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Math Help - Antiderivatives of periodic functions

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    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.
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  3. #3
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    Quote Originally Posted by tttcomrader View Post
    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)<br />

    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<br />

    And if:

    \int_{0}^{2\pi}f(t)dt=0<br />

    F(x+2 \pi)=\int_a^{x}f(t) dt + \int_x^{x+2\pi}f(t) dt<br />

    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<br />

    so:

    F(x+2 \pi)=\int_a^{x}f(t) dt=F(x)<br />

    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
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