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Can you use integration by parts?

  1. Jan 9, 2014 #1
    How would you integrate this:

    ## \int x df(x) ##

    In general, how is this solved: ## \int 1df(x) ##

    Can you use integration by parts? I tried, but kept getting 0 since I let ## 1 = u## but then ##du = 0## for later purposes. Also, if ## df(x) = u ## then I am still stumped on how to take the derivative of ##df(x)## ...is it even possible?

    Any help would be great!
     
    Last edited: Jan 9, 2014
  2. jcsd
  3. Jan 9, 2014 #2

    jedishrfu

    Staff: Mentor

    This doesn't look right is the df(x) from y=f(x) and so its really a dy? so that you have the integral ## \int x dy ##

    Can you give the whole problem and what book it came from?

    if its by parts then ##\int udv ## = uv - ## \int vdu ##

    so you dv would be df(x) and v =f(x) and u would be x so we get:

    ##\int xdf ## = xf - ## \int fdx ##
     
    Last edited: Jan 9, 2014
  4. Jan 9, 2014 #3
    If you place a problem in this homework section, you are supposed to follow the homework format. Can you please do that.

    As jedishrfu says, your problem is not correctly stated. If you are looking at ##\int xdx## that is easily solved. If by df(x) you mean f'(x), you should write it as ##\int xf'(x)dx##.

    As you've written it, it is a Riemann-Stieltjes integral; in that case we can't proceed without knowing what f(x) is.

    Please clarify.
     
  5. Jan 10, 2014 #4
    I am just wondering what forms you can transform ##\int xdy## into if y is a function of x. I am trying to manipulate another expression, but seem to be stuck once given this problem.
     
  6. Jan 11, 2014 #5

    jedishrfu

    Staff: Mentor

    Without knowing what y is as a function of x, you can only do the integration by parts rule as mentioned in earlier posts and thats it.
     
  7. Jan 11, 2014 #6
    I think you are asking how to integrate functions w.r.t. functions other than x.

    That is

    ## \int g(x)d(f(x)) ##

    The general approach I think is to take

    ## f(x)=t ##

    ## d(f(x))=dt ##

    also express x as

    [itex] x = f^{-1}(t) [/itex]

    and finally put this value of x in g(x) to get a simple integral as

    [itex] \int g(f^{-1}(t))dt [/itex]

    which can be solved by usual techniques. After solving, again substitute t=f(x) to get your answer.
     
  8. Jan 11, 2014 #7

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Of course this is provided ##f## is injective so ##f^{-1}## makes sense as a function.
     
  9. Jan 11, 2014 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If f is a differentiable function, then [itex]\int x df= \int x \frac{df}{dx}dx[/tex]

    If f is not differentiable but is increasing function then [itex]\int x df[/itex] is not a Riemann integral but is, rather, a Stieljes integral.

    (A Stieljes integral is defined in a way similar to the Riemann integral by dividing the interval into subintervals of length [itex]\Delta x[/itex] (which might vary from subinterval to subinterval), approximating the integral by the sum [itex]\sum g(x_i) \Delta x[/itex] where [itex]x_i[/itex] is a point in the ith subinterval. The definition of the Riemann integral, [itex]\int g(x)dx[/itex] takes [itex]\Delta x[/itex] to be the length of each subinterval, [itex]\alpha_{i+ 1}- \alpha_i[/itex], while the Stieljes integral, [itex]\int g(x) df[/itex], with f an increasing function (not necessarily differentiable or even continuous) [itex]\Delta x= f(\alpha_{ix+1}- f(\alpha_i)[/itex].)
     
  10. Jan 12, 2014 #9

    jedishrfu

    Staff: Mentor

    ....fixed the itex tag of HallsOfIvy post...
     
  11. Jan 18, 2014 #10
    Wow, thank you for the great explanation HoI!
     
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