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Infinite series wizardry

  1. Sep 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Check out matt grime's post in this thread (it's the last one):

    https://www.physicsforums.com/showthread.php?p=470773#post470773"

    How exactly did he know that the sum could be represented as that double integral? Also, is there a method of converting sums like that to integrals (double or otherwise) for summands other than [itex]n^{-2}[/itex] such as [itex]n^{-7}[/itex] or something?


    2. Relevant equations

    [tex]\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}dxdy=\sum_{n=1}^{\infty }\frac{1}{n^2}[/tex]

    3. The attempt at a solution

    Come to think of it, I don't even really see how that helps you, because the series expansion for the [itex]y[/itex] integral (after computing the [itex]x[/itex] integral) is the derivative of the series you're trying to find, so integrating it just brings you back to where you started. The only thing I was able to note was that:

    [tex]\sum_{n=0}^{\infty} (xy)^n=\frac{1}{1-xy}[/tex]
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Sep 1, 2011 #2

    micromass

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    What exactly are you asking?? Are you asking why

    [tex]\int_0^1\int_0^1\frac{dxdy}{1-xy}=\sum{\frac{1}{n^2}}[/tex]

    Well, you're on the right way. Just notice that

    [tex]\frac{1}{1-xy}=\sum{(xy)^{n-1}}[/tex]

    So by monotone convergence, we can switch the limit and the integral: So

    [tex]\int_0^1\int_0^1\frac{dxdy}{1-xy}=\sum\int_0^1\int_0^1 (xy)^{n-1}dxdy[/tex]

    So you only need to calculate [itex]\int_0^1\int_0^1 (xy)^{n-1}dxdy[/itex] now...
     
  4. Sep 1, 2011 #3
    Well what I'm really asking is how to convert the sum to an integral. Like If I just saw [itex]\sum_{n=1}^{\infty }\frac{1}{n^2}[/itex] somewhere, I would start with this:

    [tex]\sum_{n=1}^{\infty} (xy)^{n-1}=\frac{1}{1-xy}[/tex]

    But how did I know to start with that? It appears to have no relation to [itex]\sum_{n=1}^{\infty }\frac{1}{n^2}[/itex] ... and then all of a sudden there's a double integral with non-infinite limits.
     
  5. Sep 1, 2011 #4
    Recall the (real) geometric series,

    [itex]\frac {1}{1-x}=\sum_{n=0}^{\infty}x^n[/itex]

    This is true provided, of course, that [itex]|x|<1[/itex].

    Anytime you see a rational expression of this form, a light should go off in your head regarding the geometric series. It should be easy to see now how to prove the equivalence of your double integral and [itex]\zeta(2)[/itex].

    Not sure if this clarifies anything; let us know if you have more questions.
     
  6. Sep 1, 2011 #5
    It's not showing that the integral and sum are equivalent that's my problem, it's coming up with the integral in the first place :wink:. Let me try to rephrase this:

    Given a sum, I want to know how to write down a computable integral whose value is the value of that sum.
     
  7. Sep 3, 2011 #6
    First and last bump for this thread.
     
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