1252 
EA 1 1 
a, + Sap=1——+-—=1. 
1 é é 
The second equation may be obtained as follows. 
From the differential equation 
d 
re [zet pp (OF pe O(a) ON ee ee eN 
we may conclude 
1 
1 
1 
Wc Pp’ (2) de = — =| % (x) d [ze—* pp (2)] = 
if 
0 
0 
it 1 1 . 
=F Leet ay (ig fer oe) de 
0 
50 
] 
& ; 1 ! 
fee | (x) — ee f ge )| da —— pe Pp (1) Pp (1) == a, Lp (1). 
. r § 
Now,. the equations (4) and (5) give 
Py (ls Pp (x) [p,' (w) — y 'p -1(*)| 
wv Ss 
PO = Hp!) [Gp (2) — Fy OI 
hence 
a 
pp" (©) — 5 TE) pp (2) pj (2) — py @) gy ne (x) 
and 
n 
© 2 
= z (4) — 5 q ik (x) | = g, (©) Gy (@) — Gn (@) 4 intl (x). 
This shows that 
1 
|e [p, (2) p,' (@) — pn (©) Pr (2)] de = & ap) (1) 
0 a 
where 
1 1 
a i! 
fot! ford lt 
e 
0 0 
To obtain the second integral, the value of » being very large, 
we observe that according to equation 
Pole) = pn (2) — Pari le)e rr ers ee 
the functions 
