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Therefore,
exp
m
t
)
2
(1 + tan
2
(
y
))
.
1
√
2
πσ
t
1
2
σ
t
f
Y |t
(
y
)=
−
(tan(
y
)
−
(3.49)
π
But from
f
E|t
(
e
)=
f
Y |t
(
t
−
e
), follows that
f
E|−
1
(
e
)=
f
Y |−
1
(
−
2
−
e
)
∈
−
π,
0[
]
and
f
E|
1
(
e
)=
f
Y |
1
(
2
− e
)
∈
]0
,π
[.Thus:
exp
2
− e
)
− m
t
)
2
(1 + tan
2
(
t
π
2
σ
t
(tan(
t
π
1
√
2
πσ
t
1
f
E|t
(
e
)=
−
2
− e
)) =
(3.50)
exp
m
t
)
2
(1 + cot
2
(
e
))
.
1
√
2
πσ
t
1
2
σ
t
=
−
(cot(
e
)
−
(3.51)
Since the
f
E|t
(
e
) are disjoint,
V
R
2
(
E
)=
p
1
V
R
2
(
E
1) +
p
2
|
−
1
V
R
2
(
E
|−
1),with
1) =
π
0
1) =
0
−π
f
E|
1
(
e
)
de
and
V
R
2
(
E
f
E|−
1
(
e
)
de .
V
R
2
(
E
|
|−
(3.52)
Let us perform the following change of variable:
z
=cot(
e
)
−
m
t
.Wethen
dz
1+cot
2
(
e
)
. Taking care of the limits of the
integrals the potential can be expressed as
have cot
(
e
)
de
=
dz
,or
de
=
−
−∞
e
−z
2
/σ
t
(1 + cot
2
(
e
))
2
1+cot
2
(
e
)
1
2
πσ
t
V
R
2
(
E
|
t
)=
(
−
dz
)=
(3.53)
+
∞
+
∞
1
2
πσ
t
e
−z
2
/σ
t
(1 + (
z
+
m
t
)
2
)
dz .
=
(3.54)
−∞
√
π
z
e
−t
2
dt
(erf(
2
On the other hand, with erf(
z
)=
−∞
)=
−
1,erf(+
∞
)=
0
1),wehave:
+
∞
erf
z
σ
+
∞
1
2
πσ
2
1
4
√
πσ
1
2
√
πσ
e
−z
2
/σ
2
dz
=
=
;
(3.55)
−∞
−∞
2
πσ
2
+
∞
−∞
1
e
−z
2
/σ
2
(
z
+
m
)
2
dz
=
e
−z
2
/σ
2
+
σ
3
√
π
4
erf
z
σ
σ
2
z
2
1
2
πσ
2
mσ
2
e
−z
2
/σ
2
+
=
−
−
m
2
√
πσ
2
+
∞
σ
3
√
π
2
+
m
2
√
πσ
=
erf
z
σ
m
2
2
√
πσ
1
2
πσ
2
σ
4
√
π
+
+
=
.
−∞
(3.56)
