## Integrals of Gradshteyn and Ryzhik: 6.161 – Mellin Transforms of Theta Functions with Respect to the Lattice Parameter

We use definitions of the theta functions, shown below, from GR. Note that there is no standard notation for the theta functions.

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

\begin{align}
\tag{1a}
\label{eq:theta1e}
\theta_{1}(z|\tau) &= \theta_{1}(z,q) = 2 \sum_{n=0}^{\infty} (-1)^{n} q^{(n+1/2)^{2}} \sin[(2n+1)z] \\
\tag{1b}
\label{eq:theta1t}
&= -i \sum_{n=-\infty}^{\infty} (-1)^{n} q^{(n+1/2)^{2}} \mathrm{e}^{i(2n+1)z}
\end{align}

\begin{align}
\tag{2a}
\label{eq:theta2e}
\theta_{2}(z|\tau) &= \theta_{2}(z,q) = 2 \sum_{n=0}^{\infty} q^{(n+1/2)^{2}} \cos[(2n+1)z] \\
\tag{2b}
\label{eq:theta2t}
&= \sum_{n=-\infty}^{\infty} q^{(n+1/2)^{2}} \mathrm{e}^{i(2n+1)z}
\end{align}

\begin{align}
\tag{3a}
\label{eq:theta3e}
\theta_{3}(z|\tau) &= \theta_{3}(z,q) = 1 + 2 \sum_{n=1}^{\infty} q^{n^{2}} \cos(2nz) \\
\tag{3b}
\label{eq:theta3t}
&= \sum_{n=-\infty}^{\infty} q^{n^{2}} \mathrm{e}^{i2nz}
\end{align}

\begin{align}
\tag{4a}
\label{eq:theta4e}
\theta_{4}(z|\tau) &= \theta_{4}(z,q) = 1 + 2 \sum_{n=1}^{\infty} q^{n^{2}} \cos(2nz) \\
\tag{4b}
\label{eq:theta4t}
&= \sum_{n=-\infty}^{\infty} q^{n^{2}} \mathrm{e}^{i2nz}
\end{align}

## 6.161.1

\int_{0}^{\infty} x^{s-1} \theta_{2}(0|ix^2) dx
= 2\int_{0}^{\infty} x^{s-1} \sum_{n=0}^{\infty} \mathrm{e}^{-\pi x^{2}(n+1/2)^2} dx

\begin{align}
\int_{0}^{\infty} x^{s-1} \mathrm{e}^{-\pi x^{2}(n+1/2)^2} dx
&= \frac{1}{2(\pi (n+1/2)^2)^{s/2}} \int_{0}^{\infty} y^{-1+s/2} \mathrm{e}^{-y} dy \\
&= \frac{1}{2(\pi (n+1/2)^2)^{s/2}} \Gamma(s/2)
\end{align}

We used the substitution $$y = \pi x^{2}(n+1/2)^2$$.

Now we have
\begin{align}
\int_{0}^{\infty} x^{s-1} \theta_{2}(0|ix^2) dx &=
\frac{1}{\pi ^{s/2}} \Gamma(s/2) \sum_{n=0}^{\infty} \frac{1}{(n+1/2)^s} \\
&= \frac{2^s}{\pi ^{s/2}} \Gamma(s/2) \sum_{n=0}^{\infty} \frac{1}{(2n+1)^s} \\
&= \frac{2^s}{\pi ^{s/2}} \Gamma(s/2) (1-2^{-s}) \zeta(s)
\end{align}

## 6.161.2

\begin{align}
\int_{0}^{\infty} x^{s-1} [\theta_{3}(0|ix^2) \,- 1] dx
&= \int_{0}^{\infty} x^{s-1} \left(\left[1 + 2\sum_{n=1}^{\infty} \mathrm{e}^{-\pi x^{2} n^2} \right] -1 \right) dx \\
&= 2\sum_{n=1}^{\infty} \int_{0}^{\infty} x^{s-1} \mathrm{e}^{-\pi x^{2} n^2} dx \\
&= 2\sum_{n=1}^{\infty} \frac{\Gamma(s/2)}{2\pi^{s/2} n^{s}} \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} \sum_{n=1}^{\infty} \frac{1}{n^{s}} \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} \zeta(s)
\end{align}

We used the substitution $$y = \pi x^{2} n^2$$ and the same work as above to evaluate the integral.

