( 344 ) 



thereby at the same time, that the continuous analytic expression 

 cannot be regarded as a single analytic function. 



The same still holds for values of c, not fiillilling one of tiie above 

 inequalities, though the integral is then continuously varying with c. 



So for instance in the case n = 3, taking the stretches a ]> «i ^ a^ 

 in such a manner, that a triangle is possible having these sides, I 

 am led to conclude from the discontinuities of the first derivative 

 that in each of the following intervals 



II a — a^~\- a^'^c'^a^-]-a^ — a V c^a-\- a^ -{-a^ 



III « -j" ^1 — ^2 ^ ^' ^ ^^ ■ — "i 4~ ^'2 

 a distinct analytic function is detined by the integral. 



Some further remarks may be made. On integrating by parts 

 we find 



00 



Wn (c; aa^ . . a„_i) = 1 — « I J^ (ua) J„ [nc) J„ {na^) . , . J„ {uan-\) du 







— rtj j J"i (urtj) Jo [xa) Jq {kc) . . . Jo {uan~\) du 



or what is the same : 



1 = Wn (c; rtOj . . a,,_i) -}- Wn (a; ca^ . . a,-]) -f- Wn (a^; ac . . . a. ,-1) -|- 



Dividing both sides of Ihe equation by n -\- 1 we may interpret 

 the coming relation as follows : n -^ 1 equal or uneqnal stretches 

 being given, if n of them, taken at random, are put together to a 

 broken line, according to the rules of Pearson's problem, the proba- 

 bility is equal to , that the distance between the extremities of 



this broken line is less than the stretch that was left out. 



And from the same equation we deduce in the very particular case : 



c = a = a^ . . . =: rt„— 1 



1 



Wn {a ; a") = — — - , 



or: the rambler of Pearson's problem after walking along 71 equal 



stretches has the chance -—7 to find himself within a stretches' length 



n-\-i 



from his starting point. 



In the most general case of the problem I cannot give a practical 

 solution ; something hov/ever can be done, in the case : n very large, 

 all stretches equal, treated already by Lord Rayleigh. 



