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Integration by parts SinIntegral[x]

  1. Jan 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate the following integral exactly (no approximations) by the method of integration by parts:

    0t SinIntegral[x] dx

    2. Relevant equations

    the following hints are given:

    D[SinIntegral[x], x] = Sinc[x]; and
    SinIntegral[0] = 0

    3. The attempt at a solution

    The integration by parts forrmula is ∫abu[x]v'[x]dx = u[x]v[x]|ab-∫abv[x]u'[x]dx.

    The problem I have is that whether I let u[x]=1 and v'[x]= SinIntegral[x], or u[x]=SinIntegral[x] and v'[x]= 1, either way the integral on the right is more complex than what I started with (either ∫0txSinc[x]; or ∫0tSinIntegral[x].

    Any help is appreciated. I hope all the programming tags work here, otherwise this will look like a mess. It may still look like a mess

    So it seems to me that, although the question requires the differentiation by parts method, such method does not work because it you get more complicated integrals
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 7, 2012 #2
    Wait, maybe I undertstand it. Do you have to use integration by parts method twice? i.e., letting u[x] = SinIntegral[x], and v^\[Prime][x] = 1; u^\[Prime][x] = Sinc[x] ; v[x] = x
    Which after applying the formula would get you to tSinIntegral[t] - ∫0tSinc[x]dx.

    Then you run the integration by parts once again to find n = ∫0tSinc[x]dx.


    Then the final answer will be tSinIntegral[t] - n?

    This seems a little tedious, but the question specifically asks you to use the integration by parts formula, so I am guessing this must be it, unless I am missing something obvious... which is usually the case.
     
  4. Jan 7, 2012 #3
    nope. Still stuck
     
  5. Jan 8, 2012 #4

    SammyS

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    Do you know what Sinc(x)․x is ?
     
  6. Jan 8, 2012 #5
    Thanks Sammy. I had never encoutered the Sinc Function before. After your hint, and a little help from google, I found that Sinc[x]= (Sin[x])/x. So xSinc[x] should be Sin[x], which makes the problem much easier. I think that should have been another hint given, I'm only working in Calc II ;).
     
  7. Jan 11, 2012 #6
    Is it sin int(t)*t+cos(t)-1? since D(sin int(x))=sinc(x)=sin(x)/x, u=sin int(x). If you look a int^udv=uv-int^v*du, du is on the RHS.
     
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