An Integral Equation From Diffraction Theory

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An Integral Equation From Diffraction Theory
Arthur S Peters
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(e^-u^) /u2. A2 In order to show thct the inverse of the right hsnd side of (3 '11) csn be ey. Pressed In s form similar to th£t shovm in (2. 24) we v;ill use the fscts thct d r" AdX ^ i/j -g^ _^ u ^-/^ rT~2 /^7T~2. , '72 2, . 2 2, u^-^^ /u -A /i; -A 1/^ -a (^ -u ) md _d [^ AdA 2 /. 2 '' ^0 ^s:!^, ^^:^ "'-^ Then /"2~~2 r°° ^e^^^d^ r"". !^^ ^> r"" AdA d^. /u -a / ^ =/ e ^ -Tj ' -^^ /?:? (e^-u^)'-^" ^-^o /?:fa2-x2 ■ix r° ^ r" e^^-^c^dA r^e^^^ ^ r ^^ AdAd^ 2:j^}m^''\ ^ (jL. )^ (II. ':} . -^;;\ {~i...J-^^) :0 + K-J y;i" ' ;; ^* Cv-"?) s ■::. A r, TT - + 0- ' -J \i^ ■^ r» (-^:i--5) -^o--, ;.
'c- V ^^-"u ^^\i -A-^A V3 u . -. •Ni "^ o 2 a AH^-'-^(Xx)c!A e-s?
TTix r" iUt ^" e 00 iut It follows th£t the solution of (p. 10) is (3. 12) . ( 1^1. •^} lo n= (3. 20) Ic^i^) = cu ^'u^-a^ in (5. 15)' I^ we retein this term we ere led to s consideration of a generalized function as a possible solution of (3. 19). This is so because 5o^^^ cu = c / u -a u y 2 2 •^ u -a -1 + c end inversion gives ^ (t) - ^ ^* p-^^* '0^^^ ~ 2¥ J ^ •J —CO U - 1 ^ yj:^ QU + 2v e-i^^du c_ r"" ue-^^^cu TTi + c5{t) -a / 2 2 i/ a -u (3.


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