right)\right] $$ The limit on the right side exists because \(\int_{a}^{b} f(x) dg(x)\) exists, and now we can conclude that the limit on the left side also exists. Therefore, we have proven that the integral \(\int_{a}^{b} g(x) df(x)\) exists. Furthermore, we can determine the value of this integral by using the verified formula, which is equal to the limit of the left side of the formula: $$ \int_{a}^{b} g(x) df(x) = g(b) f(b) - g(a) f(a) - \int_{a}^{b} f(x) dg(x) $$ Thus, we have proven the existence of the integral \(\int_{a}^{b} g(x) df(x)\) and also found its value.

Key concepts

Integration theory

The concept of integration is a central part of calculus. While most students are familiar with the Riemann integral, the Riemann-Stieltjes integral is a slightly more advanced concept.
In integration theory, we aim to find the accumulation of quantities. For the Riemann-Stieltjes integral, it accumulates a function's values with respect to another function, rather than just with respect to the variable itself.

• The Riemann integral uses a linear measure, evaluating a function over an interval with respect to the variable incrementally.
• The Riemann-Stieltjes integral, on the other hand, sums the function values, but weights them using another function. This second function modifies the accumulated measure, potentially changing the focus and accumulation pace.
• As such, it provides a generalized tool that can represent real-world problems better where variations in data accumulation are not uniform.

Understanding Riemann-Stieltjes integrals can offer insights into problems involving non-uniform data or where changes depend on independent variables.

Partition of intervals

One essential part of computing integrals, especially Riemann-Stieltjes integrals, is the partition of intervals.
A partition divides the total interval \( [a, b] \) into smaller subintervals, which allows for the evaluation of a function over these chunks separately.

• Formally, a partition \( P = \{x_{0}, x_{1}, \ldots, x_{n}\} \) represents a sequence where \( a = x_{0} < x_{1} < \ldots < x_{n} = b \).
• Each \( x_{j} \) is a chosen point in the interval and helps in achieving the sum approximation to the integral by choosing a sample point \( c_{j} \) in each subinterval \( [x_{j-1}, x_{j}] \).
• As the partition becomes 'finer' (i.e., adding more points and making intervals narrower), the sum converges to the value of the integral.

Choosing appropriate partitions and sample points are crucial as they directly affect the precision of the integral approximation.

Properties of integrals

Understanding the properties of integrals is key to solving complex problems involving Riemann-Stieltjes integrals.
These properties often reflect the behaviors of sums and limits, integral operations that maintain the integrity of certain algebraic operations.

• The linearity property shows that integrals can be freely distributed over addition and pulled out scalars from integration, akin to how algebraic sums work: \( \int (cf + dg)dx = c\int fdx + d\int gdx\).
• Given two integrals within the same limits, their difference reflects the integral of their difference: \[(b)- \int_a^b f(x)g(x)dx = \int_a^b\ (f(x) - g(x))dx\].
• If you prove the existence of one integral in a pair of related functions, such as \( \int_a^b f(x) dg(x) \), you can often show the existence of another, like \( \int_a^b g(x) df(x) \) through their interrelated properties.

These properties help mathematicians and scientists handle complex integrals by allowing simplifications that can make the computations feasible and align with numerical methods.