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Math Help - Improper Integrals

  1. #16
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    ok \int_0^{1}\frac{cos(x)}{x^2}dx=\int_0^{1}\sum_{n=1  }^{\infty}\frac{(-1)^{n}x^{2n-2}}{(2n)!}dx=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n-1}}{(2n-1)\cdot{(2n)!}} which converges from 0 to 1
    But that's not the right series, did you read what TheEmptySet said?
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  2. #17
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    But that's not the right series, did you read what TheEmptySet said?
    Yes it is \frac{cos(x)}{x^2}=\frac{\sum_{n=0}^{\infty}\frac{  (-1)^{n}x^{2n}}{(2n)!}}{x^2}=\sum_{n=1}^{\infty}\fra  c{(-1)^{n}x^{2n-2}}{(2n)!}
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