Integral representations and integral transforms of some families of ...

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Integral Transforms and Special Functions 3
∫ ∞
( )
t
t
=
0 e t − 1 arctan dt.
x
(8)
In the same way, one can obtain the Laplace transform given below:
∫ ∞
( )
t t
L( ˜S(r))(x) =
e t + 1 arctan dt. (9)
x
0
II. The function S μ+1 has the following integral representation via the Bessel function J ν of
the first kind (see Cerone and Lenard [3]):
S μ+1 (r) =
√ ∫ π
∞
t μ+1/2
(2r) μ−1/2 Ɣ(μ + 1) 0 e t − 1 J μ−1/2(rt)dt (r,μ ∈ R + ). (10)
Similarly, it can be shown that (see Srivastava and Tomovski [15])
S μ (α,0) (r;{k 2/α 2 √ ∫
π ∞
t μ−1/2
}) =
(2r) μ−1/2 Ɣ(μ) 0 e t − 1 J μ−1/2(rt)dt
(
r ∈ R + ; μ> 1 )
. (11)
2
We shall also need the following known result involving the Laplace–Mellin transform [6, Vol.
I, p. 182, Entry 4.14 (9)]:
∫ ∞
0
( ρ
) [ ν
e −st t λ−1 J ν (ρt)dt = s
−λ
Ɣ(ν + λ) 1
2s Ɣ(ν + 1) 2 F 1
2 (ν+λ), 1 ]
(ν+λ+1); ν+1;−ρ2
2 s 2
( )
R(s) > |I(ρ)|; R(ν + λ) > 0 .
Making use of the known result (12) and the Legendre duplication formula for the gamma function,
it follows that
√ ∫ π ∞
t μ+1/2 (∫ ∞
e −xr
)
L(S μ+1 (r))(x) =
Ɣ(μ + 1) 0 e t − 1 0 (2r) J μ−1/2(rt)dr dt
μ−1/2
√ ∫ π
∞
t μ+1/2 (∫ ∞
)
=
e −xr r 1/2−μ J
2 μ−1/2 Ɣ(μ + 1) 0 e t μ1/2 (rt)dr dt
− 1 0
√ ∫ π
∞
t 2μ ( 1
=
2 2μ−1 xƔ(μ + 1)Ɣ (μ + 1/2) 0 e t − 1 2 F 1
2 , 1; μ + 1 2 )
2 ;−t dt
x 2
∫
2 ∞
t 2μ (
=
xƔ(2μ + 1) e t − 1 2 F 1 1, 1 2 ; μ + 1 2 )
2 ;−t dt, (13)
x 2
provided that each member of (13) exists.
Similarly, we have
L( ˜S μ+1 (r))(x) =
where ˜S μ+1 (r) is the alternating version of S μ+1 (r).
0
(12)
∫
2 ∞
t 2μ (
xƔ(2μ + 1) 0 e t + 1 2 F 1 1, 1 2 ; μ + 1 2 )
2 ;−t dt, (14)
x 2