## Theta Functions and Jacobi Elliptic Functions

While theta functions have applications in number theory and the heat equation, they are also used as helper functions in proving properties of Jacobi elliptic functions. The latter is what we will do in this post. This post will be followed by others to derive various properties of the Jacobi elliptic functions in order to evaluate integrals of them and eventually elliptic integrals.

We begin with the basic definitions of the theta functions. Note that notation for the theta functions is not standard.

## Theta Functions

$$z$$ = argument
$$\tau$$ = lattice parameter, $$\mathfrak{I}(\tau) \gt 0$$
$$q = \mathrm{e}^{i\pi \tau}$$ = nome, $$|q| \lt 1$$

a. Exponential forms

\theta_{1}(z|\tau) = \theta_{1}(z,q) = -i\sum_{n=-\infty}^{\infty} (-1)^{n} q^{(n+1/2)^{2}} \mathrm{e}^{i(2n+1)z} \\
\theta_{2}(z|\tau) = \theta_{2}(z,q) = \sum_{n=-\infty}^{\infty} q^{(n+1/2)^{2}} \mathrm{e}^{i(2n+1)z} \\
\theta_{3}(z|\tau) = \theta_{3}(z,q) = \sum_{n=-\infty}^{\infty} q^{n^{2}} \mathrm{e}^{i2nz} \\
\theta_{4}(z|\tau) = \theta_{4}(z,q) = \sum_{n=-\infty}^{\infty} (-1)^{n} q^{n^{2}} \mathrm{e}^{i2nz}

b. Trigonometric forms

\theta_{1}(z|\tau) = \theta_{1}(z,q) = 2\sum_{n=0}^{\infty} (-1)^{n} q^{(n+1/2)^{2}} \sin[(2n+1)z] \\
\theta_{2}(z|\tau) = \theta_{2}(z,q) = 2\sum_{n=0}^{\infty} q^{(n+1/2)^{2}} \cos[(2n+1)z] \\
\theta_{3}(z|\tau) = \theta_{3}(z,q) = 1 + 2\sum_{n=1}^{\infty} q^{n^{2}} \cos(2nz) \\
\theta_{4}(z|\tau) = \theta_{4}(z,q) = 1 + 2\sum_{n=1}^{\infty} (-1)^{n} q^{n^{2}} \cos(2nz)

## Jacobi Elliptic Functions as Functions of Theta Functions

The modulus $$k$$:

\sqrt{k} = \frac{\theta_{2}(0,q)}{\theta_{3}(0,q)}

The complementary modulus $$k^{\prime}$$:

\sqrt{k^{\prime}} = \frac{\theta_{4}(0,q)}{\theta_{3}(0,q)}

3 of the 12 Jacobi elliptic functions:
\begin{align}
\mathrm{sn}(u,k) &= \frac{\theta_{3}(0,q)\theta_{1}(z,q)}{\theta_{2}(0,q)\theta_{4}(z,q)} \\
\mathrm{cn}(u,k) &= \frac{\theta_{4}(0,q)\theta_{2}(z,q)}{\theta_{2}(0,q)\theta_{4}(z,q)} \\
\mathrm{dn}(u,k) &= \frac{\theta_{4}(0,q)\theta_{3}(z,q)}{\theta_{3}(0,q)\theta_{4}(z,q)}
\end{align}
where $$z$$ is the argument of the theta functions and $$u$$ is the argument of the Jacobi elliptic functions. They are related by the following equation:

z=\frac{u}{\theta_{3}^{2}(0,q)}

## Proofs of Basic Properties of Jacobi Elliptic Functions

We assume all properties of theta functions as given. For proofs of these properties, see references 2 and 4 below.

1. First “Pythagorean” Identity
\begin{align}
k^{2} + {k^{\prime}}^{2} &=
\frac{\theta_{2}^{4}(0,q)}{\theta_{3}^{4}(0,q)} + \frac{\theta_{4}^{4}(0,q)}{\theta_{3}^{4}(0,q)}
= \frac{\theta_{2}^{4}(0,q) + \theta_{4}^{4}(0,q)}{\theta_{3}^{4}(0,q)} \\
&= \frac{\theta_{3}^{4}(0,q)}{\theta_{3}^{4}(0,q)} = 1
\end{align}
We used this identity in the last step.

2. Second “Pythagorean” Identity
\begin{align}
\mathrm{cn}^{2}(u,k) + \mathrm{sn}^{2}(u,k)
&= \frac{\theta_{4}^{2}(0,q)\theta_{2}^{2}(z,q)}{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} +
\frac{\theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)}{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} \\
&= \frac{\theta_{4}^{2}(0,q)\theta_{2}^{2}(z,q) + \theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)}
{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} \\
&= \frac{[\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q) \,-\, \theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)] + \theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)}{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} = 1
\end{align}
We used this identity in the last step.

