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Series & Integrals 2

  1. May 14, 2009 #1
    1. The problem statement, all variables and given/known data

    show what integral of 1/(1-xy)dxdy = pi2/6

    dx 0 to 1
    dy 0 to 1

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 14, 2009 #2

    Mark44

    Staff: Mentor

    Doesn't look like a hard integral. What have you tried?
     
  4. May 14, 2009 #3
    [tex]\int[/tex]1/(1-xy)dx = ln(1-xy) = ln(1-y) - ln1

    [tex]\int[/tex]ln(1-y) - ln1 dy = (1-y)ln(1-y) - (1- y) - yln1

    = -ln1 - (ln1 - 1) = = -2ln1 + 1 huh
     
  5. May 14, 2009 #4
    remember

    [tex]\int[/tex]f '(x)/f(x) dx = ln[f(x)] for your 1st integral, also you're taking y constant, so treat it like one
     
  6. May 14, 2009 #5
    1/(1-xy) = sum from n= 0 to infinity of x^n y^n

    The integral is thus the sum from n = 0 to infinity of 1/(n+1)^2 which is pi^2/6
     
  7. May 14, 2009 #6

    Cyosis

    User Avatar
    Homework Helper

    The first integration isn't correct. You can see that immediately by differentiating the primitive. Use a substitution if you don't see it right away, [itex]u=1-xy,du=-ydx[/itex]. The hard part comes after this however. Hint: Use the power series of the logarithm.

    Edit: Use Count's method it's quicker and easier.
     
  8. May 15, 2009 #7
    ok so your saying

    1(1-xy) = [tex]\sum[/tex]xnyn from n = 0 to infinity

    [tex]\int\int[/tex]1/(1-xy) = [tex]\sum[/tex]1/(n+1)2 = pi2/6

    how does 1/(1-xy) go to the sum of xnyn

    and how does 1/(1-xy) got to the sum of (n+1)2 then go to pi2/2
     
  9. May 15, 2009 #8

    Cyosis

    User Avatar
    Homework Helper

    Ok let's say z=xy, then [tex] \frac{1}{1-xy}= \frac{1}{1-z}=\sum_{n=0}^\infty z^n=\sum_{n=0}^\infty (xy)^n=\sum_{n=0}^\infty x^n y^n[/tex]. Then you take the integral over the sum.

    [tex]\int_0^1 \int_0^1 \sum_{n=0}^\infty x^n y^n dxdy=\sum_{n=0}^\infty \left(\int_0^1 \int_0^1 x^n y^n dxdy\right)[/tex]. Can you see how to continue from here?

    Edit: I think you're expected to know that the series converges to pi^2/6 by heart. However http://en.wikipedia.org/wiki/Basel_problem shows you how to get to that value in case you're interested.
     
    Last edited: May 15, 2009
  10. May 15, 2009 #9
    oops i think i forgot something

    the entire problem is

    show that

    [tex]\int\int[/tex]1/(1-xy)dxdy = pi2/6

    by doing the double substitution

    x = (u - v)/[tex]\sqrt{2}[/tex] and y = (u + v)/[tex]\sqrt{2}[/tex]

    This amounts to rotating the axes cuonterclockwise through the angle pi/4
     
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