## 6.161.3

\begin{align}
\int_{0}^{\infty} x^{s-1} [1 – \theta_{4}(0|ix^2)] dx
&= \int_{0}^{\infty} x^{s-1} \left(1 – \left[1 + 2\sum_{n=1}^{\infty} (-1)^{n} \mathrm{e}^{-\pi x^{2} n^2} \right] \right) dx \\
&= -2\sum_{n=1}^{\infty} (-1)^{n} \int_{0}^{\infty} x^{s-1} \mathrm{e}^{-\pi x^{2} n^2} dx \\
&= -2\sum_{n=1}^{\infty} (-1)^{n} \frac{\Gamma(s/2)}{2\pi^{s/2} n^{s}} \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{s}} \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} (1 – 2^{1-s}) \zeta(s)
\end{align}

We used the substitution $$y = \pi x^{2} n^2$$ and the same work as above to evaluate the integral.

## 6.161.4

\begin{align}
\int_{0}^{\infty} x^{s-1} [\theta_{4}(0|ix^2) + \theta_{2}(0|ix^2) \,- \theta_{3}(0|ix^2)] dx
&= \int_{0}^{\infty} x^{s-1} ([\theta_{4}(0|ix^2) \,-1] + \theta_{2}(0|ix^2) \,- [\theta_{3}(0|ix^2) \,-1]) dx \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} (2^{1-s} – 1) \zeta(s)
+ \frac{2^s}{\pi ^{s/2}} \Gamma(s/2) (1-2^{-s}) \zeta(s)
\,- \frac{\Gamma(s/2)}{\pi^{s/2}} \zeta(s) \\
&= -\frac{\Gamma(s/2)}{\pi^{s/2}} \zeta(s) [1 – 2^{1-s} – 2^{-s} + 1 + 1] \\
&= -\frac{\Gamma(s/2)}{\pi^{s/2}} \zeta(s) (2^{-s} – 1)(2^{1-s} – 1)
\end{align}
Here we used the previous 3 results.

All Riemann zeta function expressions can be found here.

## 5.136.1

\int \mathrm{sn}\,\mathrm{cn} = -\frac{1}{k^2}\mathrm{dn}

## 5.136.2

\int \mathrm{sn}\,\mathrm{dn} = -\mathrm{cn}

## 5.136.3

\int \mathrm{cn}\,\mathrm{dn} = \mathrm{sn}

The three integrals above follow from the derivatives of the Jacobi elliptic functions.

## 5.137.1

\int \frac{\mathrm{sn}}{\mathrm{cn}^2} = \frac{1}{{k^{\prime}}^2} \frac{\mathrm{dn}}{\mathrm{cn}} = \frac{1}{{k^{\prime}}^2} \mathrm{dc}

## 5.137.2

\int \frac{\mathrm{sn}}{\mathrm{dn}^2} = -\frac{1}{{k^{\prime}}^2} \frac{\mathrm{cn}}{\mathrm{dn}}
= -\frac{1}{{k^{\prime}}^2} \mathrm{cd}

## 5.137.3

\int \frac{\mathrm{cn}}{\mathrm{sn}^2} = -\frac{\mathrm{dn}}{\mathrm{sn}} = -\mathrm{ds}

## 5.137.4

\int \frac{\mathrm{cn}}{\mathrm{dn}^2} = \frac{\mathrm{sn}}{\mathrm{dn}} = \mathrm{sd}

## 5.137.5

\int \frac{\mathrm{dn}}{\mathrm{sn}^2} = -\frac{\mathrm{cn}}{\mathrm{sn}} = -\mathrm{cs}

## 5.137.6

\int \frac{\mathrm{dn}}{\mathrm{cn}^2} = \frac{\mathrm{sn}}{\mathrm{cn}} = \mathrm{sc}

The six integrals above follow from the derivatives of the Jacobi elliptic functions.

