Summarize Timeline Top Qs Fact Check
We have the following integral representation for the full two-parameter form of Ramanujan's theta function:[ 1]
f
(
a
,
b
)
=
1
+
∫
0
∞
2
a
e
−
1
2
t
2
2
π
[
1
−
a
a
b
cosh
(
log
a
b
t
)
a
3
b
−
2
a
a
b
cosh
(
log
a
b
t
)
+
1
]
d
t
+
∫
0
∞
2
b
e
−
1
2
t
2
2
π
[
1
−
b
a
b
cosh
(
log
a
b
t
)
a
b
3
−
2
b
a
b
cosh
(
log
a
b
t
)
+
1
]
d
t
{\displaystyle {\begin{aligned}f(a,b)=1+\int _{0}^{\infty }{\frac {2ae^{-{\frac {1}{2}}t^{2}}}{\sqrt {2\pi }}}\left[{\frac {1-a{\sqrt {ab}}\cosh \left({\sqrt {\log ab}}\,t\right)}{a^{3}b-2a{\sqrt {ab}}\cosh \left({\sqrt {\log ab}}\,t\right)+1}}\right]dt+\\\int _{0}^{\infty }{\frac {2be^{-{\frac {1}{2}}t^{2}}}{\sqrt {2\pi }}}\left[{\frac {1-b{\sqrt {ab}}\cosh \left({\sqrt {\log ab}}\,t\right)}{ab^{3}-2b{\sqrt {ab}}\cosh \left({\sqrt {\log ab}}\,t\right)+1}}\right]dt\end{aligned}}}
The special cases of Ramanujan's theta functions given by φ (q ) := f (q , q ) (sequence A000122 in the OEIS ) and ψ (q ) := f (q , q 3 ) (sequence A010054 in the OEIS ) [ 2] also have the following integral representations:[ 1]
φ
(
q
)
=
1
+
∫
0
∞
e
−
1
2
t
2
2
π
[
4
q
(
1
−
q
2
cosh
(
2
log
q
t
)
)
q
4
−
2
q
2
cosh
(
2
log
q
t
)
+
1
]
d
t
ψ
(
q
)
=
∫
0
∞
2
e
−
1
2
t
2
2
π
[
1
−
q
cosh
(
log
q
t
)
q
−
2
q
cosh
(
log
q
t
)
+
1
]
d
t
{\displaystyle {\begin{aligned}\varphi (q)&=1+\int _{0}^{\infty }{\frac {e^{-{\frac {1}{2}}t^{2}}}{\sqrt {2\pi }}}\left[{\frac {4q\left(1-q^{2}\cosh \left({\sqrt {2\log q}}\,t\right)\right)}{q^{4}-2q^{2}\cosh \left({\sqrt {2\log q}}\,t\right)+1}}\right]dt\\[6pt]\psi (q)&=\int _{0}^{\infty }{\frac {2e^{-{\frac {1}{2}}t^{2}}}{\sqrt {2\pi }}}\left[{\frac {1-{\sqrt {q}}\cosh \left({\sqrt {\log q}}\,t\right)}{q-2{\sqrt {q}}\cosh \left({\sqrt {\log q}}\,t\right)+1}}\right]dt\end{aligned}}}
This leads to several special case integrals for constants defined by these functions when q := e −kπ (cf. theta function explicit values ). In particular, we have that [ 1]
φ
(
e
−
k
π
)
=
1
+
∫
0
∞
e
−
1
2
t
2
2
π
[
4
e
k
π
(
e
2
k
π
−
cos
(
2
π
k
t
)
)
e
4
k
π
−
2
e
2
k
π
cos
(
2
π
k
t
)
+
1
]
d
t
π
1
4
Γ
(
3
4
)
=
1
+
∫
0
∞
e
−
1
2
t
2
2
π
[
4
e
π
(
e
2
π
−
cos
(
2
π
t
)
)
e
4
π
−
2
e
2
π
cos
(
2
π
t
)
+
1
]
d
t
π
1
4
Γ
(
3
4
)
⋅
2
+
2
2
=
1
+
∫
0
∞
e
−
1
2
t
2
2
π
[
4
e
2
π
(
e
4
π
−
cos
(
2
π
t
)
)
e
8
π
−
2
e
4
π
cos
(
2
π
t
)
+
1
]
d
t
π
1
4
Γ
(
3
4
)
⋅
1
+
3
2
1
4
3
3
8
=
1
+
∫
0
∞
e
−
1
2
t
2
2
π
[
4
e
3
π
(
e
6
π
−
cos
(
6
π
t
)
)
e
12
π
−
2
e
6
π
cos
(
6
π
t
)
+
1
]
d
t
π
1
4
Γ
(
3
4
)
⋅
5
+
2
5
5
3
4
=
1
+
∫
0
∞
e
−
1
2
t
2
2
π
[
4
e
5
π
(
e
10
π
−
cos
(
10
π
t
)
)
e
20
π
−
2
e
10
π
cos
(
10
π
t
)
+
1
]
d
t
