# Thread: Antiderivatives of periodic functions

1. ## Antiderivatives of periodic functions

Suppose that f is $2 \pi$-periodic and let a be a fixed real number. Define $F(x) = \int^{x}_{a}f(t)dt, \forall x$

Show that F is 2\pi-period iff $\int^{2\pi}_{0}f(t)dt = 0$

My proof so far:

Assume that F is 2\pi-periodic, then F(x) = F(x+2pi) = ...

By a theorem, we know that $\int^{T}_{0}f(x)dx = \int^{a+T}_{a}f(x)dx$

So $\int^{2\pi}_{0}f(t)dt = \int^{a+2\pi}_{a}f(t)dt$

2. Originally Posted by tttcomrader
Suppose that f is $2 \pi$-periodic and let a be a fixed real number. Define $F(x) = \int^{x}_{a}f(t)dt, \forall x$

Show that F is 2\pi-period iff $\int^{2\pi}_{0}f(t)dt = 0$
(We also need the condition that $f(x)$ is integrable. Can you think of an example when this fails?)

I start you off.

1)Assume $f(x)$ is integrable and $2\pi$-periodic.

2)We will show the forward direction, i.e. $F$ is $2\pi$-periodic implies $\int_0^{2\pi}f(t)dt=0$.

3)Say that $F(x)$ is a $2\pi$-periodic function.

4)By definition it means $F(x+2\pi)=F(x)$.

5)Now #4 is true for all $x$, pick $x=a$.

6)Thus, $F(a+2\pi)=F(a)$

7)By definition of $F(x)$ we see that $\int_a^{a+2\pi}f(t)dt = \int_a^{a}f(t)dt =0$.

8)Thus, $\int_{a}^{a+2\pi}f(t)dt=0$.

9)Since $f(t)$ is $2\pi$-periodic we use the theorem that $\int_a^{a+2\pi}f(t)dt = \int_{a-a}^{a+2\pi-a}f(t)dt = \int_0^{2\pi}f(t)dt$.

10)By #9 we see that $\int_0^{2\pi}f(t)dt=0$.

Q.E.D.

3. Originally Posted by tttcomrader
Suppose that f is $2 \pi$-periodic and let a be a fixed real number. Define $F(x) = \int^{x}_{a}f(t)dt, \forall x$

Show that F is 2\pi-period iff $\int^{2\pi}_{0}f(t)dt = 0$

My proof so far:

Assume that F is 2\pi-periodic, then F(x) = F(x+2pi) = ...

By a theorem, we know that $\int^{T}_{0}f(x)dx = \int^{a+T}_{a}f(x)dx$

So $\int^{2\pi}_{0}f(t)dt = \int^{a+2\pi}_{a}f(t)dt$

$\int_{0}^{2\pi}f(t)dt=-\int_a^0 f(t) dt + \int_a^{2\pi}f(t) dt=-F(0)+F(2 \pi)
$

So if $F$ is periodic with period $2 \pi$:

$\int_{0}^{2\pi}f(t)dt=-\int_a^0 f(t) dt + \int_a^{2\pi}f(t) dt=-F(0)+F(2 \pi)=-F(2 \pi) + F(2 \pi) =0
$

And if:

$\int_{0}^{2\pi}f(t)dt=0
$

$F(x+2 \pi)=\int_a^{x}f(t) dt + \int_x^{x+2\pi}f(t) dt
$

but as $f(x)$ is periodic with period $2 \pi$ the second integral above is:

$\int_x^{x+2\pi}f(t) dt=\int_0^{2\pi}f(t) dt=0
$

so:

$F(x+2 \pi)=\int_a^{x}f(t) dt=F(x)
$

Hence $F(x)$ is periodic with period $2 \pi$

Combining these we have shown $F$ is $2\pi$-period iff $\int^{2\pi}_{0}f(t)dt = 0$

RonL