52 4. PROBLEMS IN ELECTROSTATICS AND ELECTRODYNAMICS
from which
a
.
x;
/
D
ec
5
2
2
c
3
1; 0;
c
5
c
1
2
Z
s
1
ds
1
g
3=2
.
s
/
Z
1
1
ds
ı
.
g
.
s
//
g
1=2
.
s
/
.
t s
/
:
Using the well-known form [3]
Z
dx
.Cx
2
C Bx C A/
3=2
D
2
.
2C s C B
/
q.C x
2
C Bx C A/
1=2
;
where
q D 4AC B
2
we notice from (4.4) that
s
D
B
p
B
2
4AC
2C
D
B
p
q
2C
)
p
q D 2C s
C B
and so
1
2
Z
s
1
ds
1
g
3=2
.
s
/
D
2C s
C B
qg
1=2
.
s
/
2C s C B
qg
1=2
.
s
/
ˇ
ˇ
ˇ
ˇ
1
D
p
q
qg
1=2
.
s
/
C
2
p
C
.
2C s
C B
/
2
D
1
p
qg
1=2
.
s
/
C
1
2
p
1
2
R
2
c
2
.
1
2
/
C
2
.
t
/
2
: (4.5)
e second term is
Z
1
1
ds
ı
.
g
.
s
//
g
1=2
.
s
/
.
t
s
/
and using the identity
Z
ds f
.
s
/
ı
.
g
.
s
//
D
f
.
s
/
j
g
0
.
s
/
j
ˇ
ˇ
ˇ
ˇ
s
Dg
1
.
0
/
we can evaluate
Z
1
1
ds
ı
.
g
.
s
//
g
1=2
.
s
/
.
t s
/
D
.
t s
/
j
g
0
.
s
/
j
g
1=2
.
s
/
D
1
j
g
0
.
s
/
j
g
1=2
.
s
/
:
Since
g
0
.
s
/
D
C s
2
C Bs
C A
0
D 2C s
C B D
p
q
we see that this term cancels the singularity in the first term, leaving
1
2
Z
s
1
ds
1
g
3=2
.
s
/
Z
1
1
ds
ı
.
g
.
s
//
g
1=2
.
s
/
.
t s
/
D
1
2
p
1
2
R
2
c
2
.
1
2
/
C
2
.
t
/
2
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