1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How would i evaluate an integral such as this

  1. Jun 20, 2009 #1
    1. The problem statement, all variables and given/known data
    ∫ (a^(n+1)-1)/(a-1) da


    2. Relevant equations
    i took this equation from the series of sigma with the bound of N k=0 and the argument as a^K (I'm sorry if this is confusing i don't know how to type it out and I'm not sure what all the correct terminology is).


    3. The attempt at a solution
    i really just don't know where to start with this one, i don't know what would be appropriate to use, by parts or substitution, and if it is by substitution what do i substitute?
     
  2. jcsd
  3. Jun 20, 2009 #2
    Are you saying that the original problem was

    [tex]\int \left( \sum_{k=0}^{n} {a^k} \right) da [/tex]

    ?

    In that case, you can use the fact that ∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx and apply this over and over again to your summation.

    The summation implies that n is a nonnegative integer, but if that is not the case, I don't think this function has an elementary integral.
     
  4. Jun 20, 2009 #3
    what is a non-elementary integral?
    and basically i was just trying to see what would happen if the summation was switched with an integral, because that sum is (when n approaches infinite) is 2. But that's only when the numbers increase at an integer amount, i wanted to see what it would be if it increased at every amount so i thought substituting the integral in would work. (is my logic flawed)

    EDIT:forgot to put in the fact that it only approaches two when a=1/2
     
    Last edited: Jun 20, 2009
  5. Jun 20, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [itex]a^{n+1}- 1= a^{n+1}- 1^{n+1}= (a-1)(a^n+ a^{n-1}+\cdot\cdot\cdot+ 1)[/itex]. Does that help?
     
  6. Jun 21, 2009 #5
    [tex]S=a_1+qa_1+q^2a_1+q^3a_1+...+q^na_1[/tex]

    [tex]Sq=qa_1+q^2a_1+q^3a_1+q^4a_1+...+q^{n+1}a_1[/tex]

    [tex]Sq-S=q^{n+1}a_1-a_1[/tex]

    [tex]S(q-1)=a_1(q^{n+1}-1)[/tex]

    [tex]S=\frac{a_1(q^{n+1}-1)}{q-1}[/tex]

    Is it enough?

    Regards.
     
  7. Jun 21, 2009 #6
    A non-elementary integral is one that cannot be expressed in elementary functions.

    An elementary function is one that can be written in terms of "simple" functions. A more thorough definition can be found at http://mathworld.wolfram.com/ElementaryFunction.html

    So you started with

    [tex] \lim_{n \to \infty} \sum_{k=0}^{n} {\left( \frac{1}{2} \right)^k} = 2 [/tex]

    right? I'm sorry, but I don't understand what you did afterwards.
     
  8. Jun 23, 2009 #7
    after that i replaced the sigma notation with an integral
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How would i evaluate an integral such as this
  1. How would I integrate? (Replies: 3)

Loading...