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Integration of a function that's non-zero at one point

  1. Jul 11, 2015 #1
    Say we have a function of the form [tex]f(x)=\bigg\lbrace\begin{matrix}c \qquad\quad\text{x=0}\\ 0\quad\text{elsewhere}\end{matrix}[/tex] If we then integrate this over all space i.e. [tex]\int_{-\infty}^{\infty}f(x)dx[/tex] Why does the result equal zero?
     
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  3. Jul 11, 2015 #2

    micromass

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    Loosely speaking, because the integral gives a global result and does not really care what happens locally. So changing a function in one point does not alter the integral.

    If you want a formal proof, then you'll have to start by mentioning what the definition of the integral is.
     
  4. Jul 11, 2015 #3
    Does this apply to Riemann integration?

    I was wondering because I read that one of the main reasons that the Dirac delta function doesn't make sense is the following:
    " We require that [tex]\delta (x)=\bigg\lbrace\begin{matrix}1\qquad\quad x=0\\ 0\quad\text{elsewhere}\end{matrix}[/tex] and that [tex]\int_{-\infty}^{\infty}\delta (x)=1[/tex] The problem is these to requirements cannot be satisfied simultaneously, as the integral of a function defined as above is zero".
     
  5. Jul 11, 2015 #4

    micromass

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    Yes.

    Yes, the delta function is not a function ##\mathbb{R}\rightarrow \mathbb{R}## in the classical sense of the word. Rather, it must be defined rather differently. There are many possible definitions.
     
  6. Jul 11, 2015 #5
    How does one show that if a function is defined to have a non-zero value at a single point, but is zero elsewhere, that it's Riemann integral is zero?
     
  7. Jul 11, 2015 #6

    micromass

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    Well, what is your definition of the Riemann integral? Use that definition directly.
     
  8. Jul 11, 2015 #7
    Well I tend to use the following definition, [tex]\int_{a}^{b}f(x)dx =\lim_{n\rightarrow\infty}\frac{(b-a)}{n}\sum_{i=1}^{n}f(x^{\ast}_{i})[/tex] where ##x^{\ast}_{i}\in [x_{i},x_{i+1}]##.

    I can't see how I can show it from this definition unfortunately.
     
  9. Jul 11, 2015 #8

    micromass

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    That definition is only useful for finite ##a## and ##b##. So let us do it in this case. Take ##a = -1##, ##b=1##. Then choose your partition
    [tex] -1, -1 + \frac{2}{n}, -1 + \frac{4}{n}, ..., 1 - \frac{2}{n}, 1[/tex]
    end choose your ##x^{ast}_i## arbitrary. There are essentially two different situations:

    1) ##x^*_i## is never ##0##. Then ##f(x^*_i) = 0## for each ##i##. Thus ##\sum_{i=1}^n f(x^*_i)\frac{2}{n} = 0##. So ##\int_a^b f(x)dx = 0##.
    2) There is some ##i## such that ##x^*_i = 0##. Then ##f(x^*_i) = c## and the other ##i## satisfy ##f(x^*_i) = 0##. Hence
    [tex]\sum_{i=1}^n f(x^*_i) \frac{2}{n} = \frac{2c}{n}[/tex]
    This converges to ##0## as ##n\rightarrow +\infty##. So ##\int_a^b f(x)dx = 0##.
     
  10. Jul 11, 2015 #9
    What is a better definition that takes into account infinite intervals?

    Thanks for the demonstration for the other case (I had just thought of a similar way to do it just before I received your message, so I'm glad I understood it correctly).
     
  11. Jul 11, 2015 #10
    Consider the infinite sum [itex] \sum \int_n^{n+1} f(x)dx [/itex] as n ranges over all integers.

    It reduces to a finite sum, which micromass showed you is 0.

    Look up the concept of "measure zero"
     
  12. Jul 11, 2015 #11
    So would it be correct to (formally) express a Riemann integral over an infinite interval as [tex]\int_{-\infty}^{\infty}f(x)dx=\lim_{n\rightarrow \infty}\int_{-n}^{n}f(x)dx[/tex] and then use the limit of a Riemann sum as defined before for an integral over a finite interval?!
     
  13. Jul 11, 2015 #12
    Sorry, I was a bit too loose. In this case you can do that, but that's because I know that [itex] \int_{-\infty}^{\infty} f(x)dx [/itex] exists. But generally speaking, what you should do in your integral is let the upper and lower limit go to infinity and negative infinity independently. That is, [itex] \int_{-\infty}^{\infty} f(x)d(x) = \lim_{n \rightarrow \infty, m \rightarrow -\infty} \int_{m}^n f(x)d(x) [/itex].

    If you can prove, which shouldn't be too hard, that [itex] \int_{-\infty}^{\infty} f(x)d(x) [\itex] exists (meaning the limit as shown above exists and is finite), then you can write [itex] \int_{-\infty}^{\infty} f(x)d(x) = \lim_{n \rightarrow \infty} \int_{-n}^n f(x) d(x) [/itex] because if the improper integral exists, then allowing the upper and lower limits of integration to get to infinity anyway you want will always result in the same answer.

    If the improper integral doesn't exist, then there could exist paths for the limits of integration to take which will give you different, finite answers!
     
  14. Jul 11, 2015 #13
    Incase that post was a bit confusing, when calculating the improper integral from negative infinity to positive infinity, you must calculate the riemann integral and let the upper and lower limit of integration take their own independent path to infinity and negative infinity, respectively. The integral, if it exists, will give you the same result regardless of the paths taken. If you can show the improper integral does indeed exist, then you know that it doesn't matter which path the limits of integration take, so you may choose a path that is most convenient.
     
  15. Jul 16, 2015 #14

    mathwonk

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    well how much area is under that graph? isn't it obviously zero?
     
  16. Jul 17, 2015 #15

    HallsofIvy

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    Given any partition of the interval from a to b, you can always produce a "refinement" that includes x= 0 as a "cut point" and then the only non-zero terms in the Riemann sum would be [itex](0- c)/|x_i|+ (c- 0)/|x_{i+1}[/itex] where [itex]x_i[/itex] is the largest cut point less than 0 and [itex]x_{i+1}[/itex] is the smallest cutpoint larger than 0. But that is equal to [itex]\frac{c}{x_ix_{i+2}}(x_i- x_{i+1})[/itex] and [itex]x_i- x_{i+1}[/itex] goes to 0 as we take better and better refinements.
     
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