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Integrate [itex]\int_0^1\frac{\sin(\pi x)}{1-x}dx[/itex]

  1. Jun 5, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]\int_0^1\frac{\sin(\pi x)}{1-x}dx[/itex]


    2. Relevant equations
    [itex]\int \frac{\sin (\pi x)}{1-x}=Si(\pi-\pi x)[/itex]


    3. The attempt at a solution
    I was stuck on the above integral while solving an exercise, I found out earlier on Wolfram that this integral doesn't probably have an elementary antiderivative representation so I cannot just calculate it directly by Newton-Leibniz formula. Integration by parts is not very viable in this case also. Maybe this needs the notion of contour integration to be solved? Any idea guys?
     
  2. jcsd
  3. Jun 5, 2013 #2

    haruspex

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    Have you tried expanding the denominator as a power series?
     
  4. Jun 5, 2013 #3
    Yeah, looks like it yielded some positive results finally , thanks very much for your help.
     
  5. Jun 5, 2013 #4
    Oh, I did expand the denominator as a power series and then integrate term by term by resulting series, but I can't proceed to the next step. The general formula for arbitrary k is too complicated for me.
     
  6. Jun 5, 2013 #5

    HallsofIvy

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    Then stop with some finite power for an approximation.
     
  7. Jun 5, 2013 #6

    Ray Vickson

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    If you change variables to t = 1-x you will get the value of the integral as ##I = \text{Si}(\pi),## where
    [tex] \text{Si}(x) = \int_0^x \frac{\sin(t)}{t} \, dt.[/tex] You can very easily expand ##\sin(t)## in a power series and get a very straightforward series for ##\text{Si}(x)## in powers of x. That will give you a simple-enough series for I in powers of ##\pi##.
     
  8. Jun 5, 2013 #7
    Thank HallsofIvy.
    This seems like the most plausible way to handle that imo :biggrin:. Thanks.
     
  9. Jun 5, 2013 #8

    Ray Vickson

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    Numerical integration methods such as Simpson's rule should work well also.
     
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