- #1

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## Homework Statement

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

^{2}/6

dx 0 to 1

dy 0 to 1

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- Thread starter joemama69
- Start date

- #1

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

dx 0 to 1

dy 0 to 1

- #2

Mark44

Mentor

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

- #3

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

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

- #4

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

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

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

- #5

- 1,838

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The integral is thus the sum from n = 0 to infinity of 1/(n+1)^2 which is pi^2/6

- #6

Cyosis

Homework Helper

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Edit: Use Count's method it's quicker and easier.

- #7

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1(1-xy) = [tex]\sum[/tex]x

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

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

and how does 1/(1-xy) got to the sum of (n+1)

- #8

Cyosis

Homework Helper

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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.

[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:

- #9

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the entire problem is

show that

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

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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