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

[joemama69]

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

 

[Mark44]

Doesn't look like a hard integral. What have you tried?

 

[joemama69]

\int1/(1-xy)dx = ln(1-xy) = ln(1-y) - ln1

\intln(1-y) - ln1 dy = (1-y)ln(1-y) - (1- y) - yln1

= -ln1 - (ln1 - 1) = = -2ln1 + 1 huh

 

[Firepanda]

\int1/(1-xy)dx = ln(1-xy) = ln(1-y) - ln1

\intln(1-y) - ln1 dy = (1-y)ln(1-y) - (1- y) - yln1

= -ln1 - (ln1 - 1) = = -2ln1 + 1 huh

remember

\intf '(x)/f(x) dx = ln[f(x)] for your 1st integral, also you're taking y constant, so treat it like one

 

[Count Iblis]

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

 

[Cyosis]

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, u=1-xy,du=-ydx. 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.

 

[joemama69]

ok so your saying

1(1-xy) = \sumxⁿyⁿ from n = 0 to infinity

\int\int1/(1-xy) = \sum1/(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

 

[Cyosis]

Ok let's say z=xy, then \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. Then you take the integral over the sum.

\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). 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.

 

[joemama69]

oops i think i forgot something

the entire problem is

show that

\int\int1/(1-xy)dxdy = pi²/6

by doing the double substitution

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

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