## 5.138.1

\begin{align}
\int \frac{\mathrm{cn}}{\mathrm{sn}\,\mathrm{dn}} &= \int \frac{\mathrm{cn}}{\mathrm{sn}\,\mathrm{dn}} \frac{\mathrm{dn}}{\mathrm{dn}} = \int \frac{\mathrm{cn}}{\mathrm{dn}^2} \frac{1}{\mathrm{sd}} \\
&= \int \frac{1}{w} = \ln(\mathrm{sd}) = \ln\left(\frac{\mathrm{sn}}{\mathrm{dn}}\right)
\end{align}
We used the substitution $$w=\mathrm{sd}$$.

## 5.138.2

\begin{align}
\int \frac{\mathrm{sn}}{\mathrm{cn}\,\mathrm{dn}} &= \int \frac{\mathrm{sn}}{\mathrm{cn}\,\mathrm{dn}} \frac{\mathrm{cn}}{\mathrm{cn}} = \int \frac{\mathrm{sn}}{\mathrm{cn}^2} \frac{1}{\mathrm{dc}} \\
&= \frac{1}{{k^{\prime}}^2} \int \frac{1}{w} = \frac{1}{{k^{\prime}}^2} \ln(\mathrm{dc}) = \frac{1}{{k^{\prime}}^2} \ln\left(\frac{\mathrm{dn}}{\mathrm{cn}}\right)
\end{align}
We used the substitution $$w=\mathrm{dc}$$.

## 5.138.3

\begin{align}
\int \frac{\mathrm{dn}}{\mathrm{sn}\,\mathrm{cn}} &= \int \frac{\mathrm{dn}}{\mathrm{sn}\,\mathrm{cn}} \frac{\mathrm{cn}}{\mathrm{cn}} = \int \frac{\mathrm{dn}}{\mathrm{cn}^2} \frac{1}{\mathrm{sc}} \\
&= \int \frac{1}{w} = \ln(\mathrm{sc}) = \ln\left(\frac{\mathrm{sn}}{\mathrm{cn}}\right)
\end{align}
We used the substitution $$w=\mathrm{sc}$$.

## 5.139.1

\int \frac{\mathrm{cn}\,\mathrm{dn}}{\mathrm{sn}} = \int \frac{1}{w} = \ln(\mathrm{sn})

## 5.139.2

\begin{align}
\int \frac{\mathrm{sn}\,\mathrm{dn}}{\mathrm{cn}} &= \int \frac{\mathrm{sn}\,\mathrm{dn}}{\mathrm{cn}} \frac{\mathrm{cn}}{\mathrm{cn}} = \int \frac{\mathrm{sn}\,\mathrm{dn}}{\mathrm{cn}^2} \frac{1}{\mathrm{nc}} \\
&= \int \frac{1}{w} = \ln(\mathrm{nc}) = \ln\left(\frac{1}{\mathrm{cn}}\right)
\end{align}
We used the substitution $$w=\mathrm{nc}$$.

## 5.139.3

\int \frac{\mathrm{cn}\,\mathrm{sn}}{\mathrm{dn}} = -\frac{1}{k^2} \int \frac{1}{w} = -\frac{1}{k^2} \ln(\mathrm{dn})

## Derivatives of Jacobi Elliptic Functions

Here we compute derivatives of 9 of the Jacobi elliptic functions with respect to the argument $$u$$. Three were derived here, and are reproduced below.

For convenience, we drop the argument and modulus.

