The Solution of Certain Integral Equations With Kernels K Z Zeta Z Zet
The Solution of Certain Integral Equations With Kernels K Z Zeta Z Zet
Arthur S Peters
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16) of (2. 1) can be obtained more easily without the introduction of a Hilbert-Riemann problem if we use the following technique which may be new. Introduce (2. 17) F(w) = (j} 4>(z), QZ Z - W and hence (2. 13) f"*"(0 = 7ri(0 + ^ -^ dz . If (^(z) is to satisfy (2. 1) v/e must have (2. 19) F^iO = [7ri + h(O]0(i;) + f(?) or (2. 20) HI) - Sferrf^ • The substitution of this in (2. 1) gives *«' ^ «^> In accordance with our assumption the zeros of h(l, ) + ni lie in a domain D- which, with its counte...rclockwise orientated boundary C, , is contained in D . In the integral involving F (z) in (2. 21) let C be contracted into C . Since F(v/) is analytic in D, equation (2. 21) gives h(^)M?) + f(0 = Pr^- I „. , J^L^^, . + / ^^'■^^'' h(U + 7ri r [h{z) + ^ij(z-U f [h(z) + 7ri](z - ^ {'•, :- 10 or, using (2. 19), r [h(z) + 7ri, (z - i) c ^1 from which we find ■ty\ - ^(g>^('^) 1 I f(z) dz '*'^"' " h^O + TT^ ' [h(i>)-7ri: J [h(z) + ^ij(z - O (2. 22) [h(U-7ri] f ^ - ^ F(z) dz [h(U - TTi] r [h(z) + 7rij(z - U ^1 + + If h(z) + TTi is analytic in D + C with simple zeros in E at a, k = 1, 2, .
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