Problem 36 - A Course in Calculus and Real Analysis

Question

Let \(a \in \mathbb{R}\) and let \(f, g\) : \([a, \infty) \rightarrow \mathbb{R}\) be any functions. (i) If \(f\) is differentiable, \(\int_{a}^{\infty}\left|f^{\prime}(t)\right| d t\) is convergent, \(f(x) \rightarrow 0\) as \(x \rightarrow \infty\), \(g\) is continuous, and the function \(G:[a, \infty) \rightarrow \mathbb{R}\) defined by \(G(x):=\) \(\int_{a}^{x} g(t) d t\) is bounded, then show that \(\int_{a}^{\infty} f(t) g(t) d t\) is convergent. (ii) If \(f\) is differentiable, \(\int_{a}^{\infty}\left|f^{\prime}(t)\right| d t\) is convergent, \(g\) is continuous, and \(\int_{a}^{\infty} g(t) d t\) is convergent, then show that \(\int_{a}^{\infty} f(t) g(t) d t\) is convergent. (Hint: Use Integration by Parts.)

Step-by-Step Solution

Answer

In order to prove that \(\int_{a}^{\infty} f(t) g(t) dt\) is convergent, we consider two cases: (i) We show that \(|\int_{a}^{\infty} f(t) g(t) dt| \leq \int_{a}^{\infty} (\int_{t}^{\infty} |f'(u)| du) M dt\), where M is the upper bound for G(x) in the interval [a,∞), by using the Mean Value Theorem, the properties of G(x), and Fubini's Theorem to interchange the order of integration. Since \(\int_{a}^{\infty}|f'(t)| dt\) is convergent, then \(\int_{a}^{\infty} f(t) g(t) dt\) is convergent. (ii) We use Integration by Parts and Fubini's Theorem to transform the given integral into a product of convergent integrals. Since \(\int_{a}^{\infty} g(t) dt\) is convergent and \(\int_{a}^{\infty} |f'(t)| dt\) is convergent, their product is convergent, proving that the improper integral \(\int_{a}^{\infty} f(t) g(t) dt\) is convergent.

Step 1: Define constant M

Define a constant M, which is the upper bound for G(x) in the interval [a,∞) as G(x) is a bounded function.

Step 2: Use the Mean Value Theorem

Apply the Mean Value Theorem for integrals on the interval [x, x+h], where h>0: \(f(x+h)-f(x)=f'(c)h\) for some \(c\in [x,x+h]\). Since \(|f'(c)| \leq \frac{1}{h} \int_{x}^{x+h} |f'(t)| dt\), we have \(|f(x+h)-f(x)| \leq \int_{x}^{x+h} |f'(t)| dt\). Take \(h\rightarrow \infty\), and given that \(f(x) \rightarrow 0\) as \(x \rightarrow \infty\), we get \(|f(x+h)| \leq \int_{x}^{\infty} |f'(t)| dt\).

Step 3: Use the properties of G(x)

We have \(G(x):= \int_{a}^{x} g(t) dt\). On the interval [a,∞), G(x) is bounded, so |G(x)| ≤ M for all x ≥ a.

Step 4: Prove that the integral is convergent

We want to show that \(\int_{a}^{\infty} f(t) g(t) dt\) is convergent. To do this, use the given information about f and G as follows: \(|\int_{a}^{\infty} f(t) g(t) dt| \leq \int_{a}^{\infty} |f(t)| |g(t)| dt \leq \int_{a}^{\infty} (\int_{t}^{\infty} |f'(u)| du) M dt\). Apply Fubini's Theorem to interchange the order of integration: \(\int_{a}^{\infty} (\int_{t}^{\infty} |f'(u)| du) M dt = M\int_{a}^{\infty} (\int_{a}^{u} dt) |f'(u)| du = M\int_{a}^{\infty} (u-a) |f'(u)| du\). Since \(\int_{a}^{\infty}|f'(t)| dt\) is convergent, then \(\int_{a}^{\infty} (u-a) |f'(u)| du\) is convergent. So, \(\int_{a}^{\infty} f(t) g(t) dt\) is convergent. #Case 2: Integral of g(t) converges#

Step 1: Integrate by Parts

We use integration by parts for the integral \(\int_{a}^{\infty} f(t) g(t) dt\). Let \(u=f(t)\) and \(dv=g(t)dt\). This yields \(du=f'(t)dt\) and \(v=\int_{a}^{t} g(u) du\).

Step 2: Express the integral

Applying integration by parts gives: \(\int_{a}^{\infty} f(t) g(t) dt = \lim_{b\to\infty} [f(t) (\int_{a}^{t} g(u) du)]_{a}^{b} - \lim_{b\to\infty} \int_{a}^{b} f'(t) (\int_{a}^{t} g(u) du) dt\).

