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I Sorry, how to integrate this function from step1 to step 2

  1. Nov 27, 2016 #1
    Sorry how to integrate this function from step1 to step 2(see the circle inside the picture)

    it looks too complicated
    i cant help with it
    i cannot use partial fraction to do because it looks hard to decompose it
    thank
     

    Attached Files:

  2. jcsd
  3. Nov 27, 2016 #2
    I thought partial fractional decomposition required both numerator and denominator to be polynomials?
     
  4. Nov 27, 2016 #3

    stevendaryl

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    We're trying to integrate
    [itex]\frac{z-Ru}{(R^2 + z^2 - 2Rzu)^{\frac{3}{2}}}[/itex]

    We can rewrite this as:

    [itex]\frac{R}{(2Rz)^{\frac{3}{2}}} \frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^\frac{3}{2}}[/itex]

    Now [itex]\frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^\frac{3}{2}}[/itex] is of the form

    [itex](A-u)(C-u)^\alpha[/itex]

    with [itex]A = \frac{z}{R}[/itex], [itex]C = \frac{R^2 + z^2}{2Rz}[/itex], [itex]\alpha = \frac{-3}{2}[/itex]

    Here's the general way to integrate such expressions:

    Rewrite this as:
    [itex]((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}[/itex]

    We can easily integrate these terms, to get:

    [itex]-\frac{A-C}{\alpha+1} (C-u)^{\alpha+1} - \frac{1}{\alpha+2} (C-u)^{\alpha+2}[/itex]
    [itex]= -\frac{(C-u)^{\alpha+1}}{(\alpha+1)(\alpha+2)} ((A-C)(\alpha+2) + (C-u)(\alpha+1))[/itex]
    [itex]= -\frac{(C-u)^{\alpha+1}}{(\alpha+1)(\alpha+2)} (A (\alpha+2) - C - u(\alpha+1))[/itex]

    So if you plug in our known values for [itex]A, C, \alpha[/itex], you should get the desired result (after some simplification).
     
  5. Nov 27, 2016 #4
    where is this form
    [itex](A-u)(C-u)^\alpha[/itex]
    coming from

    and Where do you got [itex]((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}[/itex] this equation from?

    thank
     
    Last edited: Nov 27, 2016
  6. Nov 27, 2016 #5
    He's just substituting there. A and C are the two terms that have r and z in it. In the first step he just rearranged it to get u by itself in both the numerator and deniminator, so that he could make the substitution.

    The second part is just adding and subtracting C, and then using the associstive property. You'll note that A - u = A + C - C - u.
     
  7. Nov 27, 2016 #6
    Honestly that's a really clever trick and I think I'm going to commit this one to memory.
     
  8. Nov 28, 2016 #7

    Integral

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    I suspect both of those relationships came from a book of integrals.
     
  9. Nov 28, 2016 #8

    stevendaryl

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    The integrand is (after factor out the factor [itex]\frac{2R}{(2Rz)^{3/2}}[/itex]):

    [itex]\frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^{3/2}}[/itex]

    That's the same as:

    [itex](\frac{z}{R} - u)(\frac{R^2 + z^2}{2Rz} - u)^{-3/2}[/itex]

    Now, let [itex]A = \frac{z}{R}, C = \frac{R^2 + z^2}{2Rz}, \alpha = -3/2[/itex], so you get

    [itex](A - u)(C - u)^{\alpha}[/itex]

    Battlemage! explained it above.
     
  10. Nov 29, 2016 #9
    But the temptation to integrate with respect to ##z## (since ##(z^2-Rzu)'=2(z-Ru)##), then integrate with respect to ##u##, and then differentiate with respect to ##z## is just irresistible...
     
  11. Nov 29, 2016 #10
    Ye gads, is this not cheating? Would that actually work, and if so is it easier?
     
  12. Nov 29, 2016 #11
    It feels so good, that it probably is...
    The integrals in the intermediate steps are much easier, in my opinion (or, maybe, more obvious). In the end you just differentiate a simple fraction, it's not too hard.
     
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