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

gudermannian-function-new

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

Thus we have:
\begin{equation}
\sec\theta = \cosh u \quad \mathrm{and} \quad \tan\theta = \sinh u
\end{equation}
Using basic trigonometry we obtain
\begin{equation}
\cot\theta = \mathrm{csch} u \,\, \cos\theta = \mathrm{sech} u \,\, \sin\theta = \tanh u \,\,
\csc\theta = \coth u
\end{equation}

We solve \(\tan\theta = \sinh u\) for \(\theta\) to obtain, \(\theta = \tan^{-1}(\sinh u) \). Differentiating yields
\begin{equation}
\frac{d\theta}{du} = \frac{\cosh u}{1 + \sinh^{2} u} = \frac{1}{\cosh u}
\end{equation}

Integrating yields
\begin{equation}
\theta = \int_{0}^{u} \frac{1}{\cosh t} dt
\end{equation}

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}

Two Additional Relations

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

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

\begin{equation}
\frac{d}{dx}\mathrm{gd}x = \mathrm{sech}(x)
\end{equation}
follows from the integral definition.

Maclaurin Series

\(\mathrm{gd}(0) = 0\) follows from the integral definition.
\begin{equation}
\mathrm{gd}^{\prime}(0) = \mathrm{sech}(0) = 1
\end{equation}

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

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

\begin{equation}
\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
\end{equation}

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
\begin{equation}
\int \mathrm{gd}x \, dx = 2\int \tan^{-1}\mathrm{e}^{x} dx – \int \frac{\pi}{2} dx
\end{equation}
Using
\begin{equation}
\tan^{-1}z = \frac{i}{2}\Big[\ln(1-iz) – \ln(1+iz)\Big]
\end{equation}
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
\begin{equation}
\int \mathrm{gd}x \, dx = i[\mathrm{Li}_{2}(-i\mathrm{e}^{x})-\mathrm{Li}_{2}(i\mathrm{e}^{x})] – \frac{\pi}{2}x
\end{equation}

References

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

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