Step 3: Apply limits

As \(f(x) \rightarrow 0\) as \(x \rightarrow \infty\), we have: \(\lim_{b\to\infty} [f(t) (\int_{a}^{t} g(u) du)]_{a}^{b} = 0\).

Step 4: Use Fubini's Theorem on the second term

For the second term, interchange order of integration using Fubini's Theorem: \(\lim_{b\to\infty} \int_{a}^{b} f'(t) (\int_{a}^{t} g(u) du) dt = \lim_{b\to\infty} \int_{a}^{b} (\int_{a}^{t} g(u) du) f'(t) dt\) \( = \int_{a}^{\infty} (\int_{a}^{t} g(u) du) f'(t) dt\).

Step 5: Prove the convergence

The given \(\int_{a}^{\infty} g(t) dt\) is convergent, and \(\int_{a}^{\infty} |f'(t)| dt\) is convergent. The product of these integrals converges, so the improper integral \(\int_{a}^{\infty} f(t) g(t) dt\) is convergent.

Key Concepts

Mean Value Theorem

The Mean Value Theorem (MVT) for integrals provides a fascinating insight that helps relate a continuous function's behavior over an interval to its behavior at a single point within that interval. For instance, if you have a function that behaves smoothly and continuously over an interval, there is at least one point where the function's derivative equals the average rate of change over that interval.

In the context of the given problem, MVT is used to draw a connection between the behavior of the function \(f\) and its derivative \(f'\). Specifically, this theorem helps in dealing with functions involved in the convergence of integrals by allowing us to bring together the potential effects of \(f\) over an infinite interval.

• The theorem states that, for a differentiable function \(f\) over \([a, b]\), there exists some \(c \in (a, b)\) where \(f(b) - f(a) = f'(c)(b - a)\).
• When extended to integrals, it describes a similar property that enables the estimation of integral values based on the integral of the absolute value of the function's derivative.

When \(|f'(t)|\) is integrable and \(f(x) \to 0\) as \(x \to \infty\), MVT helps assert the boundedness and behavior necessary for the larger problem at hand.

Integration by Parts

Integration by parts is a powerful method based on the product rule for differentiation, allowing more complex integrals to be expressed in simpler forms. This technique is essential when dealing with the product of two functions in an integral.

For the exercise, we employ integration by parts to showcase the convergence of \( \int_{a}^{\infty} f(t) g(t) dt \). Here's how it's done:

• Choose functions \(u = f(t)\) and \(dv = g(t)dt\).
• Then, \(du = f'(t)dt\) and \(v = \int g(u) du\).
• The integration by parts formula is: \( \int u \, dv = uv - \int v \, du \).

This technique shows how differentiable functions diminish over an infinite interval, with key assumptions such as \(f(x) \to 0\) helping leverage the limits effectively. It is a critical tool in proving the convergence of integrals with expressions involving function products.

Fubini's Theorem

Fubini's Theorem is a principle from calculus that allows for the interchange of the order of integration in a double integral under certain conditions. This theorem is particularly useful when the evaluation of an improper integral can become too complex if done in one go.

In the solution, Fubini's theorem is applied to change the order of integration, which simplifies the integral of interest into more manageable components. For two-variable integration or for articulating the convergence of a double iterated integral, Fubini's theorem is crucial.

• The theorem facilitates the calculation: \( \int_{a}^{\infty} \int_{t}^{\infty} \ldots \) into simpler forms, ensuring convergence.
• The use of this theorem ensures that the region of integration does not depend on the path or sequence chosen for integration.

By allowing such interchange under the convergence of functions and absolute integrability, Fubini's theorem greatly expands the power and scope of integral calculus in higher dimensions.

Bounded Functions

A bounded function is one where all its values lie within some fixed interval on its Domain. Understanding and applying the concept of bounded functions is vital when demonstrating the convergence of integrals over infinite intervals.

In this context, the presence of a bounded function \(G(x)\) ensures that our "integrating partner" function does not grow uncontrollably. This characteristic is significant because it provides assurance that any integral or expression involving \(G(x)\) remains under control and does not diverge.

• Being bounded means that there exists some constant \(M\) such that \(|G(x)| \leq M\) for all x in its domain.
• A bounded \(G(x)\) plays a crucial role in employing the properties of the other functions involved to demonstrate the overall convergence of \( \int f(t)g(t) dt \).

In sum, the bounded nature of \(G(x)\) influences the convergence results by constraining the behavior and magnitude of expressions containing it. This constraint is pivotal especially in scenarios where other involved functions have converging integrals.