( 135 ) 



where t represents twice the mean kinetic energy for one degree of 



2 



freedom or - of the mean kinetic energy of the motion of the centre 



o 



of gravity of a molecule. So we find for the mean value of the 

 potential energy of a molecule at a distance r from another molecule: 



* * UpVZ c'ülpF+ï. COS -j> 



£ = — I I r— - j/3 cos* i> -f- 1 cos <p . e -sin <p dip. sin & dd-. 







or 1 ) 



MI 2 |/3 i cos* S+~l . cos ? 



t r* < /»* 



J?= -f - S« i> C?# I cos </> 







m* 



Let us put — — = o and V 3 cos* # -f- 1 = #, then we find by par- 

 o r 3 1 



tial integration : 



* C n I 1 ) 



i? = I sin & dd- \e cx -f e~ cx ( e cx__ e —cx) 



4 J ( c# | 



o 



If we take into account that : 



J 3 ra f 



de 



ƒ /(*) <za = ƒ "{/ t» + ƒ(*-#)] *», 







then we see that instead of taking the above integral between the 

 limits and re we may take twice this integral between and -. 



a 



Now we may introduce x as new variable, writing : 



1 1 x dx 



cos & = — — |/# a — 1, and therefore — sin # d& = — — — , 



V o V o V x* — 1 



making use of the following series : 



1 [c\v 2 2c\v* 3c V ) 



e cx + e -cx ( e cx _ e -cx) — 4 _|_ + .... 



c# ( 6! 5/ 7/ ) 



we find : 



2* T 2 ^ U\v 2 2c\v 4 3cV 1 



" ~ i/3 J i^^T ("37" + ~5/~ + ~TT + • • • -j 



i 



t r* dy jc"y 2c 4 y 3 3c a ?/ 3 j 



" "~ 1/3 J Py^l (37 + ~57~^~ TT + * ' • • 



] ) I am indebted for the reduction of this integral to Prof. Dr. W. Kapteyn 

 from Utrecht, to whom I gladly express my thanks for his kind assistance. 



