Integration of periodic functions

The integration of a periodic function involves some unique properties and approaches due to the function’s repeating nature. A function $f(x)$ is said to be periodic with period $T$ if for all $x$ in the domain of $f$, $f(x + T) = f(x)$. This periodicity can significantly simplify the integration process over intervals that are multiples of the period $T$.

Properties of Integrals of Periodic Functions

(1) Integral over one period: The integral of a periodic function over one period is a constant value that does not depend on the starting point of the interval. Mathematically, this is expressed as:
$$
\int_{a}^{a+T} f(x)\,dx = \int_{0}^{T} f(x)\,dx =constant
$$
where $T$ is the period of the function and $a$ is any real number.

Proof

$\int_{0}^{T} f(x)\,dx = \int_{0}^{a} f(x)\,dx + \int_{a}^{T} f(x)\,dx$
$= \int_{0}^{a} f(x)\,dx + F(T) – F(a)$
$=\int_{0}^{a} f(x)\,dx + F(a+T) – F(a) – (F(a+T) – F(p) ) $
$=\int_{0}^{a} f(x)\,dx + \int_{a}^{a+T} f(x)\,dx – \int_{T}^{a+T} f(x)\,dx $
Now Calulate $\int_{T}^{a+T} f(x)\,dx $
Now let x = p + T
then dx= dp
$\int_{T}^{a+T} f(x)\,dx = \int_{0}^{a} f((p+T))\,dt $
Since this is periodic with period T , f(p+T) = f(p)
Therefore
$=\int_{0}^{a} f(p)\,dp $
$=\int_{0}^{a} f(x)\,dx $
Subsituting back, we have
$\int_{0}^{T} f(x)\,dx = \int_{0}^{a} f(x)\,dx + \int_{a}^{a+T} f(x)\,dx – \int_{0}^{a} f(x)\,dx=\int_{a}^{a+T} f(x)\,dx $

(2) Integral over multiple periods: If you integrate a periodic function over an interval that spans multiple periods, the result is simply the integral over one period multiplied by the number of periods covered. For an interval $[a, b]$ that spans $n$ periods:
$$
\int_{a}^{a+nT} f(x)\,dx = n \cdot \int_{a}^{a+T} f(x)\,dx,
$$
where $n = \frac{b-a}{T}$ (assuming $b-a$ is an exact multiple of $T$).

We can derive below formula based on above

$$
\int_{0}^{nT} f(x)\,dx = n \cdot \int_{0}^{T} f(x)\,dx,
$$

$$
\int_{nT}^{mT} f(x)\,dx = (m-n) \cdot \int_{0}^{T} f(x)\,dx,
$$

$$
\int_{a+nT}^{b+nT} f(x)\,dx = \int_{a}^{b} f(x)\,dx,
$$

Example 1

$$
\int_{0}^{6\pi} \sin(x)\,dx
$$

Solution
The function $f(x) = \sin(x)$, is a periodic with period $T = 2\pi$.

So
$$
\int_{0}^{6\pi} \sin(x)\,dx= 3 \times \int_{0}^{2\pi} \sin(x)\,dx= 3 \times 0 =0
$$

Solved Questions

Question 1
h(x) is continuous periodic function with period T, then the integral
$$
I= \int_{a}^{a+T} h(x)\,dx
$$
is ?
(a) equal to 2a
(b) Independent of a
(c) 4a
(d) None of these

Solution
we know by property
$$
\int_{a}^{a+T} f(x)\,dx = \int_{0}^{T} f(x)\,dx =constant
$$

Hence this is independent of a

Question 2

Assertion :If $I=\int_{0}^{T} f(x)\,dx $, then $\int_{3}^{3+3T} f(2x)\,dx =3I$
Reason:f(x) is periodic function with time period T

(a)Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
(b) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
(c)Assertion is correct but Reason is incorrect
(d)Both Assertion and Reason are incorrect

Solution
$\int_{3}^{3+3T} f(2x)\,dx $
let 2x=p, then $dx = \frac {dp}{2} $
$\int_{3}^{3+3T} f(2x)\,dx = \frac {1}{2} \int_{6}^{6+6T} f(p)\,dp$
$=\frac {1}{2} \int_{6}^{6+6T} f(x)\,dx=3 \int_{0}^{T} f(x) dx=3I $
(a) is the correct answer