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junction x ■ A, the left hand portion of the string now being regarded 

 as extending to x ■ - °°. The integral terms repiese"* interactions 

 between the ends x ■ 0, x ■ £, giving rise to a sequence of reflections 

 and transmissions. Methods have been devised for dealing approximately 

 with these interactions both when they are weak end when they are strong, 

 as in resonance situations (see Noble [5 [Ch.5]). 



Here it is of course possible to solve the coupled integral equa- 

 tions completely. First, however, we show how they may be uncoupled 

 when the kernel K(s) is even in s, as it is here. It is then possible 

 to work in a strip D which is completely symmetrical about the real axis, 

 so that Y_(-s) is a@function with the same domain R of anulyticity 

 as Z (s), while Z (-s) is aQ function. It also follows from Sec- 

 tion 10 that K (-s) - K_(s) for sED. 



Consider then (7.29) and (7.30) fcr eeD, and choose the path 

 — <""» — above s to be a straight line from -» +ia to +» +ia, a > 0. 

 Choose ~—^j — to be the image of this path in the real axis, going 

 from -» -ia to +«-ia. In (7.29) charge s to -s; the integration path 

 can be chosen to run above both s and -s, and so 



„ , % r.(-s) i f e itl Z+(t) 



<L<-> - Kjcsr + 2« J rrtntfT) dt " ° (7 - 31) 



In (7.30) change the integration variable from t to -t; (-t) runs 

 from +ia +»to +ia -«as t runs from -ia —= to -ia +» , so that the 

 path for (-t) is the same as that in (7.31). This gives 



Z+(s) i r e ltA Y.(-t) dt 

 -V a > + OsT " 2* j K.(t)(t+ n) " ° • < 7 ' 32 > 



44 



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