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Average value of a function sent to infinity

  1. Mar 17, 2006 #1
    I was wondering if it would be possible to find the average value of a function with the only condition that x is element of R. For example, could we say that f(x)=4 has an average value of 4 since no matter what values we give for a or b in the integral from a to b of f(x)/(b-a) (b is not equal to a of course), we will obtain an answer of 4? However, if this is true, could it be for other functions as well, perhaps more complicated or colourful? What conditions need to be satisfied?
     
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  3. Mar 18, 2006 #2

    benorin

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    Recall that the average value of a function which is Riemann integrable on some interval (a,b) is given by:

    [tex]f_{ave} = \frac{1}{b-a}\int_a^b f(x) dx[/tex]​

    I suppose that one could define an average value of a function over all of [tex]\mathbb{R}[/tex] by, say

    [tex]f_{ave} = \lim_{L\rightarrow \infty}\frac{1}{2L}\int_{-L}^{L} f(x) dx[/tex]​

    so that for f(x)=4 we would have


    [tex]f_{ave} = \lim_{L\rightarrow \infty}\frac{1}{2L}\int_{-L}^{L} f(x) dx = \lim_{L\rightarrow \infty}\frac{1}{2L}\int_{-L}^{L} 4 \, dx = \lim_{L\rightarrow \infty}\frac{1}{2L}\left[ 4x\right] _{x=-L}^{L} = \lim_{L\rightarrow \infty}\frac{1}{2L}\left[ 4L-(-4L)\right] = 4[/tex]

    as expected.
     
  4. Mar 18, 2006 #3

    matt grime

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    But if one defines the average as

    [tex]\frac{1}{3L}\int_{-2L}^Lf(x)dx[/tex]

    then one should check that the answers agree, or for any other way of averaging. I imagine there will be some condition on the type of f for which this makes sense.
     
  5. Mar 18, 2006 #4

    benorin

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    Sure, I took the Cauchy Principal Value of the limit whose upper and lower bounds approached their repective infinities seperately.
     
  6. Mar 18, 2006 #5

    Hurkyl

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    The bounds in the CPV don't go separately: they go together in a particular way. To go separately, you'd need something like:

    [tex]
    \lim_{\substack{a \rightarrow -\infty \\ b \rightarrow +\infty}}
    \frac{1}{b-a}\int_a^b f(x) dx[/tex]


    In general, this really does matter. For example:

    [tex]
    CPV \int_{-\infty}^{+\infty} \sin x \, dx = 0
    [/tex]

    even though

    [tex]\int_{-\infty}^{+\infty} \sin x \, dx[/tex]

    does not exist.
     
    Last edited: Mar 18, 2006
  7. Mar 18, 2006 #6

    benorin

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    Indeed, I took the CPV of such an integral.
     
  8. Mar 18, 2006 #7
    Tell me a little more of how the Cauchy Principal Value theorem works for sin(x) integrated from -infinity to positive and how it gives zero. When i evaluate it, i find that it can simply range from -2 to 2. (if you please).
     
    Last edited: Mar 18, 2006
  9. Mar 18, 2006 #8

    matt grime

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    please is such a nice word, isn't it?
     
  10. Mar 18, 2006 #9

    Hurkyl

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    By definition,

    [tex]
    CPV \int_{-\infty}^{+\infty} f(x) \, dx
    = \lim_{L \rightarrow +\infty} \int_{-L}^{L} f(x) \, dx
    [/tex]
     
  11. Mar 18, 2006 #10
    Ah, i understand now. Thanks very much!
     
    Last edited: Mar 18, 2006
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