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Math Help - Number of ways to evaluate an integral

  1. #16
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    I know I'm not being explicit and I may be coinciding with somebody's preceding post (sorry about that if it is), but it seems to me to be the type of problem you can apply polar coordinates to to get a solution.
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  2. #17
    Behold, the power of SARDINES!
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    Here is another way using the Laplace transform.

    \displaystyle f(t)=\frac{\sin(t)}{t}=\sum_{n=0}^{\infty}\frac{(-1)^nt^{2n}}{(2n+1)!}

    Then via the laplace transform we get

    \displaystyle F(s)=\int_{0}^{\infty}\frac{\sin(t)}{t}e^{-st}dt=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\int_{0}^{\infty}t^{2n}e^{-st}dt

    Notice that \displaystyle \lim_{s \to 0^{+}}F(s)=\int_{0}^{\infty}\frac{\sin(t)}{t}dt

    Now taking the transform above gives

    \displaystyle F(s)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\left( \frac{(2n)!}{s^{2n+1}}\right)=\sum_{n=0}^{\infty}\  frac{(-1)^n}{(2n+1)}\left( \frac{1}{s^{2n+1}}\right)=\tan^{-1}\left( \frac{1}{s}\right)

    This gives \displaystyle \lim_{s \to 0^+}\tan^{-1}\left( \frac{1}{s}\right)=\frac{\pi}{2}
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  3. #18
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Notice that \displaystyle \lim_{s \to 0^{+}}F(s)=\int_{0}^{\infty}\frac{\sin(t)}{t}dt
    This is actually something proof worthy, but you are right of course.
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  4. #19
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    Post Solution Details By use of Laplace

    Last edited by uniquesailor; February 16th 2011 at 02:18 PM.
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