B30 



and he calculates the value of the first integral for the case 

 a = 4tw -j- 1, the value of the second integral for the case 

 a =: 4:tu — 1 . 



In the following I give a shorter deduction for these results. 



I suppose that the two positive, otherwise arbitrary numbers ,? 

 and 7 have a for product, that / is a positive parameter and now 

 consider the integral 



-1- 00 J, 



( -~) ■ 



J 



1= \ f^e ^ 'e^^^"^d,v . 



In order to calculate (his integral, it is not necessar\ , as Stieltjks 

 does, to fall back on an integral Ibrmida treated hy Lkgkndre and 

 by Abkl. It need only be obsei-vcd that in the upper half of the 

 complex ,r-|)laiie for increasing values of r^ the modulus of the 

 integrand approaches sufficiently rapidh' to zero, to permit us to 

 equate the integral / to the sum of the residues in this upper half 

 plane, multiplied bv 2.~i 



The poles ot the integrand are the i)oles ot /^e '^ ' , that is to 

 say the points .v = — {k = 0, 1, '2, . . .), where k is prime to d. The 

 residue of such a i)ole is 



« •/ 2 I ] e '•■ z= (- // - ' Va \ -\e / 



2jr7 h^\ 



hence 



/ 



{h\ ""^ 1 r^\ fk\ 



We ought to distinguish now between the tw^o cases a = ^w -\- 1 

 and a = 4?(; — 1 . • 



For a = 4?f -|- 1 we have 



I — J = + I I and conseL;uently f \e 'M = — / I « ^ j ^ 



so that it follows from the result found for /, that 



\f\f^)sinl3ttxdx — \^-f\^^\. {a — A:XO-\-\) . . (I) 







On the other hand for a = 4iv — 1 

 I — I = — f — J and consequently fie ^ )::=-{- f I e P], 



