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Math Help - Proof with a Fourier series

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by mathieumg View Post
    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)
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  3. #3
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    Thanks a lot!
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