# Gudermannian Function

## Introduction

The Gudermannian function, named in honor of Christoph Gudermann, is a way to relate the trigonometric and hyperbolic functions without the use of complex variables. The basic definition is

\mathrm{gd} u = \int\limits_{0}^{u} \frac{1}{\cosh x} dx
\tag{1}
\label{eq:g1}

## Derivation of the Basic Definition and Basic Properties

To derive equation \eqref{eq:g1}, we begin with the hyperbola $$x^{2} – y^{2} = a^{2}$$. A point on the hyperbola has coordinates $$(x,y)$$ where

x = a \sec\theta = a \cosh u \quad \mathrm{and} \quad y = a \tan\theta = a \sinh u

Thus we have:

Using basic trigonometry we obtain

\cot\theta = \mathrm{csch} u \,\, \cos\theta = \mathrm{sech} u \,\, \sin\theta = \tanh u \,\,
\csc\theta = \coth u

We solve $$\tan\theta = \sinh u$$ for $$\theta$$ to obtain, $$\theta = \tan^{-1}(\sinh u)$$. Differentiating yields

\frac{d\theta}{du} = \frac{\cosh u}{1 + \sinh^{2} u} = \frac{1}{\cosh u}

Integrating yields

\theta = \int_{0}^{u} \frac{1}{\cosh t} dt

If we designate $$\theta$$ as the Gudermannian of $$u$$, $$\theta = \mathrm{gd} u$$, we obtain Equation \eqref{eq:g1}. Also, we can rewrite our relations between the trig and hyperbolic functions as
\begin{align}
\sec(\mathrm{gd} u) &= \cosh u, \quad \tan(\mathrm{gd} u) = \sinh u, \quad \cot(\mathrm{gd} u) = \mathrm{csch} u \\
\cos(\mathrm{gd} u) &= \mathrm{sech} u, \quad \sin(\mathrm{gd} u) = \tanh u, \quad \csc(\mathrm{gd} u) = \coth u
\end{align}

1. Let $$y = \tanh z/2$$
\begin{align}
\frac{1}{2} \mathrm{gd} u &= \frac{1}{2} \int\limits_{0}^{u} \frac{1}{\cosh z} dz
= \int\limits_{0}^{u} \frac{1}{\mathrm{e}^{z}+\mathrm{e}^{-z}} dz \\
&= \int\limits_{0}^{\tanh u/2} \frac{1}{y^{2}+1} dy \\
&= \tan^{-1}\left(\tanh\frac{u}{2} \right)
\end{align}
We can rewrite this as

\tan\left(\frac{1}{2} \mathrm{gd} u \right) = \tanh\frac{u}{2}

2. Let $$y= \mathrm{e}^{z}$$
\begin{align}
\mathrm{gd} u &= 2\int\limits_{0}^{u} \frac{1}{\mathrm{e}^{z}+\mathrm{e}^{-z}} dz
= 2\int\limits_{0}^{u} \frac{\mathrm{e}^{z}}{\mathrm{e}^{2z}+1} dz \\
&= 2\int\limits_{1}^{\mathrm{e}^{u}} \frac{1}{y^{2}+1} dy \\
&= 2\tan^{-1}\mathrm{e}^{u} \, -\frac{\pi}{2}
\tag{2}
\label{eq:g2}
\end{align}

## Derivative of the Gudermannian

\frac{d}{dx}\mathrm{gd}x = \mathrm{sech}(x)

follows from the integral definition.

## Maclaurin Series

$$\mathrm{gd}(0) = 0$$ follows from the integral definition.

\mathrm{gd}^{\prime}(0) = \mathrm{sech}(0) = 1

\mathrm{gd}^{\prime \prime}(0) = \mathrm{sech}^{\prime}(0) = -\mathrm{sech}(0)\tanh(0) = 0

\mathrm{gd}^{\prime \prime \prime}(0) = \mathrm{sech}^{\prime \prime}(0) = \mathrm{sech}(0)\tanh^{2}(0) – \mathrm{sech}^{3}(0) = -1

\mathrm{gd}x = \sum\limits_{n=0}^{\infty} \frac{x^{n}}{n!} \frac{d^{n}\mathrm{gd}(0)}{dx^{n}}
= x \, – \frac{x^{3}}{6} + \cdots

## The Gudermannian is an Odd Function

$$\mathrm{gd}(-x) = -\mathrm{gd}x$$ follows from the Maclaurin series.

## The Exponential Function and the Gudermannian

First:
\begin{align}
\mathrm{e}^{x} &= \sec(\mathrm{gd}x) + \tan(\mathrm{gd}x) \\
&= \cosh x + \sinh x \\
&= \frac{\mathrm{e}^{x} + \mathrm{e}^{-x}}{2} + \frac{\mathrm{e}^{x} – \mathrm{e}^{-x}}{2} \\
&= \mathrm{e}^{x}
\end{align}
Second:
\begin{align}
\mathrm{e}^{x} &= \frac{1+\sin(\mathrm{gd}x)}{\cos(\mathrm{gd}x)} \\
&= \sec(\mathrm{gd}x) + \tan(\mathrm{gd}x) \\
&= \mathrm{e}^{x}
\end{align}
Third:
\begin{align}
\mathrm{e}^{x} &= \tan\left(\frac{\pi}{4} + \frac{1}{2}\mathrm{gd}x \right) \\
&= \sec(\mathrm{gd}x) + \tan(\mathrm{gd}x) \\
&= \mathrm{e}^{x}
\end{align}

## Integral of the Gudermannian

Begin with Equation \eqref{eq:g2}, $$\mathrm{gd}x = 2\tan^{-1}\mathrm{e}^{x} \, – \pi/2$$ so that we have

\int \mathrm{gd}x \, dx = 2\int \tan^{-1}\mathrm{e}^{x} dx – \int \frac{\pi}{2} dx

Using

\tan^{-1}z = \frac{i}{2}\Big[\ln(1-iz) – \ln(1+iz)\Big]

and integrating this expression, yields
\begin{align}
\int \tan^{-1}\mathrm{e}^{x} dx &= \frac{i}{2}\int\Big[\ln(1-i\mathrm{e}^{x}) – \ln(1+i\mathrm{e}^{x})\Big] dx \\
&= \frac{i}{2} [-\mathrm{Li}_{2}(i\mathrm{e}^{x}) + \mathrm{Li}_{2}(-i\mathrm{e}^{x})]
\end{align}
We used the substitution $$y=-i\mathrm{e}^{x}$$ and the dilogarithm function $$\mathrm{Li}_{2}(z)$$.
Putting the pieces together, we obtain

\int \mathrm{gd}x \, dx = i[\mathrm{Li}_{2}(-i\mathrm{e}^{x})-\mathrm{Li}_{2}(i\mathrm{e}^{x})] – \frac{\pi}{2}x

## References

1. Wolfram MathWorld
2. Wikipedia
3. A Treatise on the Integral Calculus – Joseph Edwards; Chapter 3, Article 69.