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Proof for the limit of a definite integral where the integrand varies

  1. May 25, 2012 #1
    1. The problem statement, all variables and given/known data

    This problem was taken from Spivak's Calculus 3rd Edition, Problem 19-25 (d).

    Let [itex]f[/itex] be integrable on [itex][-1 , 1][/itex] and continuous at [itex]0[/itex]. Show that [tex] \lim_{h\rightarrow 0^+} {\int_{-1}^{1}{\frac{h} {h^2+x^2}}f(x)dx = {\pi}f(0).}[/tex]

    2. Relevant equations

    I already proved from part (c) that [tex] \lim_{h\rightarrow 0^+} {\int_{-1}^{1}{\frac{h} {h^2+x^2}}dx = {\pi}}[/tex]
    which is easy with arctan.

    3. The attempt at a solution

    I found a way to prove it but I am not 100% sure that it is rigorous enough (and the proof given in the answers is different).

    First consider [itex] \int_{-h}^{h}{\frac{h} {h^2+x^2}}f(x)dx. [/itex] Since [itex]f[/itex] is continuous at [itex]0[/itex], then for some [itex]\delta > 0,[/itex] [itex]f(0) - \epsilon < f(x) < f(0) + \epsilon[/itex] for all x in [itex][-\delta , \delta].[/itex] Choose [itex]0<h<\delta[/itex]. Then [tex] (f(0)-\epsilon) \int_{-h}^{h}{\frac{h} {h^2+x^2}}dx < \int_{-h}^{h}{\frac{h} {h^2+x^2}}f(x)dx < (f(0)+\epsilon) \int_{-h}^{h}{\frac{h} {h^2+x^2}}dx[/tex] [tex]{\pi}(f(0)-\epsilon) ≤ \lim_{h\rightarrow 0^+} {\int_{-h}^{h}{\frac{h} {h^2+x^2}}f(x)dx} ≤ \pi(f(0)+\epsilon) [/tex] taking the limits by part (c) and this is true for any [itex]\epsilon > 0 [/itex] so [tex] \lim_{h\rightarrow 0^+} {\int_{-h}^{h}{\frac{h} {h^2+x^2}}f(x)dx} = {\pi}f(0) [/tex]

    Now consider [itex]\int_{h+\delta'}^1{\frac{h} {h^2+x^2}}f(x)dx [/itex] for any [itex]\delta' > 0. [/itex]
    [tex] \frac{h} {h^2+1}\int_{h+\delta'}^1{f(x)dx} < \int_{h+\delta'}^1{\frac{h} {h^2+x^2}}f(x)dx < \frac{h} {h^2+(h+\delta')^2}\int_{h+\delta'}^1{f(x)dx}[/tex] (the maximum and minimum values of the fraction occur at [itex]h+\delta'[/itex] and [itex]1[/itex] respectively). Therefore, since the integral is a number, [tex]
    \lim_{h\rightarrow 0^+} {\int_{h+\delta'}^1{\frac{h} {h^2+x^2}}f(x)dx} = 0 [/tex]
    and this is true for any [itex]\delta'>0.[/itex]


    Finally consider [itex]\int_{h}^{h+\delta'}{\frac{h} {h^2+x^2}}f(x)dx. [/itex] Add the requirement that [itex]h+\delta' < \delta[/itex] so that [tex]
    \frac{h} {h^2+1}\int_{h}^{h+\delta'}{f(x)dx} < \int_{h}^{h+\delta'}{\frac{h} {h^2+x^2}}f(x)dx < (f(0)+\epsilon) \int_{h}^{h+\delta'} {\frac{h} {2h^2}dx} = \frac{\delta'}{2h}(f(0)+\epsilon) < \frac{h}{2}(f(0)-\epsilon) [/tex] if we also require that [itex]\delta'<h^2[/itex]. Since we can make this smaller than any number by choosing small enough [itex]h[/itex] and [itex]\delta'[/itex] with the given requirements, this shows that [tex] \lim_{h\rightarrow 0^+} {\int_{h}^{h+\delta'}{\frac{h} {h^2+x^2}}f(x)dx} = 0 [/tex] for any small enough [itex]\delta'[/itex].


