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Math Help - Quick Improper Integrals Question

  1. #1
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    Quick Improper Integrals Question

    "Determine whether each integral is convergent or divergent. Evaluate those that are convergent"
    \int_{-\infty}^{\infty}{\cos(\pi x)}dx=\int_{-\infty}^{0}{\cos(\pi x)}dx+\int_{0}^{\infty}{\cos(\pi x)}dx
    =\displaystyle\lim_{r\to-\infty}{\int_{r}^{0}{\cos(\pi x)}dx}+\displaystyle\lim_{t\to\infty}{\int_{0}^{t}  {\cos(\pi x)}dx}
    Sorry people, my LaTex skills break down here. Anyways,
    \int{\cos(\pi x)}dx=\frac{\sin(\pi x)}{\pi}
    But then from here I have to evaluate sine to infinity...
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Lord Voldemort View Post
    "Determine whether each integral is convergent or divergent. Evaluate those that are convergent"
    \int_{-\infty}^{\infty}{\cos(\pi x)}dx=\int_{-\infty}^{0}{\cos(\pi x)}dx+\int_{0}^{\infty}{\cos(\pi x)}dx
    =\displaystyle\lim_{r\to-\infty}{\int_{r}^{0}{\cos(\pi x)}dx}+\displaystyle\lim_{t\to\infty}{\int_{0}^{t}  {\cos(\pi x)}dx}
    Sorry people, my LaTex skills break down here. Anyways,
    \int{\cos(\pi x)}dx=\frac{\sin(\pi x)}{\pi}
    But then from here I have to evaluate sine to infinity...
    What about writing it as a sum

    \int_{0}^{\infty}\cos(\pi\cdot x)dx=\sum_{n=0}^{\infty}\int_{2 n}^{2 (n+1)}\cos(\pi x)dx
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  3. #3
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    Well, you're getting a bit ahead of me; we haven't yet learned to write functions as sums, so I don't think it would be acceptable to do on the homework. But here, can I just say that because the limit does not exist, it is divergent? It's doesn't exist for a different reason than what is usual in improper integrals that I have seen thus the reservation.
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  4. #4
    Behold, the power of SARDINES!
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    Nevermind I Think I have made an error.
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