[GeeTwo]

Originally Posted by Aren Siekmeier

Precisely ln(1+sqrt(2))/3 + sqrt(2)/15 + 2/15

Or approximately 0.52140543316
Rounded to 8 digits: 0.52140543
Rounded to 5 digits: 0.52141

A fun exercise in every integration tool I've ever learned, plus a cool rationalization trick I wasn't familiar with.

The computation took a bit to hash through some arithmetic errors and one silly differential error (chain rule!). In practice you probably wouldn't bother to do this (or use a symbolic manipulation tool like Mathematica - this is how I checked my work), and sometimes there is no closed form for the integral. To get 8 digits simply run a larger Monte Carlo simulation (a couple lines of code instead of several pages of derivation). In the end we are always limited to some finite number of digits, so numerical methods win out.

Wonderful! I've been stuck on the ln(1+√(1+y2)) terms. Didn't think to keep the ln(y) terms with them. (I did the x integrals first, so was using y where you have x-bar).

Edit - Now that I've had a chance to spend more than a few minutes on it, I see that wasn't the trick at all. I just need to go back and re-learn integration by parts.

Edit2: Gathering up my paper notes, this is what I had so far (haven't double-checked everything yet):

Originally Posted by GeeTwo

4ʃʃ(1-x)(1-y)√(x²+y²)dxdy, both integrals being over 0..1

Checking the CRC Handbook integrals table (2000 edition):

Originally Posted by CRC 2000 Handbook integral #156, ‘+’ case, letting a=y

ʃ√(x²+y²)dx = (x√(x²+y²) + y² ln(x+√(x²+y²)))/2| = (√(1+y²) + y² ln(1+√(1+y²)) - y² lny)/2

Where the | is an implied evaluation over 0..1.

Originally Posted by CRC 2000 Handbook integral #163, ‘+’ case, letting a=y

ʃx√(x²+y²)dx = (x²+y²)3/2/3| = ((1+y²)3/2 - y³)/3

• ʃ(1-y)( (√(1+y²)/2+y²ln(1+√(1+y²))/2-y²lny/2+(1+y²)3/2/3-y3/3)dy

evaluated over y=0..1. This expands to ten terms:

1. ʃ√(1+y²)dy/2
2. -ʃy√(1+y²)dy/2
3. +ʃy²ln(1+√(1+y²))dy/2
4. -ʃy³ln(1+√(1+y²))dy/2
5. -ʃy²lnydy/2
6. +ʃy³lnydy/2
7. +ʃ(1+y²)^3/2dy/3
8. -ʃy(1+y²)^3/2dy/3
9. -ʃy³dy/3
10. +ʃy⁴dy/3

I know or quickly found all but the forms with ln(1+√(1+y²)) in an online table of integrals or the CRC table: http://2000clicks.com/mathhelp/Calcu...rals.aspx#CatL

Originally Posted by Term1

ʃ√(y²+1)dy = (√(y²+1) + ln(y+√(y²+1)))/2| = √2 – 1 + ln(1+√2)

Originally Posted by Term2

-ʃ√y(y²+1)dy = -(y²+1)^3/2/3| = 1/3 - 2√2/3

Originally Posted by Term5

+ʃy²lnydy/2 = +y³(lny/3 – 1/9)| = -1/9

Originally Posted by Term6

-ʃy³lnydy/2 = +y⁴(lny/4 – 1/16)| = 1/16

Originally Posted by Term7, CRC 164

+ʃ(1+y²)^3/2dy/3 = (y(1+y²) ^3/2 + 3y(√(y²+1))/2 + 3ln(y+√(y²+1)/2)/12| = √2/6 + √2/8 + ln(1+√2)/8

Originally Posted by Term8

-ʃy(1+y²)^3/2dy/3 = -(y²+1)^5/2/15| = 1/15 - 4√2/15

Originally Posted by Term9

-ʃy³dy/3 = -y⁴/12| = -1/12

Originally Posted by Term10

+ʃy⁴dy/3 = y⁵/15| = 1/15

I have also worked through term 3, using your same u for integration by parts, and term 4 looks trivially different. Now I'm going to look for a solution that exploits the symmetry between x and y.