# Thread: Antiderivatives of periodic functions

1. ## Antiderivatives of periodic functions

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

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

My proof so far:

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

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

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

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

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

I start you off.

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

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

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

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

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

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

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

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

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

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

Q.E.D.

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

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

My proof so far:

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

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

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

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

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

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

And if:

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

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

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

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

so:

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

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

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

RonL