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Is this a valid operation (integration by parts)?

  1. Jun 16, 2014 #1
    Say I have a function,

    f(x) = x sec (f(x)) [this is just an example function, the actual problem is more complicated]

    g(x) = x f(x), then using integration by parts, I can write

    I = abg(x) dx = abx f(x) dx = (f(x) [itex]\frac{x^{2}}{2}[/itex])|[itex]^{b}_{a}[/itex]- [itex]\frac{1}{2}[/itex]ab[itex]\frac{d f(x)}{dx}[/itex] x2 dx

    Then can I write the integral as (cancelling dx from the numerator and denominator in integral on the RHS)?

    I = (f(x) [itex]\frac{x^{2}}{2}[/itex])|[itex]^{b}_{a}[/itex]- [itex]\frac{1}{2}[/itex]f(a)f(b) x2 df

    or,

    I = (f(x) [itex]\frac{x^{2}}{2}[/itex])|[itex]^{b}_{a}[/itex]- [itex]\frac{1}{2}[/itex]f(a)f(b) f2 cos2 (f) df ?

    This is simplified version of my actual problem. So any inputs will be very useful?

    PS: I have edited the originally posted problem definition as it was causing confusion.
     
    Last edited: Jun 16, 2014
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  3. Jun 16, 2014 #2

    micromass

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    So you need to find the integral

    [tex]-\int_a^b xW(-x)dx[/tex]

    where ##W## is the Lambert W-function?

    What kind of answers are acceptable? A closed form solution? A series solution?

    Wolfram alpha gives the following:

    [tex]\int - x W(-x)dx = \frac{x^2(1 - W(-x))(2W(-x)^2+1)}{8W(-x)^2}[/tex]

    which in term of your notation would be

    [tex]\frac{x^2(1 + f(x))(2f(x)^2 +1)}{8f(x)^2}[/tex]
     
  4. Jun 16, 2014 #3
    I know about the Lambert W-function. As I said, above is a simplified version of my actual problem. so I specifically need to know if my last 2 steps are valid operations. Thanks
     
  5. Jun 16, 2014 #4

    micromass

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    OK.

    I think this is ok.

    Any reason why you don't have ##f^2/e^{2f}##? You have ##x = \frac{f(x)}{e^{f(x)}}## and thus ##x^2 = \frac{f(x)^2}{e^{2f(x)}}##. No?
     
  6. Jun 16, 2014 #5
    Thanks, I have corrected the typo and also the original problem definition. Please see if its ok now.
     
  7. Jun 16, 2014 #6

    micromass

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    Seems fine to me.
     
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