Convergency of Fourier s Series. 145 



terminate, and is not discontinuous over a finite range of values 

 of the variable between a <xnd b. Thus stated the condition is 

 more general than the one given above, for it does not imply 

 that the function is continuous : it may have any number 

 whatever of discontinuous points, but not of discontinuous seg- 

 ments. The coefficients of Fourier's series are finite and deter- 

 minate, and the ?ith coefficient vanishes when n = co whenever 

 Riemann's condition as to integrability is fulfilled. For, if the 

 function integrated is never infinite, and the sum of the 

 intervals o\ 8 2 . . . S n containing the points of discontinuity can 

 be made infinitely small, the sum of the elements correspond- 

 ing to these intervals in any integral can contribute nothing 

 to the value of that integral. For this sum cannot be greater 

 than the product of the greatest value of the function, which 

 is necessarily a finite quantity, into the sum of the intervals 

 {81 + 82+ ••• + $»)> which is infinitely small. The value of 

 the integral is therefore the same as the value it would have 

 if the function were not discontinuous at the given points. 

 But we have proved that in this case the coefficients are 

 finite and determinate, and that the nth = when n— go . 



36. Now, the conditions which ensure that the coefficients 

 of the series are finite and determinate are also the conditions 

 which ensure that the portions A and C of the integral S w 

 vanish when n = co , for we have only to replace F(v) sinnvin 

 the coefficients b n by %(i#) sin(2n + l)J0 and apply the 

 reasoning of (14). Hence, whenever the coefficients of the 

 series determined by Fourier's method are finite and determi- 

 nate, the value of the series depends only upon the infinitely 

 thin strip 



and therefore the remaining condition to be fulfilled in order 

 that the series may be convergent is that this integral must 

 have a determinate value when n = co . Writing this integral 

 in the form 



FXx)C" sin (2n+ l)i<? 1 f* sin(2«+l)^ fi 



the value of the first term is F(^). Hence, if the series is to 

 be convergent, the second term on the right must vanish or 

 tend to a definite limit. In the former case the series repre- 

 sents the function for the given value of x. In the latter 

 case it does not. 



37. The general conditions under which the second term 

 in the above vanishes, or has a finite limiting value when 



I 



