98 



u" = u,' e^^t -i- e-'^i^'\ \ei^'F{t)dt\ ..... (4) 



o 



In order to determine the integral in the second member we 

 proceed in the following way. We write for it 



JS 







F (§) F (ij) e'5(?+o) dti dl 



Now F{^)F{yi) is only differing from zero if li and § differ very 

 slightly, i.e. there is for short periods a correlation of the forces F. 

 If we introduce ij = | -j- if?, it is allowed to replace ii in the exponent 

 by ^ and to split up the integral into a product of integrals according 

 to § and tfj where we may integrate from — oo to -\- cc. If then 

 we assume 



+ 00 



ƒ- 



(^)F{^\-xp)dxp = (4ö) 



which is a constant characteristic of the problem and if we 

 perform the integration towards i, then (4) is transformed after 

 substitution into 



n'z=H' g-2,3^ + ^ '—. ..... (5) 



™ kT 

 Wlien applying this equation tor i := cc, u^ = — and thus we get 



7/1 



kT 

 m 



In the same way we are able to determine the average square 

 of the distance accomplished. From [2) or by dii-ect integiation from 



(1) we get namely 



t 



u —/<„ = — /?s + iFdt 







^■^s' = (u-u,y + j p\/<[ (6) 







For the last integral we find in a quite analogous way 



If we calculate the first average with the help of (3) and (5) 

 we obtain 



