32
H. Bichsel and H. Schindler
For evaluating the second integral, we approximate dσ/dE by the Rutherford
cross section dσ R /dE ∝ 1/E 2 . Because of the rapid convergence of the integral
for sE 1, we further assume that the upper integration limit can be extended to
infinity (instead of truncating dσ/dE at E max ). Integrating by parts and substituting
z = sE yields
I 2 = ξ
∞
E 1
dE
1 − e −sE
E 2
= ξ
1 − e −sE 1
E 1
∼s
+ξs
∞
sE 1
dz
e −z
z
∼ ξs
⎛
⎜
⎝1 +
1
sE 1
dz
z
− C
⎞
⎟
⎠ ,
where C ∼ 0.577215665 is Euler’s constant. 3 Combining the two terms I 1 and I 2 ,
one obtains
F (s, x) = exp
−ξs
1 − C +
ξ
− ln sE max + β
2
,
and, applying the inverse Laplace transform,
f (, x) = L
−1
{F (s, x)} =
1
ξ
φ L (λ) ,
(2.37)
where
φ L (λ) =
1
2πi
c+i∞
c−i∞
du e
u ln u+λu .
(2.38)
is a universal function of the dimensionless variable
λ =
− −
ξ
− (1 − C) − β
2
− ln
ξ
E max
.
The maximum of φ L (λ) is located at λ ∼ −0.222782 and the most probable energy
loss is, consequently, given by
p ∼ ∼ + ξ
0.2 + β
2
+ ln κ
,
(2.39)
3
−C =
1
0
dz
e −z − 1
z
+
∞
1
dz
e −z
z
= −0.577215665 . . .
H. Bichsel and H. Schindler
For evaluating the second integral, we approximate dσ/dE by the Rutherford
cross section dσ R /dE ∝ 1/E 2 . Because of the rapid convergence of the integral
for sE 1, we further assume that the upper integration limit can be extended to
infinity (instead of truncating dσ/dE at E max ). Integrating by parts and substituting
z = sE yields
I 2 = ξ
∞
E 1
dE
1 − e −sE
E 2
= ξ
1 − e −sE 1
E 1
∼s
+ξs
∞
sE 1
dz
e −z
z
∼ ξs
⎛
⎜
⎝1 +
1
sE 1
dz
z
− C
⎞
⎟
⎠ ,
where C ∼ 0.577215665 is Euler’s constant. 3 Combining the two terms I 1 and I 2 ,
one obtains
F (s, x) = exp
−ξs
1 − C +
ξ
− ln sE max + β
2
,
and, applying the inverse Laplace transform,
f (, x) = L
−1
{F (s, x)} =
1
ξ
φ L (λ) ,
(2.37)
where
φ L (λ) =
1
2πi
c+i∞
c−i∞
du e
u ln u+λu .
(2.38)
is a universal function of the dimensionless variable
λ =
− −
ξ
− (1 − C) − β
2
− ln
ξ
E max
.
The maximum of φ L (λ) is located at λ ∼ −0.222782 and the most probable energy
loss is, consequently, given by
p ∼ ∼ + ξ
0.2 + β
2
+ ln κ
,
(2.39)
3
−C =
1
0
dz
e −z − 1
z
+
∞
1
dz
e −z
z
= −0.577215665 . . .
