An introduction to the mathematical theory of attraction ...

Examples. 253
Hence, for Ea and Eb the charges on the spheres (A) and (B), we get the
equations
oD 1 1
a = La + Lk Lie 'VE = La + Lk, sinh (nz - a) = Lk sinh (nl -a)'
Eb=-Lk Z,'. — 1
1i sinh nzM
4. An insulated conductor formed of two spheres in contact is charged to
potential L; express the charges on the two spheres by means of Eulerian
Integrals.
Investigations of most of the properties of Eulerian Integrals are given in
Williamson's Integral Calculus, Chapter VI. Some of those required in the
present case are not to be found in that treatise, and of these a brief exposition
is here supplied for the convenience of the student. For fuller information
the reader is referred to Williamson's Article on the Integral Calculus in the
F.ncyclopedia Britannica.
The second Eulerian Integral r (x) being defined by the equation
r(x) = -0 eox-1 dO;
if we assume e-~ = z, we may write!i / 1 \ a-1 1 / i \
r (x)= (log - dz; now log-=m 1in -z)
o z 1 z
when m = oo, for putting m =, we have m ( -zm) = - -- the value of
which when = O is -logz.
1 / 1 VX-l
Hlence r (x) = mx-1 ( 1 - z1 dz when n = oo.
If we assume z = ym, the integral
rl~- i \z-l 1
f m-1 (1 - z)e dz becomes mx y..-l (1 - y)-1 dy
o o
integrating by parts, we get
m-1 y)x _ 'y!ym _y)- dy = yym Y (( - 1 )-ldy,
-' (1 - )x d = - r - 1 -e 1
Ii o Ymx-l (1 l+ -YJe
and, as the part outside the integral sign vanishes at both limits, by successive
applications of this process, when m is an integer, we obtain for the right hand
member of the above equation the value
(m- 1) (n - 2).... 1 1
( + x - 1) (m x - 2)..... + 1) Jo - )-l dy.
1.2.3.... (m-l)
Hence, r ) =,.3)(a)
x (x+ 1) (x+ 2).... (x+m- 1)' (
and log r(x) = log m - -....- (b)
dwe x x x+1 x+m- '
where ms is an infinite integer.