208 



Proceedings of the Royal Irish Academy. 



First, suppose that :c is not zero. By Fourier's theorem the portion clue 

 to the term involving e''^ is simply 



TT ! .^ (7; - £) ^ </) (« - £) i , (4) 



£ being an indefinitely small positive quantity, intended to cover the case of 

 possible discontinuity. In the other portion suppose the fraction 



6-(A) - iSiX) 



C{X) + iS{\) ^^^ 



tends to a finite limit as A increases indefinitely ; this will certainly be the 

 case if the coefficients in C, S are real, and usually even if they are complex. 

 If the fraction (5) were replaced by this limit, the portion considered would 

 be zero, also by Fourier's theorem. And it is readily seen that the error due 

 to this approximation diminishes indefinitely as \ increases indefinitely. 

 For in 



J_, \C{X)+iS{X) ' 



(p (?/) die 



(6) 



(7(oo ) + ^S'(cc ) 



it is legitimate to interchange the order of integration, since the integral to 

 infinity with respect to u is uniformly convergent, owing to satisfying 

 Dirichlet's conditions, and to J" (p{u) du being convergent. And, on doing 

 so, the integral in X is at most of order h~'' (unless u = x = 0, when it is 

 finite). 



Next, suppose that x is zero. The portion due to the term involving 

 c'*^ is now v(p{e), and that due to the term involving e'^^ is 



G(oo)-iS {00) 



BO that the whole is 



C{cc ) + iS(co ) 

 2C(oo) 



TT,p(i), 



Tr.(p{e). 



(7) 



(8) 



C(x ) + iS(oo ) 



Now, suppose that all the values of A for which C(\) + iS{\) is zero 

 have negative imaginary parts. The contour in (3) may therefore be 

 deformed into a straight line, the axis of real quantities, and, dividing 

 across by 2, we thus have the equation 



Lt. 



dX 



C (A) cos A:<; + S{X) sin Xx 

 C{X)+iS{X) ' 



e'^" (p (u) du 



= '^{<}i(cc-e) + <t>{:c + e)}, x>0, 



or, 



C(oo 



6'(3c ) + hS{cc 

 Kubiect to the conditions stated. 



'P(e), X = 0, 



(9) 



