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15-1 - WKB approximation Alpha particle decay
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1
WKB approximation: particle decay
Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton
(Date: November 15, 2011)
WKB approximation
This method is named after physicists Wentzel, Kramers, and Brillouin, who all
developed it in 1926. In 1923, mathematician Harold Jeffreys had developed a general
method of approximating solutions to linear, second-order differential equations, which
includes the Schrödinger equation. But even though the Schrödinger equation was
developed two years later, Wentzel, Kramers, and Brillouin were apparently unaware of
this earlier work, so Jeffreys is often neglected credit. Early texts in quantum mechanics
contain any number of combinations of their initials, including WBK, BWK, WKBJ,
JWKB and BWKJ. The important contribution of Jeffreys, Wentzel, Kramers and
Brillouin to the method was the inclusion of the treatment of turning points, connecting
the evanescent and oscillatory solutions at either side of the turning point. For example,
this may occur in the Schrödinger equation, due to a potential energy hill.
(from http://en.wikipedia.org/wiki/WKB_approximation)
________________________________________________________________________
Gregor Wentzel (February 17, 1898 – August 12, 1978) was a German physicist known
for development of quantum mechanics. Wentzel, Hendrik Kramers, and Léon Brillouin
developed the Wentzel–Kramers–Brillouin approximation in 1926. In his early years, he
contributed to X-ray spectroscopy, but then broadened out to make contributions to
quantum mechanics, quantum electrodynamics, and meson theory.
http://en.wikipedia.org/wiki/Gregor_Wentzel
Hendrik Anthony (2 February 1894 – 24 April 1952) was a Dutch physicist who worked
with Niels Bohr to understand how electromagnetic waves interact with matter.
http://en.wikipedia.org/wiki/Hans_Kramers
Léon Nicolas Brillouin (August 7, 1889 – October 4, 1969) was a French physicist. He
made contributions to quantum mechanics, radio wave propagation in the atmosphere,
solid state physics, and information theory.
http://en.wikipedia.org/wiki/L%C3%A9on_Brillouin
________________________________________________________________________
1. Classical limit
Change in the wavelength over the distance x
x
dx
d .
When x
dx
d
.
2
In the classical domain,
dx
d
or 1
dx
d
,
which is the criterion of the classical behavior.
2. WKB approximation
The quantum wavelength does not change appreciably over the distance of one
wavelength. We start with the de Broglie wave length given by
h
p ,
)(
2
1 2 xVp
m
,
or
)]([2
2
2 xVm
h
p
,
or
)((2 xVmp .
Then we get
]
)(
[22 32
dx
xdV
m
dx
d
h
,
or
dx
xdV
p
mh
dx
xdV
p
h
h
m
dx
xdV
h
m
dx
d )()()(
3
3
2
3
2
.
When 1
dx
d
, we have
1
)(
3
dx
xdV
p
mh
(classical approximation)
3
If dV/dx is small, the momentum is large, or both, the above inequality is likely to be
satisfied. Around the turning point, p(x) = 0. |dV/dx| is very small when V(x) is a slowly
changing function of x. Now we consider the WKB approximation,
)()(
2
)(
2
22
xxV
xm
x
.
When 0V ,
ipx
ikx AeAex )( .
If the potential V is slowly varying function of x, we can assume that
)(
)(
xS
i
Aex ,
.....)(
!3
)(
!2
)(
!1
)()( 3
3
2
2
10 xSxSxSxSxS
.
____________________________________________________________________
((Mathematica))
4
WKB approximation
eq1
—2
2 m
Dx, x, 2 Vx x ∂ x;
rule1 Exp
—
S & ;
eq2 eq1 . rule1 Simplify
Sx
— 2 m ∂ 2 m Vx Sx2 — Sx
2 m
rule2
S
S0 — S1 —
2
2
S2 —
3
3
S3 —
4
4
S4 & ;
eq3 2 ∂ m 2 m Vx Sx2 — Sx;
eq4 eq3 . rule2 Expand;
list1 Tablen, Coefficienteq4, —, n, n, 0, 6
Simplify;
TableForm
0 2 m ∂ 2 m Vx S0x2
1 2 S0x S1x S0x
2 S1x2 S0x S2x S1x
3 S1x S2x 1
3
S0x S3x 1
2
S2x
4 1
12
3 S2x2 4 S1x S3x S0x S4x 2 S3x
5 1
24
4 S2x S3x 2 S1x S4x S4x
6 1
72
2 S3x2 3 S2x S4x
_______________________________________________________________________
For each power of ħ, we have
0)]('[)(22 20 xSxmVm ,
5
)(")(')('2 010 xiSxSxS ,
)(")(')(')]('[ 120
2
1 xiSxSxSxS ,
.....................................................................................................................................
