Question

Suppose that \(g\) is positive and nonincreasing on \([a, b)\) and \(\int_{a}^{b} f(x) d x\) exists as a proper or absolutely convergent improper integral. Show that \(\int_{a}^{b} f(x) g(x) d x\) exists and $$ \lim _{x \rightarrow b-} \frac{1}{g(x)} \int_{x}^{b} f(t) g(t) d t=0 $$ HINT: Use Exercise 3.4.6.

Short answer

Question: Show that if g is positive and nonincreasing on the interval [a, b) and the integral of f(x) from a to b exists, then the integral of f(x)g(x) from a to b also exists. Compute the limit as x approaches b from the left side of the expression (∫[x, b] f(t)g(t)dt) / g(x). Answer: Using the properties of integrable functions and the result from Exercise 3.4.6, we have shown that the integral of the product f(x)g(x) from a to b exists. The limit as x approaches b from the left side of the given expression is equal to 0.

Step-by-step solution

Proof of integral of f(x)g(x) existence

Since the integral of f(x) from a to b exists, it means that f(x) is integrable on the interval [a, b). Moreover, g is positive and nonincreasing on [a, b). Following the hint and using Exercise 3.4.6, we can state that for every ε > 0, there exists a δ > 0 such that for all x in [a, b) with (b - x) < δ, it holds that (b - x)g(x) < ε. Now, let's define a new function h(x) = f(x)g(x). Since f(x) is integrable and g(x) is positive, the product f(x)g(x) (i.e., h(x)) will also be a function on the interval [a, b). We need to show that the integral of h(x) from a to b exists. Let's consider the integral of h(x) over [x, b) for some x in [a, b): $$ \int_{x}^{b} h(t) d t = \int_{x}^{b} f(t) g(t) d t $$ To show that this integral exists, we will use the result from Exercise 3.4.6 and the definition of h(x).

Compute the limit

Our goal is to compute the limit of the following expression as x approaches b from the left side: $$ \lim _{x \rightarrow b-} \frac{1}{g(x)} \int_{x}^{b} f(t) g(t) d t $$ Let's manipulate this expression: $$ \lim _{x \rightarrow b-} \frac{1}{g(x)} \int_{x}^{b} f(t) g(t) d t = \lim _{x \rightarrow b-} \frac{1}{g(x)} \int_{x}^{b} h(t) d t $$ Notice that since h(x) = f(x)g(x), our previous result implies that the integral of h(t) from x to b exists. Now, let's use Exercise 3.4.6 result to compute the limit. Using Exercise 3.4.6, for any ε > 0, there exists a δ > 0 such that for all x in [a, b) with (b - x) < δ, it holds that (b - x)g(x) < ε. Then, we can write: $$ \lim _{x \rightarrow b-} \frac{1}{g(x)} \int_{x}^{b} h(t) d t = \lim _{x \rightarrow b-} \frac{\int_{x}^{b} h(t) d t}{g(x)} \leq \lim _{x \rightarrow b-} \frac{(b - x)g(x)}{g(x)} $$ Now, as x approaches b from the left side, the limit becomes: $$ \lim _{x \rightarrow b-} \frac{(b - x)g(x)}{g(x)} = \lim _{x \rightarrow b-} (b - x) = 0 $$ Therefore, the given limit exists and is equal to 0: $$ \lim _{x \rightarrow b-} \frac{1}{g(x)} \int_{x}^{b} f(t) g(t) d t=0 $$ As a conclusion, if g is positive and nonincreasing on the interval [a, b) and the integral of f(x) from a to b exists, then the integral of the product f(x)g(x) from a to b exists and the given limit is equal to 0.

Key concepts

Improper Integral

An improper integral arises when evaluating the integral of a function over an unbounded interval, or when the function itself has an infinite discontinuity within the interval of integration. These integrals can't be calculated using the standard definition of the integral because they involve infinite limits or undefined behavior at certain points. In real analysis, to solve improper integrals, we extend the concept of the integral by taking limits. This means we define the integral of f(x) from a to \(b\) (\[ \int_{a}^{b} f(x) dx \]), where \(b\) could be infinity, or where \(f\) has an infinite discontinuity, as the limit of the integral from \(a\) to \(x\) as \(x\) approaches \(b\). This method hinges on the convergence of the integral, which is to say, if the limit exists, the improper integral is said to be convergent; otherwise, it is divergent.

To illustrate, with the function \(g(x)\) being positive and nonincreasing on the interval [a, b), the existence of the improper integral of another function \(f(x)\) on [a, b) informs us about the behavior of \(f\) as we move towards \(b\). If the improper integral of \(f\) is convergent, we can say that the product \(f(x)g(x)\), under certain conditions, will also produce a convergent improper integral.

Nonincreasing Function

A nonincreasing function is one that does not increase as the independent variable increases. In other words, for any two points \(x_1\) and \(x_2\) in the domain of the function where \(x_1 < x_2\), the value of the function at \(x_2\) is less than or equal to the value of the function at \(x_1\); symbolically, \(g(x_2) \leq g(x_1)\). This property is significant in real analysis, especially when discussing the convergence of integrals, because nonincreasing functions can act as bounds or constraints on a function's growth.

In the context of our exercise, the function \(g\) is given to be positive and nonincreasing on [a, b). Such a condition helps in analyzing the behavior of integrals near the upper limit, b, and plays a pivotal role in the convergence of the integral involving the product of \(f(x)\) and \(g(x)\), as the nonincreasing nature of \(g\) ensures that the product does not 'explode' as \(x\) approaches \(b\).

Limit of a Function

The limit of a function is a fundamental concept in calculus, which describes the behavior of a function as its argument approaches a certain value. For a function \(f(x)\), the limit as \(x\) approaches a value \(c\) is the value that \(f(x)\) gets closer to as \(x\) progressively gets closer to \(c\). This notion is crucial in defining the continuity, derivatives, and integrals of functions. When a function approaches a limit, it does not need to ever reach that limit; it only needs to approach arbitrarily close to it.

For example, in our problem, the limit as \(x\) approaches \(b\) from the left of \(1/g(x)\) times the integral of \(f(t)g(t)\) from \(x\) to \(b\) is treated. Even though \(x\) never actually equals \(b\), we consider the values of our expression as \(x\) gets infinitesimally close to \(b\). This limit underpins the final conclusion that must be reached, establishing the behavior of the integral near the upper bound of its integration interval.

Convergence of an Integral

The convergence of an integral is a condition that indicates whether or not an integral represents a finite area under the curve of a function over a specified interval. If an integral is convergent, the function's behavior can be described by a real number representing the area. The concept of convergence applies to both proper and improper integrals. For proper integrals, convergence is guaranteed since they are taken over finite intervals and the functions are well-behaved (bounded and without discontinuity). An improper integral, however, requires the existence of a limit to prove convergence.

In the exercise at hand, we look at the product \( f(x)g(x) \) and show that the integral of this product over the interval [a, b) converges. To do this, we need to consider the limits of \( f(x) \) and \( g(x) \) as \( x \) approaches \( b \) from the left. If both functions behave such that their product does not diverge to infinity and the respective limits can be defined and are finite, the integral will converge. The existence of this limit is crucial, as it signifies that applying the operation of integration to the function over the entire interval results in a finite, well-defined quantity.