Let
α
,
β
>
−
1
\alpha ,\beta > - 1
and
P
λ
(
α
,
β
)
(
x
)
=
(
1
−
x
)
α
(
1
+
x
)
β
P
λ
(
α
,
β
)
(
x
)
\mathcal {P}_\lambda ^{(\alpha ,\beta )}(x) = {(1 - x)^\alpha }{(1 + x)^\beta }P_\lambda ^{(\alpha ,\beta )}(x)
, where
P
λ
(
α
,
β
)
(
x
)
P_\lambda ^{(\alpha ,\beta )}(x)
is the Jacobi function of the first kind,
λ
≥
−
(
α
+
β
+
1
)
/
2
\lambda \geq - (\alpha + \beta + 1)/2
, and
−
1
>
x
≤
1
- 1 > x \leq 1
. Let
\[
F
(
α
,
β
)
(
λ
)
=
1
2
α
+
β
+
1
⟨
f
(
x
)
,
P
λ
(
α
,
β
)
(
x
)
⟩
=
1
2
α
+
β
+
1
∫
−
1
1
f
(
x
)
P
λ
(
α
,
β
)
(
x
)
d
x
{F^{(\alpha ,\beta )}}(\lambda ) = \frac {1} {{{2^{\alpha + \beta + 1}}}}\left \langle {f(x),\mathcal {P}_\lambda ^{(\alpha ,\beta )}(x)} \right \rangle = \frac {1} {{{2^{\alpha + \beta + 1}}}}\int _{ - 1}^1 {f(x)\mathcal {P}_\lambda ^{(\alpha ,\beta )}(x)dx}
\]
whenever the integral exists. It is known that for
α
+
β
=
0
\alpha + \beta = 0
, we have (*)
\[
f
(
x
)
=
lim
n
→
∞
4
∫
0
n
F
(
α
,
β
)
(
λ
−
1
2
)
P
λ
−
1
/
2
(
β
,
α
)
(
−
x
)
λ
×
sin
π
λ
Γ
2
(
λ
+
1
/
2
)
Γ
(
λ
+
α
+
1
/
2
)
Γ
(
λ
+
β
+
1
/
2
)
d
λ
f(x) = \lim \limits _{n \to \infty } 4\int _0^n {{F^{(\alpha ,\beta )}}\left ( {\lambda - \frac {1}{2}} \right )} P_{\lambda - 1/2}^{(\beta ,\alpha )}( - x)\lambda \times \sin \pi \lambda \frac {{{\Gamma ^2}(\lambda + 1/2)}}{{\Gamma (\lambda + \alpha + 1/2)\Gamma (\lambda + \beta + 1/2)}}d\lambda
\]
almost everywhere in
[
−
1
,
1
]
[-1,1]
. In this paper, we devise a technique to continue
f
(
x
)
f(x)
analytically to the complex
z
z
-plane and locate the singularities of
f
(
z
)
f(z)
by relating them to the singularities of
\[
g
(
t
)
=
∫
0
∞
e
−
λ
t
F
(
α
,
β
)
(
λ
)
d
λ
Γ
(
λ
+
α
+
1
)
.
g(t) = \int _0^\infty {{e^{ - \lambda t}}{F^{(\alpha ,\beta )}}(\lambda )} \frac {{d\lambda }}{{\Gamma (\lambda + \alpha + 1)}}.
\]
However, this will be done in the more general case where the limit in (*) exists in the sense of Schwartz distributions and defines a generalized function
f
(
x
)
f(x)
. In this case, we pass from
f
(
x
)
f(x)
to its analytic representation
\[
f
^
(
z
)
=
1
2
π
i
⟨
f
(
x
)
,
1
x
−
z
⟩
,
z
∉
supp
f
,
\hat f(z) = \frac {1} {{2\pi i}}\left \langle {f(x),\frac {1} {{x - z}}} \right \rangle ,\quad z \notin \operatorname {supp} f,
\]
and then relate the singularities of
f
^
(
z
)
\hat f(z)
to those of
g
(
t
)
g(t)
.