Electromagnetic Waves - Introduction to Imaging from Scattered Fields - page 23

It could now be written that

( Ki

2

−− = ε ) () K ′

2

(( K

2

− i ε )

1 2 /

− K ′ )(( Ki

2

− ε )

12 /

+ K ′ )

(2.53)

i ε

KK i

�

ε

�

� -

‚ −

� -

‚ +

K + ′ + K ′

2 K 2 K

which has poles at K ′=±+ ( K ( i ε/2 K )) giving

∞

1
2

1 ( e

iK r ′

− e

− iKr ′

) KdK ′ ′

∫

Gr (, ) 0 =− lim

4 π

2

ir (2.54)

ε

� �

� i

K

-
‚

KK i
K

� ε -

‚ −

ε → 0

K + + ′ + K ′

−∞

2 ′ 2 ′

and then use Cauchy's residue theorem

∑

�

∫

fz z ( d = 2π i Residues

C

in the upper-half plane (uhp) of the complex plane and the

m − 1

1 d

d

� -

‚

Residue == − a

)! lim

( zafz − )

m

()

− 1

( m 1 z

za →

Hence, the

1
0

i ε

�

+ �

-
‚

residue = lim KK ′ − fz ()

! 2 K

KK i
K

ε

′ →+

2

Assuming Jordan's lemma holds, namely, that as R → ∞, ∫ uhp = 0.
Therefore, residue = R 1 + R 2 where

eK
Ki

−

iK r ′

′

R = lim

1

+ ( ε/ 2 KK ) + ′

KK i

ε

′ →+

2 k

e

iK ( + ( i ε /)) ( 2 Kr

K + ( i ε / 2 K ) )

R =− (2.55)

1

2 Ki + ( ε / K )

and as ε → 0, R 1 → − (1/2) e iKr .

∞

eKK

− iK r ′

′ d ′

∫ ∞

The second integral is −

[ ] [ ]

2 2
If we let K ′ = − K ′, then this integral has already been evaluated and so, R 1 = R 2

Ki + ( ε / KKKi ) + ′ + ( ε / KK ) − ′

−

1
2

1 1 1

4

e

iKr

(2.56)

Gr (, ) 0 =− 2 π iR ( += R )

1 2

4 π

2

i r π r