im PROFESSOR STOKES, ON THE DISCONTINUITY 



In order to make the left-hand member of this equsrtion agree with (18), we must have 

 c' =— 27, and therefore 



c = - 3, or 3a, or 3/3, 



a, )8 being the imaginary cube roots of - 1, of which o will be supposed equal to 



TT / . IT 



COS — + v — 1 sin — . • 



3 3 



Whichever value of c be taken, the right-hand member of equation (26) will be equal to — 9i 

 and therefore will disappear on taking the difference of any two functions cv corresponding 

 to two different values of c. This difference multiplied by an arbitrary constant will be an 

 integral of (18), and accordingly we shall have for the complete integral 



u = E ["e-'-'i^ + ae-''^)dX + F f e-^'(e'^+ /3e-^^) dX (27) 



Jo Jo 



That this expression is in fact equivalent to (19) might be verified by expanding the 

 exponentials within parentheses, and integrating term by term. 



To find the relations between £1, F and A, B, it will be sufficient to expand as far as 

 the first power of x, and equate the results. We thus get 



A + Bx= f "c-^'{(l +a)E + +li)F+S [(l -a')E + - (^)F1 '^M d\ 



which gives, since 



a»=-i8, /3^ = -«, /V^'d\ = |r(i). /■V^'xrf\ = ir(f), 



*/q Jo 

 A = ir(i){(l+a)E+{l+fi)F},\ 

 S= rQ){0+fi)E+(l+a)F\.f ^ ^ 



19. We have now to find the relations between E, F and C, D, for which purpose we 

 must compare the expressions (25), (27), supposing w indefinitely large. 



In order that the exponentials in (25), may be as large as possible, we must have 9 = — 



in the term multiplied by C, and = in the term multiplied by D. We have therefore 

 for the leading term of « 



Cfe"i^-ie**, when 0c= -; 



Dp-i^, when = 0. 



Let us now seek the leading term of u from the expression (27), taking first the case in 

 which = 0. It is evident that this must arise from the part of the integral which involves 

 e^ or in this case e"^, which is 



{E + F)f''e-'-'->-^d\. 