3. Third “Pythagorean” Identity
\begin{align}
\mathrm{dn}^{2}(u,k) + k^{2}\mathrm{sn}^{2}(u,k)
&= \frac{\theta_{4}^{2}(0,q)\theta_{3}^{2}(z,q)}{\theta_{3}^{2}(0,q)\theta_{4}^{2}(z,q)}
+ \frac{\theta_{2}^{4}(0,q)}{\theta_{3}^{4}(0,q)}\,
\frac{\theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)}{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} \\
&= \frac{\theta_{4}^{2}(0,q)\theta_{3}^{2}(z,q) + \theta_{2}^{2}(0,q)\theta_{1}^{2}(z,q)}
{\theta_{3}^{2}(0,q)\theta_{4}^{2}(z,q)} \\
&= \frac{[\theta_{3}^{2}(0,q)\theta_{4}^{2}(z,q) \,-\, \theta_{2}^{2}(0,q)\theta_{1}^{2}(z,q)]
\,+\, \theta_{2}^{2}(0,q)\theta_{1}^{2}(z,q)}
{\theta_{3}^{2}(0,q)\theta_{4}^{2}(z,q)} = 1
\end{align}
We used this identity in the last step.

## Derivations of Derivatives of the 3 Basic Jacobi Elliptic Functions

1.
\begin{align}
\frac{\partial \,\mathrm{sn}(u,k)}{\partial u}
&= \frac{\theta_{3}(0,q)}{\theta_{2}(0,q)} \frac{\partial}{\partial z} \left(\frac{\theta_{1}(z,q)}{\theta_{4}(z,q)} \right) \frac{\partial z}{\partial u} \\
&= \frac{\theta_{3}(0,q)}{\theta_{2}(0,q)} \, \frac{\theta_{4}^{2}(0,q)\theta_{2}(z,q)\theta_{3}(z,q)}{\theta_{4}^{2}(z,q)} \, \frac{1}{\theta_{3}^{2}(0,q)} \\
&= \left[\frac{\theta_{4}(0,q)\theta_{2}(z,q)}{\theta_{2}(0,q)\theta_{4}(z,q)}\right] \,
\left[\frac{\theta_{4}(0,q)\theta_{3}(z,q)}{\theta_{3}(0,q)\theta_{4}(z,q)}\right]
= \mathrm{cn}(u,k)\mathrm{dn}(u,k)
\end{align}
The derivative of the ratio of theta functions in the first line was obtained from Gradshteyn and Ryzhik 8.199(2).1.

2.

\frac{\partial}{\partial u} [\mathrm{cn}^{2}(u,k) + \mathrm{sn}^{2}(u,k) = 1]

2\mathrm{cn}(u,k)\frac{\partial}{\partial u}\mathrm{cn}(u,k) + 2\mathrm{sn}(u,k)\frac{\partial}{\partial u}\mathrm{sn}(u,k) = 0

\frac{\partial}{\partial u}\mathrm{cn}(u,k) = -\frac{\mathrm{sn}(u,k)}{\mathrm{cn}(u,k)}\frac{\partial}{\partial u}\mathrm{sn}(u,k) = -\mathrm{sn}(u,k)\mathrm{dn}(u,k)

3.

\frac{\partial}{\partial u} [\mathrm{dn}^{2}(u,k) + k^{2}\mathrm{sn}^{2}(u,k) = 1]

2\mathrm{dn}(u,k)\frac{\partial}{\partial u}\mathrm{dn}(u,k) + 2k^{2}\mathrm{sn}(u,k)\frac{\partial}{\partial u}\mathrm{sn}(u,k) = 0

\frac{\partial}{\partial u}\mathrm{dn}(u,k) = -k^{2}\frac{\mathrm{sn}(u,k)}{\mathrm{dn}(u,k)}\frac{\partial}{\partial u}\mathrm{sn}(u,k) = -k^{2}\mathrm{sn}(u,k)\mathrm{cn}(u,k)

## References

1. NIST Digital Library of Mathematical Functions
2. A Course of Modern Analysis – Whittaker and Watson
3. Gradshteyn and Ryzhik’s Table of Integrals, Series, and Products, 8th edition
4. Lectures on the Theory of Elliptic Functions – Hancock

## Integrals of Gradshteyn and Ryzhik: 3.267

We use the following definitions of the gamma and beta functions:

