iiKCL'ir iiir.oRy .ixd ornN.iiiox.iL calculus 7.u 



|)ro\i(U<l X is positive, as it is in this case. Now tlu' v.iliic of tlie 

 infmitt.' integral is known; it is \/jr/X. {'onse(|iiently 



,.-X/ 



turthcrniorc, 



•'o vT X»'« v< ^\// 2X^0 /V^ 



Integrating again !>>• parts wc get 



iC^'_J_ £I^J.M CiiILai 



^vt 2x'/V/ 2W0 /vr 



Continuing this proress, we get 



V/ XVTL 2X^(2X0= (2X0' 



1.3.5 . . . (2«-l)" 



/■ 



(82) 



+ ..+(-1)" 



(2X/)" 



(-1)" 1.3.5 ■■ ■ (2»+l) f"" g-^' 

 X 2(2X)" J, f+Wt' 



Now this series is divergent, that is, if we continue out far enough 

 in the series the terms begin to increase in value without limit. On 

 the other hand, if we stop with the nth term the error is rcprcscntcf! 

 by the integral term in (82) and this is less than 



(-1)" 1.3.5 ■.■(2«-l) 



xvr (2x/)-' • ^""^^ 



Consequently the error committed in stopping 'with any term in the 

 series is less than the value of that term. Therefore if we stop with 

 the smallest term in the series, the error is less than the smallest term 

 and decreases with increasing values of t. 

 We can therefore write the solution (77) as 



\ R^\irRt' I2\t (2\tr-(2\l)' i ^^ ' 



The first term, since \ = G/C, is simply y/C/R, the d.c. admittance 

 of the leaky cable. The divergent series shows how the current 

 approaches this final steady value. 



