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Exact Definition of Riemann Integral

  1. Nov 27, 2007 #1
    Hi I recently stumbled upon this:
    I know that the Riemann Integral is defined for every piecewise continouus curve.
    But now suppose you´re asked the following:

    you are given [tex]f(x,y)=\frac{xy^3}{(x^2+y^2)^2}[/tex] with additional Definition
    f(0,0)=0. ( It´s a textbook problem :) )

    Now surely f(x,y) is smooth everywhere except for the origin. So I tried to show that f(x,y) goes to 0 in the origin but it doesn´t goes to 1/4(easy to show by setting x=y because f(x,y)=1/4 for x=y).

    So the book asks to proof that the function [tex]g(y)=\int_{0}^{1}f(x,y)dx [/tex] is well defined on the real line.
    I would intuitively say no because of the discontinuity at the origin but otherwise the origin is a set of measure zero.
    So how does that definition go ?
  2. jcsd
  3. Nov 27, 2007 #2
    you say approaching the origin via y=x gives you 1/4, remember that limits in n-dimensions need to give the same answer no matter WHICH way you goto the origin. Now, i haven't worked with several variables in a while so i'm not going to attempt it although looking at the denomintator it looks like moving to polar coordinates would simplify matters.

    the precise definition of reimann integration can be found on wikipedia, the darboux definition is the most practical for analytic purposes but in this case all you need to know is that the single discontinuity at the origin will NOT prevent the integral from existing. Regarding measure you are correct, a function f is integrable on a set E iff f is continuous almost everywhere on E.

    hope this helps
  4. Nov 27, 2007 #3
    oh yeah i forgot to say, try approaching the origin via the parabola y=x^2.
  5. Nov 27, 2007 #4
    As you know, f(x,y) (with y held constant) is continuous except when x=y=0. This is a single point, so it is of measure zero. What else needs to be satisfied for a function to be Riemann integrable on a set [a,b]?

    As for the exact definition of the Riemann Integral, I don't think it'll help you here, what you really want is the Riemann Lebesgue Theorem, but
    A Riemann Sum
    [tex] R(f,P,T) = \sum_1^n f(\hat x_i)\Delta x_i[/tex] where [tex]P=\{x_1, \dotsc, x_n, x_{n+1}\}[/tex] is a partition of the interval [a,b] (in this sense, a partition is an ordered set of points in the interval [a,b] with [tex]x_1 = a[/tex] and [tex]x_{n+1} = b[/tex]) and [tex]T = \{\hat x_1, \dotsc, \hat x_n\}[/tex] is a set of points interspersed between the partition points of P (i.e. [tex]x_1 \leq \hat x_1 \leq x_2 \leq \dotsb \leq x_n \leq \hat x_n \leq x_{n+1}[/tex])

    We defined [tex]\Delta x_i = x_{i+1} - x_i[/tex]

    The mesh of P is the largest [tex]\Delta x_i[/tex]

    The Riemann Integral is the limit of the Riemann Sums as the mesh of P goes to 0

    In other words, although you learn in Calculus that the Riemann Sum is defined as dividing the interval up evenly and then picking a point at one of the end points of your divisions, and then the Riemann Integral is defined by the limit as the size of the divisions go to 0, the actual definition of a Riemann Sum allows you to divide the interval up however you want and place points anywhere in the divisions. You still take the integral to be the limit as the size of the divisions goes to 0.

    Anyway, like I said, this is probably completely useless to you. Look up the Riemann Lebesgue Theorem.
    Last edited: Nov 27, 2007
  6. Nov 27, 2007 #5
    Not entirely true, you're missing something. The function [tex]f(x) = 1/x[/tex] is not integrable on any interval containing 0 even though it is only discontinuous at 0.

    Edit: Mr. Brown: also keep in mind that because your function f is continuous everywhere except at (0,0), you only need to worry about g(0)
    Last edited: Nov 27, 2007
  7. Nov 27, 2007 #6
    Since it hasn't been mentioned yet, I just wanted to point out that h_y(x) = f(x,y) is actually continuous at 0 (and everywhere else). For nonzero y, the denominator is nonzero. For y = 0, h_y(x) = 0 everywhere. I think for the given problem, that's all he needed to show.

    Correction.. [tex]g(y)=\int_{0}^{1}h_y(x)dx[/tex] is well defined because h_y is continuous for fixed y.
    Last edited: Nov 27, 2007
  8. Nov 27, 2007 #7
    f needs to be bounded, unless you are considering it as an improper integral.
  9. Nov 27, 2007 #8


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    just look at the definition. this is trvially true with no high powered theory needed of any kind. the only interesting value is x=0 where the function is dead zero. done.
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