\Gamma(z) = \int\limits_{0}^{\infty} \mathrm{e}^{-t} t^{z-1} dt, \,\, \mathfrak{R}(z) \gt 0

\mathrm{B}(a,b) = \int\limits_{0}^{1} t^{a-1} (1-t)^{b-1} dt
= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}, \,\, \mathfrak{R}(a) \gt 0 \,\, \mathrm{and} \,\, \mathfrak{R}(b) \gt 0

## 3.267.1

\begin{align}
\int\limits_{0}^{1} x^{3n} (1-x^3)^{-1/3} dx
&= \frac{1}{3} \int\limits_{0}^{1} y^{n-2/3} (1-y)^{-1/3} dy
= \frac{1}{3} \mathrm{B}\left( \frac{2}{3},n+\frac{1}{3} \right) \\
&= \frac{\Gamma\left(\frac{2}{3}\right) \Gamma\left(n + \frac{1}{3}\right)}{3\Gamma(n+1)}
= \frac{2\pi}{3\sqrt{3}} \frac{\Gamma\left(n + \frac{1}{3}\right)}{\Gamma\left(\frac{1}{3}\right) \Gamma(n+1)}
\end{align}

## 3.267.2

\begin{align}
\int\limits_{0}^{1} x^{3n-1} (1-x^3)^{-1/3} dx
&= \frac{1}{3} \int\limits_{0}^{1} y^{n-1} (1-y)^{-1/3} dy
= \frac{1}{3} \mathrm{B}\left(n, \frac{2}{3} \right) \\
&= \frac{\Gamma(n)\Gamma(2/3)}{3\Gamma\left(n + \frac{2}{3}\right)}
= \frac{(n-1)! \Gamma(2/3)}{3\Gamma\left(n + \frac{2}{3}\right)}
\end{align}

## 3.267.3

\begin{align}
\int\limits_{0}^{1} x^{3n-2} (1-x^3)^{-1/3} dx
&= \frac{1}{3} \int\limits_{0}^{1} y^{n-4/3} (1-y)^{-1/3} dy
= \frac{1}{3} \mathrm{B}\left( \frac{2}{3}, n \, – \frac{1}{3} \right) \\
&= \frac{\Gamma\left(n – \frac{1}{3}\right)\Gamma(2/3)}{3\Gamma\left(n + \frac{1}{3}\right)}
\end{align}

we used the substitution $$y=x^3$$ for all three integrals.

## Integrals of Gradshteyn and Ryzhik: 3.322

We define the error function as:

\mathrm{erf}(z) = \frac{2}{\sqrt{\pi}} \int\limits_{0}^{z} \mathrm{e}^{-x^2} dx

## 3.322.1

\begin{align}
\int\limits_{u}^{\infty} \mathrm{e}^{-2ax-x^2} dx
&= \mathrm{e}^{a^2} \int\limits_{u}^{\infty} \mathrm{e}^{-(x+a)^2} dx
= \mathrm{e}^{a^2} \int\limits_{u+a}^{\infty} \mathrm{e}^{-y^2} dy \\
&= \frac{\sqrt{\pi}}{2} \mathrm{e}^{a^2} \mathrm{erf}(y) \Bigg|_{u+a}^{\infty} \\
&= \frac{\sqrt{\pi}}{2} \mathrm{e}^{a^2} \left[1-\,\mathrm{erf}(u+a)\right],\,\, u \gt 0
\end{align}
We completed the square in the argument of the exponential function and then used
the substitution $$y=x+a$$.

## 3.322.2

\int\limits_{0}^{\infty} \mathrm{e}^{-2ax-x^2} dx =
\frac{\sqrt{\pi}}{2} \mathrm{e}^{a^2} \left[1-\,\mathrm{erf}(a)\right]

follows from 3.322.1.

## 3.322.3

\mathrm{PV} \int\limits_{0}^{\infty} \mathrm{e}^{\pm i\lambda x^2} dx
= \frac{1}{\sqrt{\lambda}} \int\limits_{0}^{\infty} \mathrm{e}^{\pm i y^2} dy
= \frac{1}{2}\sqrt{\frac{\pi}{\lambda}} \mathrm{e}^{\pm i \pi /4}

we let $$y^2=\lambda x^2$$ and this problem was solved here.

