living Prohlems in the Conduction of Heat. 



605 



argument of « on the right must lie between and \tt, and 

 on the left between \ir and it. We take the path (P) of 

 fig. 1 as the standard path in the problems which follow. 



Since every element of the integral satisfies (1), the value 

 of v in (5) satisfies (1). 



Also when .i'=0, we have ?; = 0. 



And when x = a. v= -A \ da, over the path (P). (6) 



ITT 1 ex. 



Since the integrand is an odd function of ct, if we form 

 the path (Q) by taking the image of the path (P) in the real 



Fiff. 2. 



The path (Q) in the «-plane. 



axis of a, joining the ends of the path (P) and its image by 

 arcs of a circle, centre at the origin, whose radius tends to 

 infinity, we have from (6) 



v C e ~ Ka2i j 



over the path (Q), 



since the integrals over the circular arcs vanish in the limit. 

 Therefore when # = a, we have v = v . 

 Finally, when t = 0, we have 



v Csmccxda , , 



v= — 1 —. , over the path (P ). 



J 



ur i sin ua a 



Now the integrand has no infinities above the path (P), 

 and if we complete the circuit by the arc of a circle, dotted 

 in fig. 3, whose centre lies at the origin and radius tends to 

 infinity, the integral over the complete path of fig. 3 is zero. 



