306 
calculated by tables of double entry. In fact, we have the 
following formule, 
*log(1 + cot? sin *®) 4 
0 = =6f (1 — A? sin’) dix () 
oF (k’, 0) — 2r(R) ¥ (’, 0) — {u(k) — F(A) } { F(R’, OY}? 
— 4rF(hk’) — F(R) log (& sin’6) 
*log(l — | — (1 — k*sin"9) sin*p) | 
\; peg (1 — ein) Riese (2) 
we (k’,0) — 2n(k) Y(k’,0) — {u(k) — F(R) } {R(R’, O)}? 
— knF(h’) + log F(k). 
NS log (1 — &? sin?@ sin’) 
0 v (1 —#sin?¢) 
E(k) § F(R, 0) 3? — 2r(k) Y(, A). (3) 
In equation (3) if we put 6 = $7, we will have, recollect- 
ing that 
= $r(h) E(k) — 3 log k’, 
der % 
log (1 — A? sin?) » __ 7 
0 V(l Laie sin’) | dp ae log(k ) F(k). (4) 
Again, 0, being the amplitude of the semi-complete function, 
we have 
pee Pere oa 
sm 70) == rere ray 
and, 
Qk’ Vk 
¥(0,) = $r(@) 2(8) — Flog (=~); 
so that 
*log(cos’p + k’sin?9) Qh! 4/ ke! 
0 (1 — A’sin*¢) a Hog ( lak = F(k). (5) 
The values of the definite integrals (4) and (5) have been 