\frac{\partial \,\mathrm{sn}}{\partial u} = \mathrm{cn}\,\mathrm{dn}

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

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

\frac{\partial \,\mathrm{ns}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{sn}} = -\frac{\mathrm{cn}\,\mathrm{dn}}{\mathrm{sn}^2} = -\mathrm{cs}\,\mathrm{ds}

\frac{\partial \,\mathrm{nc}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{cn}} = \frac{\mathrm{dn}\,\mathrm{sn}}{\mathrm{cn}^2} = \mathrm{sc}\,\mathrm{dc}

\frac{\partial \,\mathrm{nd}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{dn}} = k^{2}\frac{\mathrm{cn}\,\mathrm{sn}}{\mathrm{dn}^2} = k^{2}\mathrm{sd}\,\mathrm{cd}

\frac{\partial \,\mathrm{cs}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{sc}}
= \frac{\partial}{\partial u}\frac{\mathrm{cn}}{\mathrm{sn}} = -\frac{\mathrm{dn}}{\mathrm{sn}^2}
= -\mathrm{dn}\,\mathrm{ns}^2

\frac{\partial \,\mathrm{sd}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{ds}}
= \frac{\partial}{\partial u}\frac{\mathrm{sn}}{\mathrm{dn}} = \frac{\mathrm{cn}}{\mathrm{dn}^2}
= \mathrm{cn}\,\mathrm{nd}^2

\frac{\partial \,\mathrm{ds}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{sd}}
= \frac{\partial}{\partial u}\frac{\mathrm{dn}}{\mathrm{sn}} = -\frac{\mathrm{cn}}{\mathrm{sn}^2}
= -\mathrm{cn}\,\mathrm{ns}^2

\frac{\partial \,\mathrm{sc}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{cs}}
= \frac{\partial}{\partial u}\frac{\mathrm{sn}}{\mathrm{cn}} = \frac{\mathrm{dn}}{\mathrm{cn}^2}
= \mathrm{nc}\,\mathrm{dc}

\frac{\partial \,\mathrm{cd}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{dc}}
= \frac{\partial}{\partial u}\frac{\mathrm{cn}}{\mathrm{dn}} = -{k^{\prime}}^2\frac{\mathrm{sn}}{\mathrm{dn}^2}
= -{k^{\prime}}^2\mathrm{sn}\,\mathrm{nd}^2

\frac{\partial \,\mathrm{dc}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{cd}}
= \frac{\partial}{\partial u}\frac{\mathrm{dn}}{\mathrm{cn}} = {k^{\prime}}^2\frac{\mathrm{sn}}{\mathrm{cn}^2}
= {k^{\prime}}^2\mathrm{sn}\,\mathrm{nc}^2

## Relationships Between Squares of Jacobi Elliptic Functions

In this post, we derive a number of relationships between squares of Jacobi elliptic functions. The derivations are trivial, but the results will be required for subsequent posts involving derivatives and integrals of Jacobi elliptic functions. The numbering scheme will also be referenced in future posts.

We drop the arguments for convenience, thus $$\mathrm{cn}(u,k)=\mathrm{cn}$$.

## The 12 Jacobi Elliptic Functions

Beginning with $$\mathrm{cn}, \mathrm{dn}, \mathrm{sn}$$ we define 9 other functions using Glaisher’s notation.

## Relationships Between Squares of Jacobi Elliptic Functions

We begin with 3 results from Theta Functions and Jacobi Elliptic Functions.

\mathrm{cn}^{2} + \mathrm{sn}^{2} = 1
\label{eq:pythag-1}
\tag{1}

\mathrm{dn}^{2} + k^{2}\mathrm{sn}^{2} = 1
\label{eq:pythag-2}
\tag{2}

k^{2} + {k^{\prime}}^{2} = 1
\label{eq:pythag-3}
\tag{3}

Divide equation \eqref{eq:pythag-1} by $$\mathrm{cn}^{2}$$ to obtain:

1 + \mathrm{sc}^{2} = \mathrm{nc}^{2}
\label{eq:pythag-4}
\tag{4}

Divide equation \eqref{eq:pythag-1} by $$\mathrm{sn}^{2}$$ to obtain:

\mathrm{cs}^{2} + 1 = \mathrm{ns}^{2}
\label{eq:pythag-5}
\tag{5}

Divide equation \eqref{eq:pythag-2} by $$\mathrm{dn}^{2}$$ to obtain:

