[oo]
T = [pi][sigma]^2 | z[psi](z)dz. (35)
_/0

In Laplace's notation the second member of (34), multiplied by 2[pi],
is represented by H.

As Laplace has shown, the values for K and T may also be expressed in
terms of the function [phi], with which we started. Integrating by
parts, we get
_ _
/ /
| [psi](z)dz = z[psi](z) + (1/3)z^3[PI](z) + 1/3 | z^3[phi](z)dz,
_/ _/
_ _
/ /
| z[psi](z)dz = 1/2z^2[psi](z) + (1/8)z^4[PI](z) + 1/8 | z^4 [phi](z)dz.
_/ _/

In all cases to which it is necessary to have regard the integrated
terms vanish at both limits, and we may write
_ _ _ _
/ [oo] 1 / [oo] / [oo] 1 / [oo]
| [psi](z)dz = -- | z^3 [phi](z)dz, | z[psi](z)dz = -- | z^4 [phi](z)dz; (36)
_/0 3 _/0 _/0 8 _/0

so that
_ _
2[pi] / [oo] [pi] / [oo]
K0 = ----- | z^3[phi](z)dz, T0 = ---- | z^4 [phi](z)dz. (37)
3 _/0 8 _/0

A few examples of these formulae will promote an intelligent
comprehension of the subject. One of the simplest suppositions open to
us is that

[phi](f) = e^([beta] f). (38)

From this we obtain

[Pi](z) = [beta]^(-1) e^([beta] z), [psi](z) = [beta]^(-3)([beta]z + 1) e^(-[beta] z), (39)

K0 = 4[pi][beta]^(-4), T0 = 3[pi][beta]^(-5). (40)

The range of the attractive force is mathematically infinite, but
practically of the order [beta]^(-1), and we see that T is of higher
order in this small quantity than K. That K is in all cases of the
fourth order and T of the fifth order in the range of the forces is
obvious from (37) without integration.

An apparently simple example would be to suppose [phi](z) = z^n. We
get

z^(n+1) z^(n+3)
[PI](z) = - -------, [psi](z) = ---------,
n+1 n+3 . n+1

2[pi]z^(n+4) |[oo]
K0 =