Proving the Limit of an Integral Involving a Periodic Function: An SEO-Friendly Guide

In this comprehensive guide, we will explore the proof of a statement involving the limit of an integral with respect to a periodic function. The goal is to show that the following expression holds true:

limn→∞∫abf(x)g(nx)dx (1/c)∫0cg(t)dt∫abf(x)dx

Step-by-Step Proof Outline

Step 1: Change of Variables

The first step in our proof involves the change of variables technique. We define:

u nx

and correspondingly:

du n dx

and thus:

dx du/n

Using this transformation, the limits of integration change as follows:

When x a, u na When x b, u nb

This allows us to rewrite the integral as:

∫abf(x)g(nx)dx (1/n)∫nanbf(u/n)g(u)du

Step 2: Express the Integral

Now, let's express the integral in a more simplified form:

∫abf(x)g(nx)dx (1/n)∫nanbf(u/n)g(u)du

Step 3: Analyze the Limit as n → ∞

As n approaches infinity, we observe the behavior of f(u/n). If 0 is within the interval [a, b], f(u/n) approaches the value of f(0). Otherwise, it approaches the value of f at the corresponding edge. Since f(x) is piecewise continuous, we can use the dominated convergence theorem.

Step 4: Use the Periodicity of g(x)

Since g(x) is periodic with period c, we evaluate the integral over one period and then sum the results for the entire interval [na, nb].

∫nanbg(u)du Σk?na/c??nb/c?∫0cg(t)dt

The number of periods k in the interval [na, nb] is approximately:

(nb - na)/c (nb - a)/c

Step 5: Combine the Results

By combining steps 1-4, we arrive at the following expression:

∫nanbf(u/n)g(u)du ≈ ((nb - na)/c)∫0cg(t)dt∫abf(x)dx

Step 6: Conclude the Limit

Finally, as n → ∞, the limit can be computed as:

limn→∞(1/n)∫nanbf(u/n)g(u)du (1/c)∫0cg(t)dt∫abf(x)dx

Therefore, we have shown that:

limn→∞∫abf(x)g(nx)dx (1/c)∫0cg(t)dt∫abf(x)dx

Which is what we set out to prove.

Alternative Approach Using Fourier Series

In addition to the above proof, we can also prove the statement using Fourier series. For a periodic function g(x) with period c, we can express it as a Fourier series:

g(x) Σk-∞∞qke^(2πikx/c)

Then, we can rewrite the integral as:

limn→∞∫abf(x)g(nx)dx limn→∞∫abf(x)Σk-∞∞qke^(2πiknx/c)dx

Using the linearity of the integral, we get:

limn→∞Σk-∞∞qk∫abf(x)e^(2πiknx/c)dx

The Riemann-Lebesgue Lemma tells us that:

limn→∞∫abf(x)e^(2πipx)dx 0

for p ≠ 0, so the above sum reduces to:

q0∫abf(x)dx

For the right-hand side of the original equation, we have:

(1/c)∫0cg(t)dt∫abf(x)dx (1/c)∫0cΣk-∞∞qke^(2πikt/c)dt∫abf(x)dx

Using the linearity of the integral, we get:

(1/c)Σk-∞∞qk∫0ce^(2πikt/c)dt∫abf(x)dx

The orthogonality of exponentials with different periods gives:

∫0ce^(2πikt/c)dt 0 for k≠0, and ∫0cdt c

Thus, the right-hand side simplifies to:

(1/c)q0c∫abf(x)dx q0∫abf(x)dx

Both sides are equal, thus we have proven:

limn→∞∫abf(x)g(nx)dx (1/c)∫0cg(t)dt∫abf(x)dx