8 KAEL PEAKSON 



Hence putting 2 Jy = z, 



.8 



} (v) 



\J a {z)\ -e jl 64 . 57( ,z+ 192 25g 



(50n-57)n z 10 _ (1892 -2l25n + 270n*)n z 12 __ 

 + 1800 1024 103,680 ~ 4096 ~~ e C ' 



1 2 



= e~ taz {l — a 4 z 4 — a 6 z 6 — a s z a — a 10 z 10 — . a 12 .z 12 — etc.}, 



let us write, for brevity. The a's are then known coefficients. 



Now by (iii) 



1 [* 

 $n (r 2 ) = o- M ^> (w) Ro (mZ)}" <^- 



Z7TJ o 



But we know that* : 



I" ue-* nuH2 J (ur)du = -^ i e- r2/nP (vi). 



Write : ^nP = 0-* (vii). 



Thus: \^ ue^ 1 ^ J {ur)du = \e-^ l,r2 (viii). 



Jo c 



Differentiate (viii) s times with regard to cr 2 : 

 Hence, if /3— — 2o- 2 /r 2 , we have: 



L 



_ j 2« + l -iw 2 <r 2 7- / \ 7 



2s — \ u e 2 J (ur) du 

 Jo 



2 25+1 d s /l 



f)2S + l 



= — ^+2 l as ' sa y- 



gl# 



r 2 ' 

 We have therefore : 



N_ o - \r^ M N *=^ l 2s 2* s+1 a 2s i 



2 



d s 1 1 

 where it remains to evaluate i w = -^ ( 5 e ltf 





We find 



1 /r\ 6 -r2 /2(r 2 / r 2 1 r 4 



^-§W 8 ' l 2 - 2 ? + 4,- 



*~ 16 W \ <r 2 + 4o- 4 8^/' 



1 /r\ 10 _ r 2/2n-2 / r 2 r 4 r 6 1 r s 



i e = -— (-) e r/2<r (24-48- 2 +18 --2^- + — ^- 



8 32 \cr/ \ cr 2 cr 4 o- 6 16o- ! 



* Gray and Mathews, loc. cit. Eqn. (162), p, 78 