(a) Derivation of )(0 xS
)()]([2)]('[ 220 xpxVmxS ,
where
)]([2)(2 xVmxp ,
or
)()('0 xpxS ,
or
x
x
dxxpxS
0
)()(0 .
Since )()( xkxp ,
x
x
dxxkxS
0
)()(0 .
(b) Derivation of )(1 xS
)(")(')('2 010 xiSxSxS ,
)('
)('
2)('2
)("
)('
0
0
0
0
1 xS
xS
dx
d
i
xS
xiS
xS ,
which is independent of sign.
)](ln[
2
)]('ln[
2
)(')( 011 xk
i
xS
i
dxxSxS ,
or
6
2/1
1 )](ln[)](ln[2
1
)( xkxkxiS ,
or
)(
1)(1
xk
e xiS
.
(c) Derivation of )(2 xS
)(")(')(')]('[ 120
2
1 xiSxSxSxS ,
)('
)]('[)("
)('
0
2
11
2 xS
xSxiS
xS
.
Then the WKB solution is given by
...)](ln[
2
)(
.....)(
!3
)(
!2
)(
!1
)()(
0
3
3
2
2
10
xkidxxk
xSxSxSxSxS
x
x
The wave function has the form
)]})(exp["])(exp['))]{(ln(
2
1
exp[)(
00
x
x
x
x
dxxkiBdxxkiAxkx ,
or
])(exp[
)(
])(exp[
)(
])(exp[
)(
'
])(exp[
)(
'
)(
00
00
x
x
x
x
x
x
x
x
dxxki
xk
B
dxxki
xk
A
dxxki
xk
B
dxxki
xk
A
x
.
where we put
'A
A ,
'B
B .
3. The probability current density
7
We now consider the case of B = 0.
])(exp[
)(
)(
0
x
x
dxxki
xk
A
x .
The probability is obtained as
mv
A
xk
A
xxxP
22
)(
)()(*)( ,
where mvxk )( .
The probability current density is
2
2
2
A
mmv
A
vvJ
.
Fig.
2avdtJadt , or 2vJ .
4. WKB approximation near the turning points
We consider the potential energy V(x) and the energy shown in the following figure.
The inadequacy of the WKB approximation near the turning point is evident, since
0)( xk implies an unphysical divergence of )(x .
(a) V(x): increasing function of x around the turning point x = a
8
Vx
x
aO
E
(i) For x>>a where V(x)>,
])(exp[
)(
])(exp[
)(
)( 11
x
a
x
a
I dxx
x
B
dxx
x
A
x
,
where A1 and B1 are constants, and
)(21)( xVmx
,
(ii) For x<a where V(x)<,
])(sin[
)(
])(cos[
)(
)( 22
a
x
a
x
II dxxk
xk
B
dxxk
xk
A
x ,
where
)(2
1
)( xVmxk
.
________________________________________________________________________
(b) V(x): decreasing function of x around the turning point
9
Vx
x
bO
E
(i) For x<<b where V(x)>,
))(exp(
)(
))(exp(
)(
)(
b
x
b
x
I dxx
x
B
dxx
x
A
x
,
with
)(21)( xVmx
,
(ii) For x>b where V(x)<,
)))(sin(
)(
))(cos(
)(
2
)(
x
b
x
b
II dxxk
xk
B
dxxk
xk
A
x
where
)(2
1
)( xVmxk
.
_______________________________________________________________________
5. Exact solution of wave function around the turning point x= a
10
Vx
x
aO
E
The Schrödinger equation is given by
)()(
2 2
22
xxV
dx
d
m
,
or
0])([
2 2
22
xV
dx
d
m
,
where is the energy of a particle with a mass m. We assume that
)()( axgExV ,
in the vicinity of x =a, where g>0. Then the Schrödinger equation isexpressed by
0)(
2
22
2
axgm
dx
d
.
