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Math Help - series expansion of the cosine integral

  1. #1
    Super Member Random Variable's Avatar
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    series expansion of the cosine integral

    \text{Ci} (x) = -\int_{x}^{\infty} \frac{\cos t}{t} \ dt

     = -\int_{x}^{\infty} \frac{1}{t} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n!)} \ t^{2n} \ dt

     = -\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!} \int_{x}^{\infty} t^{2n-1} \ dt (since the series expansion of the cosine function converges uniformly)

     = -\int^{\infty}_{x} \frac{dt}{t} -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n)!} \int_{x}^{\infty} t^{2n-1} \ dt

     = -\ln t\Big|^{\infty}_{x} - \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} \ t^{2n} \Big|_{x}^{\infty}

     \lim_{t \to \infty} \Big(-\ln t - \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} t^{2n} \Big) + \ln x + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} \ x^{2n}


    I thought this had to be incorrect, but Maple says that the limit does indeed evaluate to Euler's Constant. Why?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: series expansion of the cosine integral

    Quote Originally Posted by Random Variable View Post
    \text{Ci} (x) = -\int_{x}^{\infty} \frac{\cos t}{t} \ dt =

    ...

    = \lim_{t \to \infty} \Big(-\ln t - \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} t^{2n} \Big) + \ln x + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} \ x^{2n}


    I thought this had to be incorrect, but Maple says that the limit does indeed evaluate to Euler's Constant. Why?
    The problem is the computation of the limit...

    \lambda=\lim_{t \rightarrow \infty} \{\ln t + \int_{0}^{t} \frac{\cos \tau-1}{\tau}\ d\tau\} (1)

    Applying a fundamental property of the Laplace Transform first You compute [I obtained the esult some time ago]...

    \mathcal{L}\{\frac{\cos \tau-1}{\tau} \}= \int_{s}^{\infty}\frac{1}{\sigma\ (1+\sigma^{2})}\ d \sigma = \frac{1}{2}\ \ln \frac{1+s^{2}}{s^{2}} (2)

    Then You find...

    \mathcal{L}\{\int_{0}^{t} \frac{\cos \tau-1}{\tau} \}= \frac{1}{2 s}\ \ln \frac{1+s^{2}}{s^{2}} = \frac{1}{s}\ \{\ln \sqrt{1+s^{2}} - \ln s\} (3)

    Third step is the application of the 'final value theorem' [remember that is \mathcal{L}\{\ln t \}= - \frac{\gamma- \ln s}{s} obtaining...

    \lambda= \lim_{s \rightarrow 0} \{- \gamma + \ln s + \ln \sqrt{1+s^{2}} - \ln s \}= - \gamma (4)



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  3. #3
    Super Member Random Variable's Avatar
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    Re: series expansion of the cosine integral

    What do we need to show to justify the use of the FVT?

    EDIT: Do we just need to show that the the original limit exists?
    Last edited by Random Variable; December 22nd 2011 at 01:47 PM.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: series expansion of the cosine integral

    Quote Originally Posted by Random Variable View Post
    What do we need to show to justify the use of the FVT?

    EDIT: Do we just need to show that the the original limit exists?
    The final value theorem exptablishes that \lim_{t \rightarrow \infty} h(t)= \lim_{s \rightarrow 0} s\ H(s) can be applied if H(s) has no poles with real part greater than 0 and no poles in s=i\ \omega with \omega \ne 0...



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    Last edited by chisigma; December 24th 2011 at 12:37 AM.
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  5. #5
    Super Member Random Variable's Avatar
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    Re: series expansion of the cosine integral

    Wouldn't those be the conditions if we didn't know h(t)$? Although in this case those conditions are obviously satisfied. The article on Wikipedia suggests that we only need to show that \lim_{t \to \infty} h(t) is finite (which is easier said than done for this problem).
    Last edited by Random Variable; December 23rd 2011 at 10:48 AM.
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: series expansion of the cosine integral

    Quote Originally Posted by Random Variable View Post
    Wouldn't those be the conditions if we didn't know h(t)$? Although in this case those conditions are obviously satisfied. The article on Wikipedia suggests that we only need to show that \lim_{t \to \infty} h(t) is finite (which is easier said than done for this problem).
    What is the article of Wikipedia You talk about?... if is is this one...

    Laplace transform - Wikipedia, the free encyclopedia

    ... it is clearly explained that if h(t) has finite limit, then the FVT supplies this limit only if s\ H(s) has all poles in the left half plane. If the article is another, then it is possible that something different is written because in my opinion Wikipedia is not completely reliable...



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