# Proof with a Fourier series

• May 29th 2009, 01:13 PM
mathieumg
Proof with a Fourier series
I am given:

$F(x) = \displaystyle\int^x_0 f(t)\,dt$ where f is an even function and F is its Fourier series.

I have to prove that F is an odd function but I don't know where to start, however I am pretty sure I will need integral substitution.
• May 29th 2009, 01:39 PM
TheEmptySet
Quote:

Originally Posted by mathieumg
I am given:

$F(x) = \displaystyle\int^x_0 f(t)\,dt$ where f is an even function and F is its Fourier series.

I have to prove that F is an odd function but I don't know where to start, however I am pretty sure I will need integral substitution.

so we need to show that
$F(-x)=-F(x)$

$F(-x)=\int_{0}^{-x}f(t)dt$ let $u=-t \implies du=-dt$
$F(-x)=\int_{0}^{-x}f(t)dt=\int_{0}^{x}f(-u)(-du)$
but $f(t)$ is even so $f(-u)=f(u)$
$F(-x)=\int_{0}^{-x}f(t)dt=\int_{0}^{x}f(-u)(-du)=-\int_{0}^{x}f(u)du=-F(x)$