Here we put
)(
2
3/1
2
ax
mg
z
.
((Note))
dz
dmg
dz
d
dx
dz
dx
d
3/1
2
2
,
11
2
23/2
22
2 2
dz
dmg
dx
d
.
Then we get
0)(
)(
2
2
zz
dz
zd .
The solution of this equation is given by
)()(2)( 21 zBCzACz ii ,
where we use 2C1 instead of C1. The asymptotic form of the Airy function Ai(z) for large
|z| is given by
)
4
cos(||)( 4/12/1
zzAi , for z<0
and
ezzAi
4/12/1
2
1
)( , for z>0
where
2/3||
3
2
z
12
-10 -8 -6 -4 -2 2 4
z
0.5
1.0
Aiz
Fig. Plot of the Ai(z) (red) and its asymptotic form (blue) as a function of z for z<0.
The asymptotic form of the Airy function Bi(z) for large |z|,
)
4
sin(||)( 4/12/1
zzBi , for z<0
ezzBi
4/12/1)( , for z>0
with
2/3||
3
2
z
13
-10 -8 -6 -4 -2 2
z
-0.5
0.5
1.0
1.5
2.0
Biz
Fig. Plot of the Bi(z) (red) and its asymptotic form (blue) as a function of z for z<0.
Here we note that
For z<0,
2/1
3/1
22
||
2
)(
2
)( z
mg
xag
m
xk
.
Then we have
2/3
2/3
2/1
2
2/1
2
||
3
2
)(
2
3
2
2
)(
z
xa
mg
dxxa
mg
dxxk
a
x
a
x
For z>0,
2/1
3/1
22
||
2
)(
2
)( z
mg
axg
m
x
,
we have
14
2/3
2/3
2/1
2
2/1
2
||
3
2
)(
2
3
2
2
)(
z
ax
mg
dxax
mg
dxx
x
a
x
a
______________________________________________________________________
6. Connection formula (I; upward)
Vx
x
aO
E
15
Fig. Connection formula (I; upward). The condition 1
dx
d
for the WKB
approximation is not satisfied in the vicinity of x = a. We need the asymptotic
form the exact solution of the Schrödinger equation.
(i) Asymptotic form for z<0 (x<a)
The asymptotic form of the wave function for z<0 (x<a) can be expressed by
)]
4
)(sin(
)(
1
)
4
)(cos(
)(
1
2[
2
)
4
sin(||)
4
cos(||2)()(2
2
1
6/1
2
2/1
4/12/1
2
4/12/1
121
a
x
a
x
ii
dxxk
xk
C
dxxk
xk
C
mg
zCzCzBCzAC
(1)
where
2/3||
3
2
)( zdxxk
a
x
, 2/1
3/1
2
||
2
)( z
mg
xk
.
16
(ii) The asymptotic form for z>0;
The asymptotic form of the wave function for z>0 (x>a) can be expressed by
)])(exp(
)(
1
))(exp(
)(
1
[
2
)()(2
2
1
6/1
2
2/1
4/12/1
2
4/12/1
121
x
a
x
a
ii
dxx
x
C
dxx
x
C
mg
ezCezCzBCzAC
(2)
where
2/3||
3
2
)( zdxx
x
a
, 2/1
3/1
2
||
2
)( z
mg
x
.
From Eqs.(1) and (2) we have the connection rule (I; upward) as follows.
])(exp[
)(
])(exp[
)(
]
4
)(sin[
)(
]
4
)(cos[
)(
2
x
a
x
a
a
x
a
x
dxx
x
B
dxx
x
A
dxxk
xk
B
dxxk
xk
A
(I; upward)
at the boundary of x = a.
Vx
x
aO
E
where C1 = A and C2 = B.
17
______________________________________________________________________
7. Exact solution of wave function around the turning point x= b
Vx
x
bO
E
The Schrödinger equation is given by
)()(
2 2
22
xxV
dx
d
m
,
or
0])([
2 2
22
xV
dx
d
m
,
where is the energy of a particle with a mass m. We assume that
)()( bxgxV ,
in the vicinity of x =b, where g>0. The Schrödinger equation is expressed by
0)(
2
22
2
bxgm
dx
d
.
Here we put
)(
2
3/1
2
bx
mg
z
.
Then we get
18
0)(
)(
2
2
zz
dz
zd .
