3.3 Solutions 243
and minima are smeared out, just as in the case of optical diffraction from a
diffuse boundary of objects characterized by a slow varying refractive index.
3.117 f(θ)=−(2μ/q^2 )
∫∞
0 V(r)sin(qr)rdr
Integrate by parts
∫∞0V(r)sin(qr)rdr=V(r)[
1
q^2sin qr−r
qcos qr]∞
0−∫∞
0dV
dr(
1
q^2sinqr−r
qcosqr)
drThe first term on the right hand side vanishes at both limits becauseV(∞)=
0, Therefore:
∫∞0V(r)sin(qr)rdr=−1
q^2∫∞
0dV
drsinqrdr+1
q∫∞
0dV
drrcosqrdrEvaluate the second integral by parts
1
q∫∞
0dV
drrcos(qr)dr=1
q[
dV
dr(
r
qsinqr+cosqr
q^2)]∞
0−1
q∫∞
0(
r
qsinqr+cosqr
q^2)
d^2 V
dr^2drNow the term(
1
q^2)
r(dV
dr)
sinqr∣
∣∞
0 vanishes at both the limits because it is
expected that (dV/dr)r=∞=0.
Integrating by parts again
(
1
q^3)∫
cosqrd^2 V
dr^2dr=(
1
q^3)
cosqrdV
dr∣
∣
∣
∣
∞0+
(
1
q^2)∫∞
0(
dV
dr)
sinqrdr
(
1
q)∫∞
0(
dV
dr)
rcosqrdr=(
1
q^3)(
dV
dr)
cosqr∣
∣
∣
∣
∞0−(
1
q^2)∫∞
0(
d^2 V
dr^2)
rsinqrdr−(
1
q^3)(
dV
dr)
cosqr∣∣
∣
∣
∞0−
(
1
q^2)∫∞
0(
dV
dr)
sinqrdr.The first and third terms on the right hand side get cancelled
∫∞0V(r)sin(qr)rdr=−(
1
q^2)∫ (
d^2 V
dr^2+
2
rdV
dr)
sin(qr)rdr.Now for spherically symmetric potential∇^2 V=d^2 V
dr^2+
(
2
r)
dV
dr.
Furthermore by Poisson’s equation: