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Mr. W. Williams on the 



may, outside the strip B, assign to F(6 + x) any continuous 

 finite values at pleasure. 



16. Since within the strip B the range of integration is 

 infinitely small, we may replace F(0-f#) by F (x) 4 &¥' (x) . 

 We then get, putting \6 for sin \Q, 



« ¥(x) C h sin (2w + 1)40^ F'(«) C h . , 9 .^ ia>M 



S„= -^/ — i(9 3^ + ~^r sin (2n+l) 4050, 



the 



right because 



The value of 



which reduces to the first term on the 



integral of the term involving F'(#) is zero. 



S n when 7i = co is therefore the same as the value it would have 



ifF(6 + x) remained constant throughout and equal to F(#). 



Hence S Q0 =F(^) by (12). 



17. If we change the limits of integration in S» from — it 

 and 7r to — 7r and 0, or and it respectively, we can evaluate 

 the integral exactly as before. For since the portion taken 

 between — it and — h, or between h and 7r, vanishes when 

 n = co y the value of the integral depends only upon the infi- 

 nitely thin strip taken between —h and 0, or and h. Hence, 

 replacing F(0 + a?) in this strip by F(V) + 0F'O), it follows, 

 as before, that the value of the integral is the same as the 

 value it would have if F(0 + #) remained constant throughout 

 and equal to F(#). Hence in this case S<»=JF(aj) by the 

 latter portion of (12). 



From this it follows that in the original integral taken 

 between + 7T, F(0 + #) may change abruptly in value or 

 experience a discontinuity when = 0; for we can break up 

 the integral into two portions at the point = 0, and evaluate 



each portion by the above as if the other were absent. If 

 F(0+#) is discontinuous when = 0, it will have different 

 values at that point according to whether 6 attains the value 

 zero from the negative or from the positive side. 



Thus, let 6 have a small numerical value 8, and let 0A= — 8, 

 OB = 8, A A' = F [x - 8) , BB' = F (x + 8) . Then when 8 vanishes, 



F(x-— 8) becomes 

 F(* + 0) or OB". 



FU-0) or OA", and F(a? + 8) becomes 

 If, then, we evaluate each of the above 

 portions as if the other were absent we get JOA" or ^F(x— 0) 

 for the first portion, and JOB" or JF(# + 0) for the second. 



