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Interesting Integral problem

  1. May 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose [tex]f[/tex] is continuous and [tex]F(x)=\int_a^x f(t)dt[/tex] bounded on [tex][a,b)[/tex]. Given [tex]g>0, g'\geq 0[/tex] and [tex]g'[/tex] locally integrable on [tex][a,b)[/tex] and [tex]lim_{ x\rightarrow b^-} g(x) =[/tex] infinity. prove
    for p>1
    [tex]\displaystyle{lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} \int^x_a f(t)g(t)dt = 0}[/tex]

    2. Relevant equations



    3. The attempt at a solution
    If you know [tex]lim_{ x\rightarrow b^-} g(x) =\infty[/tex], don't you also know [tex]lim_{ x\rightarrow b^-} \frac{1}{[g(x)]} = 0[/tex] and therefore [tex]lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p}=0[/tex]...so we're done? But the hint says to use parts so I dont know :(
    Any help would be greatly appreciated
     
  2. jcsd
  3. May 8, 2010 #2

    lanedance

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    thats only the case if the integral is bounded, so you should try & show that...
     
  4. May 8, 2010 #3
    How would I show an integral is bounded? If I know obviously [tex]\int_a^x f(t)dt[/tex] is bounded, how would I know anything about the g function. The information given basically denies that its bounded right? Since the limit is infinity? I'm so lost!
     
  5. May 8, 2010 #4

    lanedance

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    have you tried the suggested hint?
     
  6. May 8, 2010 #5

    lanedance

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    also, bounded would show it, but I think a lesser constraiint may also work, either way I would try simplifying things by using integration by parts
     
  7. May 8, 2010 #6
    When I tried parts, i let u= g(t) and dv = f(t)dt so du = g'(t)dt and v = F(t) and I got


    [tex]\displaystyle{lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} \int^x_a f(t)g(t)dt = lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} (g(x)F(x) -g(a)F(a)- \int^x_a F(t)g'(t)dt)}[/tex]

    [tex]\displaystyle{= lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} (g(x)F(x) - \int^x_a F(t)g'(t)dt)}[/tex]
    since F(a) = 0

    Can I reduce further...?
     
  8. May 8, 2010 #7
    i see it now, using parts helped a lot, thanks

    i'm suppose to follow it up with this addition:

    assume further that [tex]\int^b_a f(t)dt[/tex] converges and show [tex]lim_{x \rightarrow b^-} \frac{1}{g(x)} \int_a^x f(t)g(t)dt = 0[/tex]


    i used parts again and came down to the fact:

    [tex]lim_{x \rightarrow b^-} \frac{1}{g(x)} \int_a^x f(t)g(t)dt = lim_{x \rightarrow b^-} \frac{1}{g(x)} \int_a^x F(t)g'(t)dt[/tex]


    how do i show, for sure, this =0?
     
  9. May 9, 2010 #8

    lanedance

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    how about usng the fact that F is bounded, g'>0 and setting up an inequality which squeezes your expression to zero?
     
  10. May 9, 2010 #9
    The entire expression, or just the integral part? Could I do:

    [tex]lim_{x\rightarrow b^-} \frac{1}{g(x)} \leq lim_{x\rightarrow b^-} \frac{1}{g(x)} \int^x_a F(t)g'(t)dt \leq lim_{x\rightarrow b^-} \int^x_a M g'(t)dt[/tex]

    where the right hand side becomes

    [tex]=lim_{x\rightarrow b^-} \frac{1}{g(x)}M(g(x)-g(a))[/tex]
    [tex]=lim_{x\rightarrow b^-} M(1-\frac{g(a)}{g(x)})[/tex]

    But how does this squeeze to 0? or am i looking at it wrong?
     
  11. May 9, 2010 #10

    lanedance

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    the 2nd half of the equality is looking good, but i think you dropped a factor of p somewhere, maybe you could assume [itex]p = 1+\epsilon, \epsilon>0[/itex]

    you may need to tighten it up by accounting for the case when M<0, ie by starting from [itex] \exists M > 0 : |F(x)|<M,\forall x \in [a,b)[/itex]
     
  12. May 9, 2010 #11
    but there's no p in the second part of the problem...?
     
  13. May 9, 2010 #12
    This is the second part...
     
  14. May 9, 2010 #13

    lanedance

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    ok sorry, i missed the 2nd part, so you're ok with the first bit now?

    so if i integrate by parts I get a little different:
    [tex] \lim_{x \to b^-} \frac{1}{g(x)}\int_a^x dt f(t)g(t) [/tex]
    [tex] = \lim_{x \to b^-}\frac{1}{g(x)}(F(t)g(t)|_a^x - \int_a^x dt F(t)g'(t))[/tex]
    [tex] = \lim_{x \to b^-}(F(x)-\frac{F(a)g(a)}{g(x)} - \frac{1}{g(x)}\int_a^x dt F(t)g'(t))[/tex]

    using convergence of F as x->b, then
    [tex] = F(b) - \lim_{x \to b^-}\frac{1}{g(x)}\int_a^x dt F(t)g'(t)[/tex]
    so working this line, i guess you need to try & show the limit of the end expression is F(b)...
     
    Last edited: May 9, 2010
  15. May 9, 2010 #14

    lanedance

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    i haven't totally worked it out, but it may help to try and show for any a<x0<b
    [tex] \lim_{x \to b^-}\frac{1}{g(x)}\int_a^{x_0} dt F(t)g'(t) = 0[/tex]
    noting that as g' is locally integrable, it means g is defined on the interval [a,b), ie. it doesnt explode anywhere

    then try splitting the integral into separate portions & considering the continuity of the function f on a small region near to b. (though i could see this might get tricky with double limits....)
     
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