The solution of this equation is given by
)()(2)( 21 zBCzACz ii .
We note the following.
(i) For z<0 (x>b)
k(x) is expressed by
2/1
3/1
22
||
2
)(
2
)( z
mg
bxg
m
xk
,
2/32/3
2/1
2
2/1
2
||
3
2
)(
2
3
22
)( zbx
mg
dxbx
mg
dxxk
x
b
x
b
.
(ii) For z>0 (x<b), where >V(x)
(x) is expressed by
2/1
3/1
2
22
2
)(
2
)]([
2
)(
z
mg
xbg
m
xV
m
x
2/32/3
2/1
2
2/1
2 3
2
)(
2
3
22
)( zxb
mg
dxxb
mg
dxx
b
x
b
x
__________________________________________________________________
8. Connection formula-II (downward)
19
Fig. Connection formula (II; downward). The condition 1
dx
d
for the WKB
approximation is not satisfied in the vicinity of x = b. We need the asymptotic
form the exact solution of the Schrödinger equation.
The asymptotic form for z<0;
)]
4
)(sin(
)(
1
)
4
)(cos(
)(
1
2[
2
)
4
sin(||)
4
cos(||2)()(2
2
1
6/1
2
2/1
4/12/1
2
4/12/1
121
x
b
x
b
ii
dxxk
xk
C
dxxk
xk
C
mg
zCzCzBCzAC
The asymptotic form for z>0;
20
)])(exp(
)(
1
))(exp(
)(
1
[
2
)()(2
2
1
6/1
2
2/1
4/12/1
2
4/12/1
121
b
x
b
x
ii
dxx
x
C
dxx
x
C
mg
ezCezCzBCzAC
Then we have the connection formula (II; downward) as
]
4
)(sin[
)(
]
4
)(cos[
)(
2
])(exp[
)(
])(exp[
)(
x
b
x
b
b
x
b
x
dxxk
xk
B
dxxk
xk
A
dxx
x
B
dxx
x
A
(II, downward)
with
Vx
x
bO
E
where C1 = A and C2 = B.
9. Tunneling probability
We apply the connection formula to find the tunneling probability. In order that the
WKB approximation apply within a barrier, it is necessary that the potential V(x) does not
change so rapidly. Suppose that a particle (energy and mass m) penetrates into a barrier
shown in the figure. There are three regions, I, II, and III.
21
ba
I II III
Vx
E
Fig. The connection formula I (upward) is used at x = a and the connection formula II
(downward) is used at x = b.
For x>b, (region III)
]
4
)(sin[
)(
]
4
)(cos[
)(
)]
4
)((exp[
)(
1
1
1
1
1
1
x
b
x
b
x
b
III
dxxk
xk
iA
dxxk
xk
A
dxxki
xk
A
(we consider on the wave propagating along the positive x axis), where
))((
2
)(
21
xV
m
xk
.
The connection formula (II, downward) is applied to the boundary between the regions
III and II.
]
4
)(sin[
)(2
]
4
)(cos[
)(
])(exp[
)(2
])(exp[
)(2
1
1
1
1
x
b
x
b
b
x
b
x
dxxk
xk
B
dxxk
xk
A
dxx
x
B
dxx
x
A
(II, downward)
Here we get
22
iAB 2 .
Then we get the wave function of the region II,
]
4
)(sin[
)(
]
4
)(cos[
)(
])(exp[
)(
])(exp[
)(2
1
1
1
1
x
b
x
b
III
b
x
b
x
II
dxxk
xk
iA
dxxk
xk
A
dxx
x
iA
dxx
x
A
or
])(exp[
2)(
])(exp[
1
)(
])()(exp[
)(
])()(exp[
)(2
x
a
x
a
x
a
b
a
b
a
x
a
II
dxx
r
x
A
dxx
rx
iA
dxxdxx
x
iA
dxxdxx
x
A
where
])([
2
)(
2
xVmx
,
and
])(exp[
b
a
dxxr ,
Next, the connection formula (I; upward) is applied to the boundary between the regions
II and I.
23
)])(exp(
)(
))(exp(
)(
)]
4
)(sin(
)(
)
4
)(cos(
)(
2
2
2
2
2
x
a
x
a
a
x
a
x
dxx
x
D
dxx
x
C
dxxk
xk
D
dxxk
xk
C
(I; upward)
Here we get
r
iAC
1
,
r
A
D
2
.
