102
L. Igumnov et al.
W
L
σ (α, s) =
1
1 − M L (s)
e
−b(s)α
.
The function G
L
(α, s) = e
−b(s)α original was found in (Korovaytseva et al.
2017) for the case of exponential relaxation kernel M(τ ) = ae
−ϑτ . Also restricting
ourselves by this case, we represent function W
L
σ (α, s) transform as follows:
W
L
σ (α, s) = F
L
(α, s + ϑ), F
L
(α, s) = f (s)e
−d(s)α
,
f (s) =
1
1 − ˜
M L (s)
, d(s) =
s − ϑ
1 − ˜
M L (s)
, ˜
M
L
(s) =
a
s
.
We expand the function f (s) in series in the vicinity of infinitely distant point s
of the plane:
f (s) =
1
1 − ˜
M L (s)
=
∞
k=0
˜
M
L
(s)
k =
∞
k=0
a
k
s k .
Expansion in series in powers of s of the function F
L
0 (α, s) = e
−d(s)α was carried
out in (Korovaytseva et al. 2017) and had the form
F
L
0 (α, s) = e
−(c 1 a−ϑ)α e
−sα Q
L
(α, s), Q
L
(α, s) = 1
+
∞
l=0
g l+1 (α)
s l+1 =
∞
l=0
g l (α)
s l , g 0 (α) = 1,
g l (α) =
l
m=1
(−1)
m
α
m h l−m,m
m!
, h k,m+1 =
k
n=0
h k−n,m e n (m ≥ 1), h k,1 = e k ,
e n = d n+1 a
n+1
, c n =
(2n − 1)!!
2 n n!
, d n =
2n + 1
2(n + 1)
a − ϑ
c n .
Then,
F
L
(α, s) = e
−(c 1 a−ϑ)α e
−sα
1 +
∞
m=0
d m+1 (α)
s m+1
,
where k + l = m, d m (α) =
m
k=0
a
k g m−k (α).
So,
F(α, τ ) = e
−(c 1 a−ϑ)α
δ(τ − α) +
∞
m=0
d m+1 (α)
m!
(τ − α)
m
H (τ − α),
L. Igumnov et al.
W
L
σ (α, s) =
1
1 − M L (s)
e
−b(s)α
.
The function G
L
(α, s) = e
−b(s)α original was found in (Korovaytseva et al.
2017) for the case of exponential relaxation kernel M(τ ) = ae
−ϑτ . Also restricting
ourselves by this case, we represent function W
L
σ (α, s) transform as follows:
W
L
σ (α, s) = F
L
(α, s + ϑ), F
L
(α, s) = f (s)e
−d(s)α
,
f (s) =
1
1 − ˜
M L (s)
, d(s) =
s − ϑ
1 − ˜
M L (s)
, ˜
M
L
(s) =
a
s
.
We expand the function f (s) in series in the vicinity of infinitely distant point s
of the plane:
f (s) =
1
1 − ˜
M L (s)
=
∞
k=0
˜
M
L
(s)
k =
∞
k=0
a
k
s k .
Expansion in series in powers of s of the function F
L
0 (α, s) = e
−d(s)α was carried
out in (Korovaytseva et al. 2017) and had the form
F
L
0 (α, s) = e
−(c 1 a−ϑ)α e
−sα Q
L
(α, s), Q
L
(α, s) = 1
+
∞
l=0
g l+1 (α)
s l+1 =
∞
l=0
g l (α)
s l , g 0 (α) = 1,
g l (α) =
l
m=1
(−1)
m
α
m h l−m,m
m!
, h k,m+1 =
k
n=0
h k−n,m e n (m ≥ 1), h k,1 = e k ,
e n = d n+1 a
n+1
, c n =
(2n − 1)!!
2 n n!
, d n =
2n + 1
2(n + 1)
a − ϑ
c n .
Then,
F
L
(α, s) = e
−(c 1 a−ϑ)α e
−sα
1 +
∞
m=0
d m+1 (α)
s m+1
,
where k + l = m, d m (α) =
m
k=0
a
k g m−k (α).
So,
F(α, τ ) = e
−(c 1 a−ϑ)α
δ(τ − α) +
∞
m=0
d m+1 (α)
m!
(τ − α)
m
H (τ − α),
