The Ramanujan theta function is defined as
f
(
a
,
b
)
=
∑
n
=
−
∞
∞
a
n
(
n
+
1
)
2
b
n
(
n
−
1
)
2
{\displaystyle f(a,b)=\sum _{n=-\infty }^{\infty }a^{\frac {n(n+1)}{2}}\;b^{\frac {n(n-1)}{2}}}
for |ab . The Jacobi triple product identity then takes the form
f
(
a
,
b
)
=
(
−
a
;
a
b
)
∞
(
−
b
;
a
b
)
∞
(
a
b
;
a
b
)
∞
.
{\displaystyle f(a,b)=(-a;ab)_{\infty }\;(-b;ab)_{\infty }\;(ab;ab)_{\infty }.}
Here, the expression
(
a
;
q
)
n
{\displaystyle (a;q)_{n}}
q -Pochhammer symbol
φ
(
q
)
=
f
(
q
,
q
)
=
∑
n
=
−
∞
∞
q
n
2
=
(
−
q
;
q
2
)
∞
2
(
q
2
;
q
2
)
∞
{\displaystyle \varphi (q)=f(q,q)=\sum _{n=-\infty }^{\infty }q^{n^{2}}={\left(-q;q^{2}\right)_{\infty }^{2}\left(q^{2};q^{2}\right)_{\infty }}}
and
ψ
(
q
)
=
f
(
q
,
q
3
)
=
∑
n
=
0
∞
q
n
(
n
+
1
)
2
=
(
q
2
;
q
2
)
∞
(
−
q
;
q
)
∞
{\displaystyle \psi (q)=f\left(q,q^{3}\right)=\sum _{n=0}^{\infty }q^{\frac {n(n+1)}{2}}={\left(q^{2};q^{2}\right)_{\infty }}{(-q;q)_{\infty }}}
and
f
(
−
q
)
=
f
(
−
q
,
−
q
2
)
=
∑
n
=
−
∞
∞
(
−
1
)
n
q
n
(
3
n
−
1
)
2
=
(
q
;
q
)
∞
{\displaystyle f(-q)=f\left(-q,-q^{2}\right)=\sum _{n=-\infty }^{\infty }(-1)^{n}q^{\frac {n(3n-1)}{2}}=(q;q)_{\infty }}
This last being the Euler function , which is closely related to the Dedekind eta function . The Jacobi theta function may be written in terms of the Ramanujan theta function as:
ϑ
00
(
w
,
q
)
=
f
(
q
w
2
,
q
w
−
2
)
{\displaystyle \vartheta _{00}(w,q)=f\left(qw^{2},qw^{-2}\right)}
Integral representations
edit
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 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}
The special cases of Ramanujan's theta functions given by φ (q ) := f (q , q )OEIS : A000122 ψ (q ) := f (q , q 3 )OEIS : A010054 [ 2] [ 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 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}}}
Application in string theory
edit
Bailey, W. N. (1935). Generalized Hypergeometric Series . Cambridge Tracts in Mathematics and Mathematical Physics. Vol. 32. Cambridge: Cambridge University Press. Gasper, George; Rahman, Mizan (2004). Basic Hypergeometric Series . Encyclopedia of Mathematics and Its Applications. Vol. 96 (2nd ed.). Cambridge: Cambridge University Press. ISBN 0-521-83357-4 "Ramanujan function" , Encyclopedia of Mathematics EMS Press , 2001 [1994]Kaku, Michio (1994). Hyperspace: A Scientific Odyssey Through Parallel Universes, Time Warps, and the Tenth Dimension . Oxford: Oxford University Press. ISBN 0-19-286189-1 Weisstein, Eric W. "Ramanujan Theta Functions" . MathWorld 
![{\displaystyle 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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d833a850026ea49ed8e8cc14e0c5c319ff52dace)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d958638c9c5c21543a41751082961c3959ca16b)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3031f986c5ca7b5f2e2295b80296f8011cec02b)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37ae89a40223719a2eb7a92b180785ac7bbe090d)