Then we have the wave function of the region I,
)]}
4
)((exp[)]
4
)(({exp[
4)(
)]}
4
)((exp[)]
4
)(({exp[
1
)(
]
4
)(sin[
)(2
]
4
)(cos[
1
)(
2
22
2
22
2
2
2
2
2
a
x
a
x
a
x
a
x
a
x
a
x
I
dxxkidxxki
ir
xk
A
dxxkidxxki
rxk
iAdxxkr
xk
A
dxxk
rxk
iA
or
)]}
4
)((exp[))
4
1
()]
4
)((exp[)
4
1
{(
)(
)]}
4
)((exp[)
1
4
()]
4
)((exp[)
1
4
{(
)(
22
2
22
2
x
a
x
a
a
x
a
x
I
dxxki
r
r
dxxki
r
rxk
iA
dxxki
r
r
dxxki
r
r
xk
iA
The first term corresponds to that of the reflected wave and the second term corresponds
to that of the incident wave. Then the tunneling probability is
24
))(2exp(
)
4
1
(
1 2
2
b
a
dxxr
r
r
T ,
where
))(exp(
b
a
dxxr .
9. -particle decay: quantum tunneling
r1 r2
Coulomb repusion
Nuclear binding
Vr
rO
E
Fig. Gamov's model for the potential energy of an alpha particle in a radioactive
nucleus.
25
-2
0
2
4
6
8
10
r
nu
cl
ea
r
po
te
nt
ia
lM
eV
Re yar
Ea
Fig. The tunneling of a particle from the 238U (Z = 92). The kinetic energy 4.2 MeV.
http://demonstrations.wolfram.com/GamowModelForAlphaDecayTheGeigerNutt
allLaw/
For r1<r<r2.
)(21)( rVmr
.
At r = r2,
20
2
1
4
2
r
eZ
.
The tunneling probability is
])(2exp[
1
2
r
r
drreP ,
where
26
])1([arccos
2
])(arccos[
2
1
2
)(
2
)(
2
1
2
1
2
1
2
121
2
1
2
2
2
1
2
1
2
1
r
r
r
r
r
r
r
m
rrr
r
r
r
m
dr
r
rm
drrV
m
drr
r
r
r
r
r
r
where m is the mass of -particle (= 4.001506179125 u). fm = 10-15 m (fermi).
The quantity P gives the probability that in one trial an particle will penetrate the
barrier. The number of trials per second could estimated to be
12r
v
N ,
if it were assumed that a particle is bouncing back and forth with velocity v inside the
nucleus of diameter 2r1. Then the probability per second that nucleus will decay by
emitting a particle, called the decay rate R, would be
2
12
e
r
v
R .
((Example))
We consider the particle emission from 238U nucleus (Z = 92), which emits a K =
4.2 MeV particle. The a particle is contained inside the nuclear radius r1 = 7.0 fm (fm =
10-15 m).
(i) The distance r2:
From the relation
20
2
4
2
r
Ze
K
,
we get
r2 = 63.08 fm.
(ii) The velocity of a particle inside the nucleus, v:
27
From the relation
2
1 2
1
vmK ,
where m is the mass of the a particle; m = 4.001506179 u, we get
v = 1.42318 x 107 m/s.
(iii) The value of :
])(arccos[
2
121
2
1
2 rrrr
r
r
mK
=51.8796.
(iv) The decay rate R:
2
12
e
r
v
R = 8.813 x 10-25.
((Mathematica))
28
Clear"Global`";
rule1 u 1.660538782 1027, eV 1.602176487 1019,
qe 1.602176487 1019, c 2.99792458 108,
— 1.05457162853 1034, 0 8.854187817 1012,
MeV 1.602176487 1013, Ma 4.001506179125 u,
fm 1015, Z1 92, r1 7 fm , K1 4.2 MeV;
eq0 K1
2 Z1 qe2
4 0 r
. rule1
6.729141013
4.245021026
r
eq01 Solveeq0, r; r2 r . eq011
6.308421014
r2
fm
. rule1
63.0842
eq1
1
2
Ma v2 K1 . rule1; eq2 Solveeq1, v;
v1 v . eq22
1.42318107
2 Ma K1
—
r2 ArcCos r1
r2
r1 r2 r1 .
