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Bringing limit under indefinite integral

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Given:[itex]lim_{n\rightarrow ∞}[/itex] [itex]\int^{a^n}_{1}[/itex] [itex]\frac{t^{1/n}}{(1+t)t}[/itex] dt=[itex]\int^{∞}_{1}[/itex] [itex]\frac{1}{(1+t)t}[/itex] dt

    a - Natural number.

    I need to prove that I can bring limit under the integral sign.

    2. Relevant equations


    3. The attempt at a solution
    I've got this so far:
    | [itex]\int^{a^n}_{1}[/itex] [itex]\frac{t^{1/n}}{(1+t)t}[/itex] dt-[itex]\int^{∞}_{1}[/itex] [itex]\frac{1}{(1+t)t}[/itex] dt|[itex]\stackrel{?}{\rightarrow}[/itex] 0 while n→∞

    | [itex]\int^{a^n}_{1}[/itex] [itex]\frac{t^{1/n}}{(1+t)t}[/itex] dt-[itex]\int^{∞}_{1}[/itex] [itex]\frac{1}{(1+t)t}[/itex] dt|= [did everything I could and wound up with following]=|[itex]\int^{a^n}_{1}[/itex] [itex]\frac{t^{1/n}-1}{(1+t)t}[/itex] dt|

    Now I need to either find a function g(t) so f(t)≤g(t) and [itex]\int^{a^n}_{1}[/itex] g(t) dt →0. This is basically the place where I'm stuck and need your help.

    p.s. I meant definite integral in the caption.
     
    Last edited: Jun 19, 2012
  2. jcsd
  3. Jun 19, 2012 #2
    How do you take the limit as x goes to infinity of an expression made up of a,n and t?
    Still, using the fundamental theorem of calculus will help here.
     
  4. Jun 19, 2012 #3
    Oh, I new I had made some mistakes writing the post. It was meant to be n -> infinity.

    And I can't see how can the fundamental theorem of calculus help.
     
  5. Jun 19, 2012 #4
    In general, let's take the functions f(x,n), a(x,n) and b(x,n), and write this:
    [tex]\int_{a(x,n)}^{b(x,n)}f(x,n)dx[/tex]
    If F is the antiderivative of f, then we obtain this is equal to
    [tex]F(b(x,n),n)-F(a(x,n),n)[/tex]
    Now, taking a(x,n)=1, we have
    [tex]F(b(x,n),n)-F(1,n)[/tex]

    It shouldn't be hard from there.
     
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