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Quick series and integral Q

  1. Feb 9, 2009 #1
    Hi all. I've just come across two issues in some problems I have been doing and although I know the results, I can't remember the reason why or how to show it. Hoping someone can point me in the right direction.

    1. The problem statement, all variables and given/known data

    1: lim[n->∞] f(g(n)) = f(lim[n->∞]g(n))

    2. Integral terminals [0,a]: ∫dx/(x(a-x))

    3. The attempt at a solution

    1: I'm pretty sure this is true given f is continuous. I'm sure there's some series theorem that establishes it but I haven't been able to find anything in my notes.

    2: From memory I think this integral comes out to be something nice like pi or pi/2, but I couldn't figure it out with trig substitution.

    Thanks
     
  2. jcsd
  3. Feb 9, 2009 #2

    HallsofIvy

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    That's pretty much the definition of continuous!

    What "trig substitution"? Do it by "partial fractions":
    [tex]\frac{1}{x(a-x)}= \frac{1}{a}\left(\frac{1}{x}+ \frac{1}{a- x}\right)[/tex]
    [tex]\int_0^a \frac{dx}{x(a-x)}= \frac{1}{a}\left(\int_0^a \frac{dx}{x}+ \int_0^a\frac{dx}{a-x}\right)[/tex]

    [tex]= \frac{1}{a}\left(ln(x)- ln(a-x)\right)= \frac{1}{a}ln\left(\frac{x}{a-x}\right)[/tex]
    and, since that does not exist at x= 0 or x= a, the integral does not exist.
     
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