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Thread: Integration

  1. #1
    Senior Member pankaj's Avatar
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    Integration

    Given $\displaystyle f(x)$ is an odd function and defined everywhere,periodic with period $\displaystyle 2$ and integrable on every interval.

    Let $\displaystyle g(x)=\int_{0}^xf(t)dt$.

    Then prove that $\displaystyle g(2n)=0$ for every integer $\displaystyle n$.
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  2. #2
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    Quote Originally Posted by pankaj View Post
    Given $\displaystyle f(x)$ is an odd function and defined everywhere,periodic with period $\displaystyle 2$ and integrable on every interval.

    Let $\displaystyle g(x)=\int_{0}^xf(t)dt$.

    Then prove that $\displaystyle g(2n)=0$ for every integer $\displaystyle n$.
    Given that $\displaystyle f$ is defined everywhere and integrable on every interval we can consider

    $\displaystyle I = \int _{-1}^{1} f(t) dt$

    $\displaystyle f$ being odd implies (easy to prove if you're not allowed to state) $\displaystyle I = 0$

    Now you can see that a simple substitution should give you that $\displaystyle I = g(2)$.

    Using the periodicity of f you should can show that any integral of f over an interval of length 2 has the same value. Therefore

    $\displaystyle g(2) = \int_0^2 f(t) dt = \int_{2}^4 f(t) dt = \int_4^6 f(t) = \ldots = \int_{2n-2}^{2n} f(t) dt = 0 $

    Adding together these integrals gives g(2n) = 0

    Hope this helps.
    Last edited by pomp; Sep 8th 2009 at 09:59 AM.
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