{\displaystyle {\begin{aligned}\varphi \left(e^{-k\pi }\right)&=1+\int _{0}^{\infty }{\frac {e^{-{\frac {1}{2}}t^{2}}}{\sqrt {2\pi }}}\left[{\frac {4e^{k\pi }\left(e^{2k\pi }-\cos \left({\sqrt {2\pi k}}\,t\right)\right)}{e^{4k\pi }-2e^{2k\pi }\cos \left({\sqrt {2\pi k}}\,t\right)+1}}\right]dt\\[6pt]{\frac {\pi ^{\frac {1}{4}}}{\Gamma \left({\frac {3}{4}}\right)}}&=1+\int _{0}^{\infty }{\frac {e^{-{\frac {1}{2}}t^{2}}}{\sqrt {2\pi }}}\left[{\frac {4e^{\pi }\left(e^{2\pi }-\cos \left({\sqrt {2\pi }}\,t\right)\right)}{e^{4\pi }-2e^{2\pi }\cos \left({\sqrt {2\pi }}\,t\right)+1}}\right]dt\\[6pt]{\frac {\pi ^{\frac {1}{4}}}{\Gamma \left({\frac {3}{4}}\right)}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}&=1+\int _{0}^{\infty }{\frac {e^{-{\frac {1}{2}}t^{2}}}{\sqrt {2\pi }}}\left[{\frac {4e^{2\pi }\left(e^{4\pi }-\cos \left(2{\sqrt {\pi }}\,t\right)\right)}{e^{8\pi }-2e^{4\pi }\cos \left(2{\sqrt {\pi }}\,t\right)+1}}\right]dt\\[6pt]{\frac {\pi ^{\frac {1}{4}}}{\Gamma \left({\frac {3}{4}}\right)}}\cdot {\frac {\sqrt {1+{\sqrt {3}}}}{2^{\frac {1}{4}}3^{\frac {3}{8}}}}&=1+\int _{0}^{\infty }{\frac {e^{-{\frac {1}{2}}t^{2}}}{\sqrt {2\pi }}}\left[{\frac {4e^{3\pi }\left(e^{6\pi }-\cos \left({\sqrt {6\pi }}\,t\right)\right)}{e^{12\pi }-2e^{6\pi }\cos \left({\sqrt {6\pi }}\,t\right)+1}}\right]dt\\[6pt]{\frac {\pi ^{\frac {1}{4}}}{\Gamma \left({\frac {3}{4}}\right)}}\cdot {\frac {\sqrt {5+2{\sqrt {5}}}}{5^{\frac {3}{4}}}}&=1+\int _{0}^{\infty }{\frac {e^{-{\frac {1}{2}}t^{2}}}{\sqrt {2\pi }}}\left[{\frac {4e^{5\pi }\left(e^{10\pi }-\cos \left({\sqrt {10\pi }}\,t\right)\right)}{e^{20\pi }-2e^{10\pi }\cos \left({\sqrt {10\pi }}\,t\right)+1}}\right]dt\end{aligned}}}
and that
ψ
(
e
−
k
π
)
=
∫
0
∞
e
−
1
2
t
2
2
π
[
cos
(
k
π
t
)
−
e
k
π
2
cos
(
k
π
t
)
−
cosh
k
π
2
]
d
t
π
1
4
Γ
(
3
4
)
⋅
e
π
8
2
5
8
=
∫
0
∞
e
−
1
2
t
2
2
π
[
cos
(
π
t
)
−
e
π
2
cos
(
π
t
)
−
cosh
π
2
]
d
t
π
1
4
Γ
(
3
4
)
⋅
e
π
4
2
5
4
=
∫
0
∞
e
−
1
2
t
2
2
π
[
cos
(
2
π
t
)
−
e
π
cos
(
2
π
t
)
−
cosh
π
]
d
t
π
1
4
Γ
(
3
4
)
⋅
1
+
2
4
e
π
16
2
7
16
=
∫
0
∞
e
−
1
2
t
2
2
π
[
cos
(
π
2
t
)
−
e
π
4
cos
(
π
2
t
)
−
cosh
π
4
]
d
t
{\displaystyle {\begin{aligned}\psi \left(e^{-k\pi }\right)&=\int _{0}^{\infty }{\frac {e^{-{\frac {1}{2}}t^{2}}}{\sqrt {2\pi }}}\left[{\frac {\cos \left({\sqrt {k\pi }}\,t\right)-e^{\frac {k\pi }{2}}}{\cos \left({\sqrt {k\pi }}\,t\right)-\cosh {\frac {k\pi }{2}}}}\right]dt\\[6pt]{\frac {\pi ^{\frac {1}{4}}}{\Gamma \left({\frac {3}{4}}\right)}}\cdot {\frac {e^{\frac {\pi }{8}}}{2^{\frac {5}{8}}}}&=\int _{0}^{\infty }{\frac {e^{-{\frac {1}{2}}t^{2}}}{\sqrt {2\pi }}}\left[{\frac {\cos \left({\sqrt {\pi }}\,t\right)-e^{\frac {\pi }{2}}}{\cos \left({\sqrt {\pi }}\,t\right)-\cosh {\frac {\pi }{2}}}}\right]dt\\[6pt]{\frac {\pi ^{\frac {1}{4}}}{\Gamma \left({\frac {3}{4}}\right)}}\cdot {\frac {e^{\frac {\pi }{4}}}{2^{\frac {5}{4}}}}&=\int _{0}^{\infty }{\frac {e^{-{\frac {1}{2}}t^{2}}}{\sqrt {2\pi }}}\left[{\frac {\cos \left({\sqrt {2\pi }}\,t\right)-e^{\pi }}{\cos \left({\sqrt {2\pi }}\,t\right)-\cosh \pi }}\right]dt\\[6pt]{\frac {\pi ^{\frac {1}{4}}}{\Gamma \left({\frac {3}{4}}\right)}}\cdot {\frac {{\sqrt[{4}]{1+{\sqrt {2}}}}\,e^{\frac {\pi }{16}}}{2^{\frac {7}{16}}}}&=\int _{0}^{\infty }{\frac {e^{-{\frac {1}{2}}t^{2}}}{\sqrt {2\pi }}}\left[{\frac {\cos \left({\sqrt {\frac {\pi }{2}}}\,t\right)-e^{\frac {\pi }{4}}}{\cos \left({\sqrt {\frac {\pi }{2}}}\,t\right)-\cosh {\frac {\pi }{4}}}}\right]dt\end{aligned}}}