518 DR. E. W. HOBSON ON A TYPE OF SPHERICAL HARMONICS 
Next take the path to be from 0 to — co along the real axis, along a semicircle of 
infinite radius to + co , from + co along the real axis to 1, and from 1 along a 
circular arc whose centre is the origin to the point —. If the real part of n — m is 
negative, the part of the integral taken along the infinite semicircle is zero; we have 
z 
then, writing h — e l7T . e n , in the first integral, h — e u , in the second integral, and 
h = e _lf h in the third integral, 
Q/' (cos 0 -f 0 . t) 
3 ~ „ / . . „ , i \ cos 7H7T 
— e ’ mm . 2'". II (in — ^) II (— -sm 
in® Q | | Q—(.n+m+l)iir _ 
Jo e (,s+i/ * r ‘ (2 cosh u - 2 cos 6) m+i du ~ f 0 
An+h)u 
An + 1) n 
(2 cosh u + 2 cos 6) 
—(u+D 
rrf+i 
du 
ie 
(m+i) 2m / 9 
(2 cos ^ — 2 cos 0) 
A+t cj -4>y 
If the real part of m lies between ± ^. and if the real parts of n + m + 1, in — n 
are positive, both the formulae we have found for Q/' (cos 6 -f- 0 . i) hold. Multiply 
the first expression by e~ mm , and the second by e mm , and then add; we find 
2 cos 1Y17T. Q/' (cos 6 + 0 . i) 
, ~ „ / . . „ , . , cos mir . . f f 
7r 
2 cosh (n + u 
o (2 cosh u — 2 cos 6 ) m+ * 
du 
| g—(«—»! + l)lJT I 
J 0 
2 cosh (n + J) u 
(2 cosh u + 2 cos 9) m+ - 
i du 
(116), 
which holds, provided the real part of in lies between and those of n + in + 1, 
in — n are positive. 
If m = 0, we have 
Q„ (cos 0 -J- 0 . 
" cosh + I) u f cosh {n + I) u , 
o (2 cosh ic —2 cosh 6 )* J 0 (2 cosh u + 2 cos 6) h 
(117), 
which holds provided the real part of n is between 0 and — 1. 
It is to be remembered that Q J( (cos 6 + 0 . t) = Q„ (cos 6) — ~ P„ (cos 6). 
