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Maclaurin Series for an Integral

  1. Nov 15, 2006 #1
    For my homework in my class, I'm supposed to find the first 3 non-zero terms of the Maclaurin series of f(x) = Integral sin(t)/t dt evaluated from 0 to x. I'm fairly sure that sint/t dt has no integral, so I'm lost in my search for the solution. Can you guys point me in the right direction?
     
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  3. Nov 15, 2006 #2
    [tex] \sin x = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1!}x^{2n+1} [/tex]


    [tex] \frac{\sin x}{x} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1!}x^{2n} [/tex]

    [tex] \int_{0}^{x} \frac{\sin x}{x}\; dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1!(2n+1)}x^{2n+1} [/tex]
     
    Last edited: Nov 15, 2006
  4. Nov 15, 2006 #3

    Hurkyl

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    Why do you think that? You're integrating a continuous, bounded function over a finite interval -- that's like the poster child for being integrable!

    The integral most surely exists. The function

    [tex]g(x) := \int_0^x \frac{\sin t}{t} \, dt[/tex]

    is perfectly well-defined. It just might be that there isn't a "simpler" way to represent g(x). But it doesn't matter; this form is more than adequate for solving the problem at hand.
     
    Last edited: Nov 15, 2006
  5. Nov 15, 2006 #4
    I'm sorry, I didn't make myself clear. What I meant when I said that sint/t dt has no derivative is that the indefinite integral sint/t has no simple function to represent it. I looked up the integral, and wolfram just calls it sine integral.
     
  6. Nov 16, 2006 #5

    Hurkyl

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    You do see that this form is adequate for the problem, though, right? If not... then where do you see an obstacle?
     
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