rule1
51.8796
R1
v1
2 r1
Exp2 . rule1
8.812821025
29
9. Bound state: Bohr-Sommerfeld condition.
ab
I II III
Vx
E
For x<b (region I), the unnormalized wave function is
])(exp[
)(
1
b
x
I dxx
x
,
Using the connection rule (II; downward)
]
4
)(sin[
)(2
]
4
)(cos[
)(
])(exp[
)(2
])(exp[
)(2
x
b
x
b
II
b
x
b
x
dxxk
xk
B
dxxk
xk
A
dxx
x
B
dxx
x
A
I
(II; downward)
we get
2A , B = 0.
Then we have
)]
4
)(cos[
)(
2
x
b
II dxxk
xk
. for b<x<a
This may also be written as
30
]
4
)(sin[])(cos[
)(
2
]
4
)(cos[])(sin[
)(
2
]
4
)()(sin[
)(
2
]
24
)()(cos[
)(
2
]
4
)()(cos[
)(
2
]
4
)(cos[
)(
2
a
x
a
b
a
x
a
b
a
x
a
b
a
x
a
b
a
x
a
b
x
b
II
dxxkdxxk
xk
dxxkdxxk
xk
dxxkdxxk
xk
dxxkdxxk
xk
dxxkdxxk
xk
dxxk
xk
Here we use the connection rule (I, upward),
])(exp[
)(
])(exp[
)(
]
4
)(sin[
)(
]
4
)(cos[
)(
2
x
a
x
a
a
x
a
x
dxx
x
B
dxx
x
A
dxxk
xk
B
dxxk
xk
A
(I; upward)
From this we have
]
4
))(sin[])(cos[
)(
2
]
4
))(cos[])(sin[
)(
2
a
x
a
b
a
x
a
b
II
dxxkdxxk
xk
dxxkdxxk
xk
with
])(sin[
a
b
dxxkA , ])(cos[2
a
b
dxxkB .
Since III should have such a form
31
])(exp[
)(
x
a
III dxx
x
A
.
for x>a. Then we need the condition that
0])(cos[2
a
b
dxxkB ,
or
)
2
1
()( ndxxk
a
b
,
or
)
2
1
()( ndxxp
a
b
,
where n = 0, 1, 2, ... (Bohr-Sommerfeld condition).
10. Simple harmonics
We consider a simple harmonics,
22
00
22 2)
2
1
(2))((2)( xxmxmmxVmxp ,
where
2
0
0
2
m
x .
Then we get
0
2
0
0
2
0
0
0
22
00
2
2
1
4
22)(
00
0
m
m
x
mdxxxmdxxp
xx
x
.
When
)
2
1
()(
0
0
ndxxp
x
x
,
32
we have
)
2
1
(
0
n ,
or
)
2
1
( n
_______________________________________________________________________
APPENDIX
Connection formula
)(2
1
)( xVmxk
)(21)( xVmx
(i) Connection formula at x = a (upward)
33
)])(exp(
)(
))(exp(
)(
)]
4
)(sin(
)(
)
4
)(cos(
)(
2
1
1
1
1
x
a
x
a
a
x
a
x
dxx
x
B
dxx
x
A
dxxk
xk
B
dxxk
xk
A
formula I (upward)
(ii) Connection formula at x = b (downward)
)]
4
)(sin(
)(2
)
4
)(cos(
)(
)])(exp(
)(2
))(exp(
)(2
2
2
2
2
x
b
x
b
b
x
b
x
dxxk
xk
D
dxxk
xk
C
dxx
x
D
dxx
x
C
formula II (downward)
(i) Connection formula at x = b
]
4
)(sin[
)(2
]
4
)(cos[
)(
])(exp[
)(2
])(exp[
)(2
1
1
1
1
x
b
x
b
b
x
b
x
dxxk
xk
D
dxxk
xk
C
dxx
x
D
dxx
x
C
formula II (downward,)
(ii) Connection formula at x = a
34
])(exp[
)(
])(exp[
)(
]
4
)(sin[
)(
]
4
)(cos[
)(
2
2
2
2
2
x
a
x
a
a
x
a
x
dxx
x
B
dxx
x
A
dxxk
xk
B
dxxk
xk
A
formula I (upward)
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