    Combining all three results shows that [tex] \lim_{h\rightarrow 0^+} {\int_{-1}^{1}{\frac{h} {h^2+x^2}}f(x)dx} = \lim_{h\rightarrow 0^+} {\int_{-h}^{h}{\frac{h} {h^2+x^2}}f(x)dx} + 2\lim_{h\rightarrow 0^+} {\int_{h+\delta'}^1{\frac{h} {h^2+x^2}}f(x)dx} + 2\lim_{h\rightarrow 0^+} {\int_{h}^{h+\delta'}{\frac{h} {h^2+x^2}}f(x)dx} = {\pi}f(0) + 0 + 0 = {\pi}f(0) [/tex] (using symmetry).

    QED


    Ok that was long and maybe confusing, I'm sorry... but that's the proof I could find. Can you tell me if there are any gaps or incorrect assumptions in the proof? Thanks!
     
  2. jcsd
  3. May 26, 2012 #2
    I see that for [itex] \lim_{h\rightarrow 0^+} {\int_{-1}^{1}{\frac{h} {h^2+x^2}}dx = {\pi}} [/itex] , but [itex] \int_{-h}^{h}{\frac{h} {h^2+x^2}}dx= {\pi}/2 [/itex] I believe? If you visualize [itex] \lim_{h\rightarrow 0^+} [/itex] as [itex] \lim_{n\rightarrow \inf} [/itex] with all the h=1/n, you can evaluate as a pointwise convergent function, that only has value at x=0 as n-> inf, and if you normalize [itex] {\frac{h} {h^2+x^2}} [/itex] over ℝ , you'll get a delta function, which gives useful properties in terms of integration. Nascent delta function comes to mind for this case
     
  4. May 26, 2012 #3
    Oh you're right! Then I must've completely messed-up haha (although I still got the correct answer)

    In this case this would mean that another [itex]{\frac {\pi} {2}}f(0)[/itex] must appear from somewhere... And yet intuitively the other integrals should converge to 0, shouldn't they?
     
  5. May 26, 2012 #4
    Replace your limits of integration with h^2, (or 1/n^2) ... i loaaaaaaaathe one sided limits, and prefer sequences of 1/n that will converge to 0+.
    Remember youre dealing with a function that pointwise will converge to 0 for all non zero x, but go to infinity for x=0. if you redefine youre h/(h^2+x^2) as a delta function by normalizng the integral over R ( ie delta(X)=lim n-> inf n/(pi*(1+(nx)^2)), youll get integral delta(x)*f(x)=f(0), and multiplying the pi over from the normalized delta function gives pi*f(0)
     
  6. May 26, 2012 #5
    hmmm I'm not really familiar with the delta function. Why does the delta function work with those kinds of limits? And also, since the book doesn't mention it yet, I suppose he expects a solution using an epsilon-delta argument.
     
  7. May 26, 2012 #6
    Delta function is 0 for all non zero x, but goes to infinity at x=0. It's interesting because when integrated over all R, it's 1, and pops up alot in physics. What material are you covering in this chapter? Anything having to do with cauchys principal value theorem?
     
  8. May 26, 2012 #7
    No it's actually a chapter on Indefinite Integration. But the problems tend to get hard and more off-topic after the first few. I haven't heard of the Cauchy principal value theorem :uhh:
    In the answers he uses an epsilon-delta argument but his proof was confusing to me so I tried finding a similar argument that would be more clear to me.
     
  9. May 26, 2012 #8
    Is there any way you could post the solution he gave? I could help explain what he is doing. If not, there's a whole complex analysis way of solving this also.
     
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