492 
DR. E. W. HOBSON ON A TYPE OF SPHERICAL HARMONICS 
now suppose the real part of m-f -\ is positive, then provided the real part of p. is positive 
the figure is as in Art. 24 ; we may take the path of integration to be two straight lines 
on opposite sides of the straight line joining the points 2 , —, and two indefinitely small 
circles round these points ; the integrals round these circles will vanish. If the real 
part of fx had been negative, so that the line joining 2 , —, were on the left of 0, the 
z 
path of the integral (50) must have been placed so that 0 was on its left hand, 
and thus we could not have reduced the integral to integrals along the line joining 
2 , —; it is therefore essential in what follows that the real part of p, be supposed to 
z 
be positive. 
We have now 
r(«+, i/z-) - f 3 
j h H ~ m ( I - 2 fih + h 2 ) m -> dh = (e-' 2 '"*- 1 - 1)J Ji n ~ m ( 1 - 2 fill + A*)" 
dli, 
where the phases of 1 — hz, 1 — — in the integral on the right-hand side, are 27 t — f3, 
Z 
and a respectively, we thus have 
= 
—. (/x 3 - i)~ im (V- w (i - 2nil + wy^dh. 
— 2 ) J 1/2 
2 * n (- *) n (m - *) 
Let h = /x -f- (p 3 — 1 ) 5 cos \j), then 
dh — — v/p 3 — 1 sin \p dxjj 
1 — 2 fih + Id = — (p, 3 — 1) sin 3 xp = e +ur (p 3 —■ 1) sin 3 \p 
since the phase of 1 — 2 fill + Id is a — (3 -j- 27 t, which is 2\ -f- n, and the phase of 
p. 3 — 1 is 2A.; thus 
M = ^nf-iVncA-i) f 0 (/* + \/? - 1 cos sill 2 ,,i V' # • ( 67 )- 
Or, using equation (19). we have 
