Integral of 1/(1-xy): Pi2/6 from 0 to 1

For a more detailed explanation please see rotated-positionf
  • #1

Homework Statement

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

dx 0 to 1
dy 0 to 1

Homework Equations

The Attempt at a Solution

  • #2
Doesn't look like a hard integral. What have you tried?
  • #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
  • #4
[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


[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
  • #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
  • #6
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.
  • #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
  • #8
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 shows you how to get to that value in case you're interested.
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  • #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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