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Passing an integral through an infinite sum

  1. Aug 30, 2013 #1
    1. The problem statement, all variables and given/known data
    I want to show that
    $$
    \tan^{-1}(x)=\sum\limits_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}x^{2n+1}.
    $$

    2. Relevant equations
    I start with
    $$
    \int\frac{1}{1+x^{2}}dx.
    $$

    3. The attempt at a solution
    I want to be able to do the following:
    $$
    \int\frac{1}{1+x^{2}}dx=\int\sum\limits_{n=0}^{\infty}(-1)^{n}x^{2n}dx=\sum\limits_{n=0}^{\infty}\int (-1)^{n}x^{2n}dx
    $$
    but I am afraid that the infinite sum might create problems. Can anyone take a look? Thanks!
     
  2. jcsd
  3. Aug 31, 2013 #2

    jfizzix

    User Avatar
    Science Advisor
    Gold Member

    The order of the sum and integral can be switched, since they are both sums, so that's fine.

    So it seems that [itex]\frac{1}{1+x^2}[/itex] can be expressed as an infinite geometric series so long as [itex]|x^{2}|\leq 1[/itex].

    Where
    [itex]\frac{1}{1-r}=\sum_{i=0}^{∞} r^{i}[/itex]
    when [itex]|r|\leq 1[/itex]

    we can say that

    [itex]\frac{1}{1+x^2}=\sum_{i=0}^{∞} (-x^2)^{i}=\sum_{i=0}^{∞} (-1)^{i}(x)^{2i}[/itex]
    when [itex]|x^{2}|\leq 1[/itex]

    Hope this helps:)
     
  4. Aug 31, 2013 #3
    They cannot always be switched. Only under certain conditions is that statement valid. While this can be done so here it is bad habit to think the statement as always true.

    To reference http://en.wikipedia.org/wiki/Fubini's_theorem
     
  5. Aug 31, 2013 #4
    Thanks guys! That helps a lot.
     
  6. Aug 31, 2013 #5

    Zondrina

    User Avatar
    Homework Helper

    Have you heard about uniform convergence? Long story short, it allows us to verify if the following relationships actually hold :

    $$\int_{a}^{b} \sum_{n=0}^{∞} a_nx^n = \sum_{n=0}^{∞} \int_{a}^{b} a_nx^n$$
    $$\frac{d}{dx} \sum_{n=0}^{∞} a_nx^n = \sum_{n=0}^{∞} \frac{d}{dx} a_nx^n$$
    $$lim_{x→a} \sum_{n=0}^{∞} a_nx^n = \sum_{n=0}^{∞} lim_{x→a} a_nx^n$$
     
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