## Integrals of Gradshteyn and Ryzhik: 3.321

We define the error function as:

\mathrm{erf}(z) = \frac{2}{\sqrt{\pi}} \int\limits_{0}^{z} \mathrm{e}^{-x^2} dx

## 3.321.2

\int\limits_{0}^{u} \mathrm{e}^{-q^{2}x^{2}} dx = \frac{1}{q} \int\limits_{0}^{qu} \mathrm{e}^{-z^{2}} dz
= \frac{\sqrt{\pi}}{2q} \mathrm{erf}(z) \Big|_{0}^{qu}
= \frac{\sqrt{\pi}}{2q} \mathrm{erf}(qu),\,\,q \gt 0

using the substitution $$z=qx$$

## 3.321.3

\int\limits_{0}^{\infty} \mathrm{e}^{-q^{2}x^{2}} dx = \frac{\sqrt{\pi}}{2q},\,\,q \gt 0

follows from 3.321.2

## 3.321.4

I = \int\limits_{0}^{u} x\mathrm{e}^{-q^{2}x^{2}} dx
= \frac{\sqrt{\pi}}{2q} \,x\, \mathrm{erf}(qx) \,- \frac{\sqrt{\pi}}{2q} \int \mathrm{erf}(qx) dx

using integration by parts.
\begin{align}
\int \mathrm{erf}(qx) dx &= \frac{1}{q} \int \mathrm{erf}(y) dy \\
&= \frac{1}{q} \left[y\,\mathrm{erf}(y) – \frac{2}{\sqrt{\pi}} \int y\,\mathrm{e}^{-y^2} dy \right] \\
&= \frac{1}{q} y\,\mathrm{erf}(y) + \frac{1}{q\sqrt{\pi}} \mathrm{e}^{-y^2} \\
&= x\,\mathrm{erf}(qx) + \frac{1}{q\sqrt{\pi}} \mathrm{e}^{-q^{2}x^{2}}
\end{align}
using integration by parts and the substitution $$y=qx$$

I = \frac{\sqrt{\pi}}{2q} x \,\mathrm{erf}(qx) \,- \frac{\sqrt{\pi}}{2q} x \,\mathrm{erf}(qx)
\,- \frac{1}{2q^2} \mathrm{e}^{-q^{2}x^{2}} \Big|_{0}^{u}
= \frac{1}{2q^2} \left[1 – \mathrm{e}^{-q^{2}u^{2}} \right]

## 3.321.5

\begin{align}
I &= \int\limits_{0}^{u} x^{2} \mathrm{e}^{-q^{2}x^{2}} dx
= \frac{-1}{2q^2} x \mathrm{e}^{-q^{2}x^{2}} + \frac{1}{2q^2} \int\mathrm{e}^{-q^{2}x^{2}} dx \\
&= \frac{1}{2q^3} \left[-qx \mathrm{e}^{-q^{2}x^{2}} + \frac{\sqrt{\pi}}{2} \mathrm{erf}(qx) \right]\Big|_{0}^{u} \\
&= \frac{1}{2q^3} \left[\frac{\sqrt{\pi}}{2} \mathrm{erf}(qu) \,-\, qu \mathrm{e}^{-q^{2}x^{2}} \right]
\end{align}
We used integration by parts, 3.321.4, and 3.321.2.

## 3.321.6

\begin{align}
I &= \int\limits_{0}^{u} x^{3} \mathrm{e}^{-q^{2}x^{2}} dx
= \frac{\sqrt{\pi}}{4q^3}x\,\mathrm{erf}(qx) \,- \frac{x^2}{2q^2}\mathrm{e}^{-q^{2}x^{2}}
– \frac{\sqrt{\pi}}{4q^3} \int \mathrm{erf}(qx) dx + \frac{1}{2q^2} \int x\,\mathrm{e}^{-q^{2}x^{2}} dx \\
&= \frac{\sqrt{\pi}}{4q^3}\,x\,\mathrm{erf}(qx) \,- \frac{x^2}{2q^2}\mathrm{e}^{-q^{2}x^{2}}
– \frac{\sqrt{\pi}}{4q^3}\,x\,\mathrm{erf}(qx) \,- \frac{1}{4q^4}\mathrm{e}^{-q^{2}x^{2}}
– \frac{1}{4q^4}\mathrm{e}^{-q^{2}x^{2}} \\
&= \frac{1}{2q^4} (-q^{2}x^{2} – 1)\mathrm{e}^{-q^{2}x^{2}}\Big|_{0}^{u} \\
&= \frac{1}{2q^4} \left[1-(1+q^{2}u^{2}) \mathrm{e}^{-q^{2}u^{2}} \right]
\end{align}
using integration by parts, 3.321.5, the work done in the solution of 3.321.4 and the result 3.321.4.

