Question

Suppose that \(g\) is an integrable periodic function with period \(T\) (a) If \(0 \leq a \leq T,\) show that $$\int_{0}^{T} g(x) d x=\int_{a}^{a+T} g(x) d x$$ Hint: Show first that \(\int_{0}^{a} g(x) d x=\int_{T}^{a+T} g(x) d x .\) Consider the change of variable \(s=\) \(x-T\) in the second integral. (b) Show that for any value of \(a,\) not necessarily in \(0 \leq a \leq T\) $$\int_{0}^{T} g(x) d x=\int_{a}^{a+T} g(x) d x$$ (c) Show that for any values of \(a\) and \(b\), $$\int_{a}^{a+T} g(x) d x=\int_{b}^{b+T} g(x) d x$$

Short answer

Based on the solution provided above, here is a short answer to the given exercise: We proved that for a periodic function g(x) with period T: 1. The integral over a whole period from 0 to T is equal to the integral from a to a + T, for any value of a (not just for 0 <= a <= T), by showing the relationship between the integrals with different limits of integration and using the properties of periodic functions. 2. Extending the proof for any value of a, we showed that the integral from a to a + T is equal to the integral from b to b + T, for any values of a and b, by equating both integrals to the integral from 0 to T and using the results from part (b).

Step-by-step solution

Relate the given integrals

Since we want to prove that $$\int_{0}^{T} g(x) dx =\int_{a}^{a+T} g(x) dx,$$ let's break the integral from 0 to T into two parts, from 0 to a and from a to T: $$\int_{0}^{T} g(x) dx =\int_{0}^{a} g(x) dx + \int_{a}^{T} g(x)dx.$$

Show the hint relation

We need to show that $$\int_{0}^{a} g(x) dx=\int_{T}^{a+T} g(x) dx .$$ To do this, we will use the change of variable s = x - T, so \(x = s + T\) and \(dx = ds\). The limits of integration change as well which gives us: $$\int_{0}^{a} g(x) dx = \int_{-T}^{-T+a} g(s+T) ds.$$ Since g is a periodic function with period T, we have g(s+T) = g(s), so the integral becomes $$\int_{-T}^{-T+a} g(s+T) ds = \int_{-T}^{-T+a} g(s) ds.$$ We notice that this last integral has exactly the same integrand as the one we want to prove but with different limits of integration, since the change of variables was just a translation. Therefore, we have shown that $$\int_{0}^{a} g(x)dx = \int_{-T}^{-T+a} g(s) ds = \int_{T}^{a+T} g(x) dx.$$

Conclude the proof for (a)

Refer back to Step 1 where we broke the integral from 0 to T into two parts, and substitute the relation we just found in Step 2: $$\int_{0}^{T} g(x) dx = \int_{0}^{a} g(x) dx + \int_{a}^{T} g(x) dx = \int_{T}^{a+T} g(x) dx + \int_{a}^{T} g(x) dx.$$ Now we have shown that $$\int_{0}^{T} g(x) dx = \int_{a}^{a+T} g(x) dx.$$ (b)

Change a, not necessarily in 0

We can now modify the proof to hold for any value of a, not just when 0 <= a <= T. If a is not in this range, a can be written as \(a = Tq + a'\), where q is an integer and 0 <= a' <= T. Now, we have $$\int_{a}^{a+T} g(x) dx = \int_{Tq+a'}^{Tq+a'+T} g(x) dx.$$ From the result in part (a), we know that $$\int_{0}^{T} g(x) dx = \int_{a'}^{a'+T} g(x) dx.$$ We can apply this result to the integral above, with \(a'\) taking the place of \(a\): $$\int_{Tq+a'}^{Tq+a'+T} g(x) dx = \int_{0}^{T} g(x) dx.$$ Hence, we have shown that for any value of a, $$\int_{0}^{T} g(x) dx = \int_{a}^{a+T} g(x) dx.$$ (c)

Show the integral is the same for any a and b

Lastly, we want to show that for any values of a and b, $$\int_{a}^{a+T} g(x) dx = \int_{b}^{b+T} g(x) dx.$$ Using the result from part (b), we know that $$\int_{0}^{T} g(x) dx = \int_{a}^{a+T} g(x) dx$$ and $$\int_{0}^{T} g(x) dx = \int_{b}^{b+T} g(x) dx.$$ Since both of these integrals are equal to the integral from 0 to T, it follows that they are equal to each other: $$\int_{a}^{a+T} g(x) dx = \int_{b}^{b+T} g(x) dx.$$ This concludes the proof for part (c) and the whole exercise.

Key concepts

Periodic Functions

Periodic functions are fascinating because they repeat their values at regular intervals. Imagine a wave traveling across the ocean. Every crest and trough occurs at predictable intervals, much like a periodic function. Mathematically, a function \( g(x) \) is periodic if there exists a positive number \( T \) (known as the period) such that:

• \( g(x + T) = g(x) \) for all values of \( x \).

This means that the function behaves the same over every interval of length \( T \), providing a repetitive pattern. In this exercise, the periodic nature of \( g(x) \) with period \( T \) simplifies integrals because it's possible to slide these intervals along the x-axis without changing the integral's value. The inherent repetition in periodic functions beautifully ties into how these integrals are evaluated over different intervals.

Change of Variables

The change of variables is a powerful technique in calculus that simplifies the evaluation of integrals. It works by transforming the variable of integration to make the integral easier to work with.
A classic example is the simple substitution when calculating the area under a curve. By switching variables, we can sometimes turn a complex problem into a straightforward one.
In the exercise, we use the substitution \( s = x - T \). This modifies our limits of integration and the function, but due to the periodic nature (\( g(s + T) = g(s) \)), the integrals remain consistent. This technique aligns integrals over new intervals in such a way that the fundamental properties of the periodic function ensure consistent results, a cornerstone of integral calculus.

Definite Integrals

Definite integrals represent the accumulation of quantities, like area under a curve, between specific bounds. The integral \( \int_{a}^{b} g(x) \, dx \) calculates this accumulation from \( a \) to \( b \).
For periodic functions, definite integrals take an interesting twist. The periodic property ensures that integrating over one full period yields the same result, regardless of where you start the interval.
This exercise illustrates the intriguing property of definite integrals over periods. Regardless of where you start on the x-axis, as long as the length of the interval is a full period, the integral remains unchanged. This reiterates the balance and repetition inherent in periodic functions, which is beautifully captured through the concept of a definite integral in integral calculus.
Understanding definite integrals in the context of periodic functions reveals how nature connects repetition and symmetry in mathematics.