Методы математической физики - 151 стр.

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Γ
µ
1
2
=
Z
0
1
t
e
t
dt =
π .
2m + 1 1
2m(2m 2) . . . 2
(2m + 1)!
2
2m+1
m!
Γ
µ
2m + 3
2
=
(2m + 1)!
2
2m+1
m!
π .
J
1/2
(x) =
r
2
πx
X
m=0
(1)
m
x
2m+1
(2m + 1)!
=
r
2
πx
sin x .
ν = 1/2
J
n+1/2
(x) = (1)
n
x
n
r
2x
π
µ
1
x
d
dx
n
sin x
x
,
ν = 1/2
J
1/2
(x) =
X
m=0
(1)
m
Γ(m + 1) Γ(m + 1/2)
³
x
2
´
2m1/2
.
Γ
µ
2m + 1
2
=
(2m)!
2
2m
m!
π ,
J
1/2
(x) =
r
2
πx
X
m=0
(1)
m
x
2m
(2m)!
=
r
2
πx
cos x .