## 3.321.7

\begin{align}
I &= \int\limits_{0}^{u} x^{4} \mathrm{e}^{-q^{2}x^{2}} dx
= -\frac{1}{2q^4}x\mathrm{e}^{-q^{2}x^{2}} – \frac{1}{2q^4}q^{2}x^{3}\mathrm{e}^{-q^{2}x^{2}}
+ \frac{1}{2q^4}\int \mathrm{e}^{-q^{2}x^{2}} dx + \frac{q^{2}}{2q^4}\int x^{2}\mathrm{e}^{-q^{2}x^{2}} dx \\
&= \frac{1}{2q^5}\left[-qx\,\mathrm{e}^{-q^{2}x^{2}} – q^{3}x^{3}\,\mathrm{e}^{-q^{2}x^{2}}
+ \frac{\sqrt{\pi}}{2} \mathrm{erf}(qx) + \frac{\sqrt{\pi}}{4} \mathrm{erf}(qx)
-\frac{1}{2}qx\,\mathrm{e}^{-q^{2}x^{2}} \right]\Big|_{0}^{u} \\
&= \frac{1}{2q^5} \left[\frac{3\sqrt{\pi}}{4}\mathrm{erf}(qx)
– qx\,\mathrm{e}^{-q^{2}x^{2}}\left(\frac{3}{2} + q^{2}x^{2}\right) \right]\Big|_{0}^{u} \\
&= \frac{1}{2q^5} \left[\frac{3\sqrt{\pi}}{4}\mathrm{erf}(qu)
– qu\,\mathrm{e}^{-q^{2}u^{2}}\left(\frac{3}{2} + q^{2}u^{2}\right) \right]
\end{align}
using integration by parts, 3.321.6, 3.321.2, and 3.321.5.

## Properties of Ultra Gamma Function by Kuldeep Singh Gehlot

In this paper we study the integral of type

{}_{\delta , a}\Gamma_{\rho , b}(x) = \Gamma(\delta , a; \rho , b)(x) = \int\limits_{0}^{\infty} t^{x-1} \mathrm{exp}\left( -\frac{t^{\delta}}{a}-\frac{1}{b t^{\rho}} \right) dt

Different authors called this integral by different names like ultra gamma function, generalized gamma function, Kratzel integral, inverse Gaussian integral, reaction-rate probability integral, Bessel integral etc. We prove several identities and recurrence relation of above said integral, we called this integral as Four Parameter Gamma Function. Also we evaluate relation between Four Parameter Gamma Function, p-k Gamma Function and Classical Gamma Function. With some conditions we can evaluate Four Parameter Gamma Function in term of Hypergeometric function.

The entire paper is available here.

## Legendre-type relations for generalized complete elliptic integrals by Shingo Takeuchi

Legendre’s relation for the complete elliptic integrals of the first and second kinds is generalized. The proof depends on an application of the generalized trigonometric functions and is alternative to the proof for Elliott’s identity.

The entire paper is available here.

## On The Extended Incomplete Pochhammer Symbols and Hypergeometric Functions by Rakesh Kumar Parmar, R.K. Raina

In this paper, we first introduce certain forms of extended incomplete Pochhammer symbols which are then used to define families of extended incomplete generalized hypergeometric functions. For these functions, we investigate various properties including the integral representations, derivative formula, certain generating function and fractional integrals (and derivatives) relationships. Some special cases of the main results are also deduced.

The entire paper is available here.

## A Collection of Free Math Books from Folkscanomy Mathematics: Books of a Mathematic Nature

See Folkscanomy Mathematics: Books of a Mathematic Nature. Here, you can find such books as The Theory of Functions by E.C. Titchmarsh and A Treatise on the Theory of Bessel Functions by G.N. Watson.

## Some formulae for products of Fubini polynomials with applications by Levent Kargın

In this paper we evaluate sums and integrals of products of Fubini polynomials and have new explicit formulas for Fubini polynomials and numbers. As a consequence of these results new explicit formulas for p-Bernoulli numbers and Apostol-Bernoulli functions are given. Besides, integrals of products of Apostol-Bernoulli functions are derived.

The entire paper is available here.

## Higher order generalized geometric polynomials by Levent Kargin, Bayram Çekim

According to generalized Mellin derivative (Kargin), we introduce a new family of polynomials called higher order generalized geometric polynomials. We obtain some properties of them.We discuss their connections to degenerate Bernoulli and Euler polynomials. Furthermore, we find new formulas for the Carlitz’s (Carlitz) and Howard’s (Howard2) finite sums. Finally, we evaluate several series in closed forms, one of which has the coefficients include values of the Riemann zeta function. Moreover, we calculate some integrals in terms of generalized geometric polynomials.

The entire paper is available here.