Question

Prove that if \(g\) is a piecewise continuous \(2 \pi\)-periodic function on \(\mathbb{R}\), then for every real \(a\) $$ \int_{-\pi+a}^{\pi+a} g(t) d t=\int_{-\pi}^{\pi} g(t) d t . $$

Short answer

\( \int_{-\pi+a}^{\pi+a} g(t) \, dt = \int_{-\pi}^{\pi} g(t) \, dt \) because \( g \) is \( 2\pi \)-periodic.

Step-by-step solution

Understand the setting

We need to analyze a function \( g \) which is piecewise continuous and \( 2\pi \)-periodic. This means that for any real number \( t \), \( g(t + 2\pi) = g(t) \). The integral of interest is from \( -\pi + a \) to \( \pi + a \).

Use the periodicity property

Given that \( g(t + 2\pi) = g(t) \), we realize that translating the interval over which we integrate by \( 2\pi \) does not change the value of the integral, provided the length of the interval is \( 2\pi \). Our task is to show that: \[ \int_{-\pi+a}^{\pi+a} g(t) \, dt = \int_{-\pi}^{\pi} g(t) \, dt. \] Using periodicity, we know that \( g(t) \) on \([-\pi + a, \pi + a]\) has the same values as \( g(t - a) \) on \([-\pi, \pi]\) because it's \( 2\pi \)-periodic.

Change of variables

Let \( u = t - a \), then \( du = dt \). When \( t = -\pi + a \), \( u = -\pi \), and when \( t = \pi + a \), \( u = \pi \). Therefore, the integral: \[ \int_{-\pi+a}^{\pi+a} g(t) \, dt = \int_{-\pi}^{\pi} g(u + a) \, du. \]Given \( g(u + a) = g(u) \) (because \( g \) is \( 2\pi \)-periodic), this becomes: \[ \int_{-\pi}^{\pi} g(u) \, du = \int_{-\pi}^{\pi} g(t) \, dt. \]

Conclusion

From the change of variables and the periodicity of function \( g \), the integral over any \( 2\pi \)-interval has the same value. Thus, the statement is proven: \[ \int_{-\pi+a}^{\pi+a} g(t) \, dt = \int_{-\pi}^{\pi} g(t) \, dt. \]

Key concepts

Periodicity

Periodicity is a vital concept in mathematics, especially when dealing with functions that repeat over specific intervals. A function is said to be periodic if it repeats its values in regular intervals. For a function \( g \) to be \( 2\pi \)-periodic, it means that the function's value repeats every \( 2\pi \) units. This can be expressed as \( g(t + 2\pi) = g(t) \) for any real number \( t \). This property allows us to confidently shift intervals over which we compute integrals without altering the result, provided the intervals' length remains consistent.
Understanding periodicity helps simplify problems by leveraging the repetitive nature of functions. In our exercise, knowing that \( g \) is \( 2\pi \)-periodic allows us to equate the integral from \( -\pi + a \) to \( \pi + a \) with the integral from \( -\pi \) to \( \pi \). This is because each segment of \( 2\pi \) within the function shares identical values due to its repeating nature.

Change of Variables

The change of variables is a helpful technique in integral calculus that simplifies the evaluation of integrals by substituting variables. This method involves replacing a variable with another one, making the integral easier to solve. In our specific problem, we use the substitution \( t = u + a \). When \( t = -\pi + a \), then \( u = -\pi \). Similarly, when \( t = \pi + a \), then \( u = \pi \), simplifying the limits of integration.
This substitution is crucial because it allows us to shift the integration interval back to \( -\pi \) to \( \pi \), a standard interval for periodic functions. By transforming the integral, we leverage the periodicity of \( g \) to equate the two integrals from the original improper intervals to a consistent, manageable one. With \( g(u + a) = g(u) \), thanks to its periodic nature, the process becomes seamless.

Integral Calculus

Integral calculus is a fundamental branch of mathematics focused on finding the total amount accumulated by integration. It encompasses techniques for calculating areas, volumes, and other quantities through integration. In our exercise, the objective is to show that the integral of a piecewise continuous \(2\pi\)-periodic function \(g\) over any \(2\pi\) interval is the same, regardless of the interval's location.
The need to prove this equality brings the concept of definite integrals into the picture. A definite integral represents the net area between the function and the x-axis over a specific interval. Periodic functions with properties like \( g \) often simplify evaluations due to their structured repetition. By recognizing the overarching principles of integral calculus, we efficiently demonstrate that the shifted and original integrals yield the same result, confirming the property's robustness.