斯科惹函数 (Scorers functions)是下列方程的两个解
Scorers Gi function Scorers Hi Function
y
″
(
x
)
−
x
y
(
x
)
=
1
π
{\displaystyle y''(x)-x\ y(x)={\frac {1}{\pi }}}
G
i
(
x
)
=
1
π
∫
0
∞
sin
(
t
3
3
+
x
t
)
d
t
,
{\displaystyle \mathrm {Gi} (x)={\frac {1}{\pi }}\int _{0}^{\infty }\sin \left({\frac {t^{3}}{3}}+xt\right)\,dt,}
H
i
(
x
)
=
1
π
∫
0
∞
exp
(
−
t
3
3
+
x
t
)
d
t
.
{\displaystyle \mathrm {Hi} (x)={\frac {1}{\pi }}\int _{0}^{\infty }\exp \left(-{\frac {t^{3}}{3}}+xt\right)\,dt.}
也可以通过艾里函数 定义：
G
i
(
x
)
=
B
i
(
x
)
∫
x
∞
A
i
(
t
)
d
t
+
A
i
(
x
)
∫
0
x
B
i
(
t
)
d
t
,
H
i
(
x
)
=
B
i
(
x
)
∫
−
∞
x
A
i
(
t
)
d
t
−
A
i
(
x
)
∫
−
∞
x
B
i
(
t
)
d
t
.
{\displaystyle {\begin{aligned}\mathrm {Gi} (x)&{}=\mathrm {Bi} (x)\int _{x}^{\infty }\mathrm {Ai} (t)\,dt+\mathrm {Ai} (x)\int _{0}^{x}\mathrm {Bi} (t)\,dt,\\\mathrm {Hi} (x)&{}=\mathrm {Bi} (x)\int _{-\infty }^{x}\mathrm {Ai} (t)\,dt-\mathrm {Ai} (x)\int _{-\infty }^{x}\mathrm {Bi} (t)\,dt.\end{aligned}}}
G
i
(
z
)
=
∑
k
=
0
∞
c
o
s
(
(
2
k
−
1
)
∗
π
3
)
Γ
(
k
+
1
3
)
∗
(
3
1
/
3
∗
z
)
k
k
!
{\displaystyle Gi(z)=\sum _{k=0}^{\infty }cos({\frac {(2k-1)*\pi }{3}})\Gamma ({\frac {k+1}{3}})*{\frac {(3^{1/3}*z)^{k}}{k!}}}
H
i
(
z
)
=
3
−
2
/
3
π
∑
k
=
0
∞
Γ
(
(
2
k
+
1
)
∗
π
3
)
(
3
1
/
3
∗
z
)
k
k
!
{\displaystyle Hi(z)={\frac {3^{-2/3}}{\pi }}\sum _{k=0}^{\infty }\Gamma ({\frac {(2k+1)*\pi }{3}}{\bigr )}{\frac {(3^{1/3}*z)^{k}}{k!}}}
Olver, F. W. J., Scorer functions , Olver, Frank W. J. ; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W. (编), NIST Handbook of Mathematical Functions , Cambridge University Press, 2010, ISBN 978-0521192255MR 2723248 Scorer, R. S., Numerical evaluation of integrals of the form
I
=
∫
x
1
x
2
f
(
x
)
e
i
ϕ
(
x
)
d
x
{\displaystyle I=\int _{x_{1}}^{x_{2}}f(x)e^{i\phi (x)}dx}
G
i
(
z
)
=
1
π
∫
0
∞
s
i
n
(
u
z
+
1
3
u
3
)
d
u
{\displaystyle {\rm {Gi}}(z)={\frac {1}{\pi }}\int _{0}^{\infty }{\rm {sin}}\left(uz+{\frac {1}{3}}u^{3}\right)du}
3 : 107–112, ISSN 0033-5614 MR 0037604 doi:10.1093/qjmam/3.1.107 
