A Problem in Conduction of Heat. 163 



fig. 1, and we take the expressions 



e -ha.H e ia(x-x') g a ^ 



Z7TJ 



and 



'IttJ 



according- as x x , 



The path (P) must have the argument of a. at infinity on 



the right hand between and — , and on the left between 



— and 7r, so that e~ kaH may vanish there : and for the same 



reason we have chosen the different expressions for the 



integral as x x' . 



Associate with this solution 



Ae 1 "* e~ ka2t dcx, 



fr 



over the same path (P). 



The quantity A is determined by the surface condition (3), 

 and is given bv 



A=s _*±«.«~ ..... (5) 



h — la 



We have thus obtained a solution of (1) and (3) in the form 

 of the sum of two complex integrals, and we proceed to show 

 that this solution corresponds to a source at (<</) in our solid. 

 Our solution is given by the equation 



>'= ./ { [*-***■€-*&-*> da— ^e- ka2t e^ x+ ^ J ^^da\, (6) 



the integrals being taken over the path (P), and the first 



integral being altered when x>x\ as explained above. 



This may be written 



if if ih f e i<x+x') 



e -ka2t e -;*u-s) da + _L \ e -ican e ia(x+z') da _ <_ e -^t — -—da. (7) 





 The a-plane. 



Since there is no pole of any of these integrals within the 

 closed circuit of fig. 2, and the parts contributed by the 



31 2 



