14 Axel S6derblom, 



ce qui donne 



i? 3 ) = 0.1335 0000 i?' 3 (0 = 1.3700 0000 i? 3 (s,) = 4.1203 7037 



41 



5 = 



V0.1335 V4.1203 7037 + 0.1335 + 0.685 

 60 



1.7550 0730 



60 19 

 + 60 



de sorte que 



1 1 .7550 0730 £ 



4.s 3 + 



108 



Calculer la valeur de I'integrale I I = I 2 : 

 On a 



1. 



9* = —% 



9s = 



5 

 108 



1 



log 1 92 

 log <? a 1 g 2 



log C 4 $ 



0.2218 487 — 1 

 0.6197 887 — 3 

 0.8593 662 — 5 



tog </a 



log C 3 ^ 3 



log c 5 I <7a 1 .9s 



log c M | $ | = 0.2402 728 — 6 

 log c 1 g\g 3 = 0.0989 436 — 6 

 log c s , | g 2 \ g\ = 0.4882 835 — 7 

 log c 9 *gl = 0.4065 550 — 8 



0.6655 462 — 2 

 0.9173 582 — 4 

 0.5169 435 — 5 

 0.5870 604 — 6 

 0.6855 640 — 8 

 log c 9A | | (7 3 = 0.6830 168 — 8 



tog c 6>2 $ 

 log c s,i9$ 



Ainsi, on a, a l'aide de la formule (10), 



1 = 1 



c 3 g 3 = 0.0008 2672 

 c 4 gl = 0.0000 7233 8 

 C^gl = 0.0000 0386 42 



c 1 g\g 3 = 0.0000 0125 58 

 c 8A gt = 0.0000 0004 848 

 ^,2$ = 0.0000 0002 550 

 1.0009 0425 2 

 — 0.0042 0164 2 



c 2 <7 2 = — 0.0041 6666 6 

 c. g 2 g 3 = — 0.0000 3288 1 

 c 6tl gl = _ 0.0000 0173 89 

 c 8 ,2#2#3 = — 0.0000 0030 781 

 C9.i#2#3 = — 0.0000 0004 82 

 — 0.0042 0164 19 



l x = I 2 = 0.9967 0261 



