Proof with a Fourier series

**mathieumg** · May 29th 2009, 12:13 PM

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.

**TheEmptySet** · May 29th 2009, 12:39 PM

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
$F(-x)=-F(x)$

so let start with

$F(-x)=\int_{0}^{-x}f(t)dt$ 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 $f(-u)=f(u)$

$F(-x)=\int_{0}^{-x}f(t)dt=\int_{0}^{x}f(-u)(-du)=-\int_{0}^{x}f(u)du=-F(x)$

**mathieumg** · May 29th 2009, 12:54 PM

Thanks a lot!

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