\mathrm{cd}^{2} + \mathrm{sd}^{2} = \mathrm{nd}^{2}
\label{eq:pythag-6}
\tag{6}

Divide equation \eqref{eq:pythag-2} by $$\mathrm{dn}^{2}$$ to obtain:

1 + k^{2}\mathrm{sd}^{2} = \mathrm{nd}^{2}
\label{eq:pythag-7}
\tag{7}

Divide equation \eqref{eq:pythag-2} by $$\mathrm{cn}^{2}$$ to obtain:

\mathrm{dc}^{2} + k^{2}\mathrm{sc}^{2} = \mathrm{nc}^{2}
\label{eq:pythag-8}
\tag{8}

Divide equation \eqref{eq:pythag-2} by $$\mathrm{sn}^{2}$$ to obtain:

\mathrm{ds}^{2} + k^{2} = \mathrm{ns}^{2}
\label{eq:pythag-9}
\tag{9}

Eliminate $$\mathrm{nd}^{2}$$ from equations \eqref{eq:pythag-6} and \eqref{eq:pythag-7} and use equation \eqref{eq:pythag-3} to obtain:

\mathrm{cd}^{2} + {k^{\prime}}^{2}\mathrm{sd}^{2} = 1
\label{eq:pythag-10}
\tag{10}

Eliminate $$\mathrm{sd}^{2}$$ from equations \eqref{eq:pythag-6} and \eqref{eq:pythag-7} and use equation \eqref{eq:pythag-3} to obtain:

1 = k^{2}\mathrm{cd}^{2} + {k^{\prime}}^{2}\mathrm{nd}^{2}
\label{eq:pythag-11}
\tag{11}

Eliminate $$\mathrm{nc}^{2}$$ from equations \eqref{eq:pythag-4} and \eqref{eq:pythag-8} and use equation \eqref{eq:pythag-3} to obtain:

1 + {k^{\prime}}^{2}\mathrm{sc}^{2} = \mathrm{dc}^{2}
\label{eq:pythag-12}
\tag{12}

Eliminate $$\mathrm{sc}^{2}$$ from equations \eqref{eq:pythag-4} and \eqref{eq:pythag-8} and use equation \eqref{eq:pythag-3} to obtain:

\mathrm{dc}^{2} = k^{2} + {k^{\prime}}^{2}\mathrm{nc}^{2}
\label{eq:pythag-13}
\tag{13}

Eliminate $$\mathrm{ns}^{2}$$ from equations \eqref{eq:pythag-5} and \eqref{eq:pythag-9} and use equation \eqref{eq:pythag-3} to obtain:

\mathrm{cs}^{2} + {k^{\prime}}^{2} = \mathrm{ds}^{2}
\label{eq:pythag-14}
\tag{14}

Multiply equations \eqref{eq:pythag-11} by $$\mathrm{ns}^{2}$$ to obtain:

\mathrm{ds}^{2} = k^{2}\mathrm{cs}^{2} + {k^{\prime}}^{2}\mathrm{ns}^{2}
\label{eq:pythag-15}
\tag{15}

Eliminate $$\mathrm{sn}^{2}$$ from equations \eqref{eq:pythag-1} and \eqref{eq:pythag-2} and use equation \eqref{eq:pythag-3} to obtain:

\mathrm{dn}^{2} = k^{2}\mathrm{cn}^{2} + {k^{\prime}}^{2}
\label{eq:pythag-16}
\tag{16}

Eliminate 1 from equations \eqref{eq:pythag-1} and \eqref{eq:pythag-2} and use equation \eqref{eq:pythag-3} to obtain:

\mathrm{dn}^{2} = \mathrm{cn}^{2} + {k^{\prime}}^{2}\mathrm{sn}^{2}
\label{eq:pythag-17}
\tag{17}

## References

1. Handbook of Elliptic Integrals for Engineers and Scientists – Byrd and Friedman

## 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.