130_notes.dvi

the rest of the contour to give zero. That’s easy to do since the integrand goes to zero at infinity on
the real axis and the exponentials go to zero either at positive or negative infinity for the imaginary
part ofk. Examine the integral

∮ k

(k+iμ)(k−iμ)e

ikrdkaround a contour in the upper half plane as

shown below.

Real Axis

Imaginary

ιμ

−ιμ

pole

The pole inside the contour is atk=iμ. The residue at the pole is
[
keikr
k+iμ

]

k=iμ

=

iμe−μr

iμ+iμ

=

1

2

e−μr.

The integrand goes to zero exponentially on the semicircle at infinity so only the real axis contributes
to the integral along the contour. The integral along the real axiscan be manipulated to do the
whole problem for us.

∮

k

(k+iμ)(k−iμ)

eikrdk = 2πi

[

keikr

k+iμ

]

k=iμ

∫∞

−∞

k

(k+iμ)(k−iμ)

eikrdk = 2πi

iμe−μr

iμ+iμ

∫^0

−∞

k

(k+iμ)(k−iμ)

eikrdk+

∫∞

0

k

(k+iμ)(k−iμ)

eikrdk = 2πi

1

2

e−μr

k′ = −k

∫^0

∞

−k′

(−k′+iμ)(−k′−iμ)

e−ik

′r

(−dk′) +

∫∞

0

k

(k+iμ)(k−iμ)

eikrdk = πie−μr

−

∫∞

0

k′

k′^2 +μ^2

e−ik

′r

dk′+

∫∞

0

k

k^2 +μ^2

eikrdk = πie−μr