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Series expansion of Sine Integral Si(x)

  1. May 14, 2013 #1
    This is not really a homework, I am trying to expand Si(x) into a series.


    The series expansion of Si(x) is given in articles:
    [tex]Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^{\infty}\frac {(-1)^k x^{2k+1}}{(2k+1)(2k+1)!}[/tex]

    This is my work, I just cannot get the right answer:
    [tex]Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^m \frac{sin \theta_i}{\theta_i}\Delta\theta_i[/tex]

    [tex]\sin \theta=\sum_0^{\infty}\frac{(-1)^k \theta^{2k+1}}{(2k+1)!}\Rightarrow\;\frac{\sin\theta}{\theta}=\sum_0^{\infty}\frac{(-1)^k \theta^{2k}}{(2k+1)!}[/tex]

    [tex]\Delta \theta_i = \frac x m\;\hbox{ and }\; \theta_i= i\Delta \theta_i =i\frac x m[/tex]

    [tex]\Rightarrow\;Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^m \frac{sin \theta_i}{\theta_i}\Delta\theta_i\;=\;\sum_0^m\sum_0^{\infty}\frac{(-1)^k\left (i\frac x m\right )^{2k}}{(2k+1)!}\frac x m\;=\;\sum_0^m\sum_0^{\infty}\frac{(-1)^ki^{2k} x^{2k+1}}{m^{2k+1}(2k+1)!} [/tex]

    I don't know how to get the answer, please help.
     
    Last edited: May 14, 2013
  2. jcsd
  3. May 14, 2013 #2

    Dick

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    I have no idea what all the delta thetas are all about. You don't need to derive the formula for integrating a power. You have a series for ##\sin(\theta)##, divide it by ##\theta## and integrate term by term.
     
  4. May 14, 2013 #3
    I missed the subscript i, ##\Delta\theta_i## is just the series representation of ##d\theta##. Where I chop the 0<##\theta##<x into m sections.

    I want to verify and derive the equation given from articles.
     
    Last edited: May 14, 2013
  5. May 14, 2013 #4

    Dick

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    You don't have to chop into sections. Just integrate the power series term by term!
     
  6. May 14, 2013 #5
    I got it!!! It's just that easy!!! Thanks.
     
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