62 



Proceedings of the Royal Irish Academy. 



For 



I p = Lt _ 



Lt 1 



-I 



+1 



-ILJ 



ts+ieK(t)dt 



x S+i6 



yi-s-ie^^dy 



d,e 



\8+i9 



«i * 2 . . . tf) K^) . . . K{t p ) dt, ...dt p 



Lt 1 



-I 



d% 



" dy . , 

 o y 



*5+t8 I y-l-5-ieip(y)dy 



1 ft,t 2 ... t p x\s+ie 



dd 



y 



Kit,) . . . K[t p )dh ...dt p 



The transformation 

 gives us 



y = t y t z . . . t p xer 



I p = Lt ±- 



|-^oo lit 



-I 



de\ ... 



<»i . . . dt p Kit,) . . . K(t p ) 



eT{s+io)ip(t 1 t i ...t p xe-r)dT. 



log 



t\tz . . .tpZ 



We may prove as before that the function e TS \p (^ 2 . . . tpXe- T ) remains 

 absolutely integrable as r approaches oo. Integrating first with regard to 

 we get 



/„ 



Ltlf 



f.-J 



dt, . . . dtpKit,) . . . K{t p ) 



1"K 



SUIT? 

 t\tj...tpX T 



c$T\p itiU . . . tpXe-r)dT. 



There is here a difficulty concerning the neighbourhood of the values 

 ti = 0, t t = 0, . . . t n = 0, which give - oo in the lower limit, because the 

 function e T *\p i^U. . . . t p xe-r) is not absolutely integrable in this direction. 

 Taking b a constant negative quantity we see that 



C"8i/, ifit t . . . tjpse-^dr 



Tt If f ' f+«> sin tE 



j= I - J o • • • J o * ■ ■ ■ J** K ^) • ■ • K <tp)\ b —r 



is equal to 



| o . . . f -K^i) ■ • . ff &) ^(t£ . . . ipX^A . . .dt p = ^(.r). 



Putting tttt . . . tpj ; = p, we have now to consider the integral 



T \ b sin t£ , , , . , 

 J=\ erm P e-r)dr. 



.1 log " T 



Put X = 'y, and suppose - (s - 1)X > b > - sX, and - q\ > log - > - iq + 1)A, 



