From Osher Doctorow
Now let's get down to the brass tacks of integrating. I notice that
the subsections of the thread are getting scattered in different orders
because I digressed from numbering, so maybe this part 3 will bring
them back together. Several others are way after this - one page down
on my browser.
We have:
1) f(x,y) = exp(-x-y-kxy)[(1 + kx)(1 + ky) - k]
2) Dxy[P(X-->Y)(x,y)] = f(x,y) + Dxy[R(x,y)]
3) R(x,y) = I1[I2 f(x,y) dy]dx, I1 integral from x to infinity, I2 from
y to infinity
So all we need for (2) by integration is to evaluate (3) by integration
and then do a double (mixed) partial differentiation.
Let's look at the arguments in (1) and notice that:
4) (1 + kx)(1 + ky) - k = (1 - k) + kx + ky + k^2xy = (1 - k) + k(x +
y + kxy)
Therefore, notice that the argument of exp in (1) is -(k + y + kxy),
and that appears in (4) without the negative sign, so define:
5) z = x + y + kxy
which converts (4) to:
6) (1 - k) + kz
and converts the exponential in (1) to:
7) exp(-z)
and finally:
8) f(x,y) = exp(-z)[(1 - k) + kz]
Now let's proceed into integration by noticing that:
9) I2[f(x,y)]dy = (1 - k)I2[exp(-z)dy + kI2[z exp(-z)]dy
So what is dy relative to dz? Well, z = x + y + kxy, and in
integrating with respect to y, x is fixed, so dz = dy + kxdy = (1 +
kx)dy, so dy = dz/(1 + kx), and substituting into the second term on
the right hand side of (9) yields for it:
10) kI2[z exp(-z)]dy = [k/(1 + kx)]I2(z exp(-z)) dz
The term in (10) and the first right hand side term of (9) are both
easy integrals, the second using the equation:
11) I[u exp(au)]du = (1/a^2)(au - 1)exp(au) + C (C const of
integration)
and with a = -1, (11) is:
12) I2[u exp(au)]du = -(u + 1)exp(-u) + C
Don't forget to multiply by k/(1 + kx) from (10).
Then integrate with respect to x, that is to say apply I1...dx.
Osher Doctorow
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