# Gudermannian Function

A Treatise on the Integral Calculus: With Applications, Examples and Problems by Joseph Edwards was mentioned in Inside Interesting Integrals by Paul Nahin. While perusing the book I came across the Gudermannian fuction, which I had not heard of before. This function is essentially a means of relating the trigonometric functions to the hyperbolic functions without the use of complex variables.

From the following references, we have the basic definition of the Gudermannian:

$$\mathrm{gd} u = \int_{0}^{u} \frac{1}{\cosh t} dt$$

Click to enlarge.

Wolfram MathWorld

The references above do not provide a derivation, so let us do so here using the geometric reasoning of Edwards.

We begin with the following diagram:

P is a point on the hyperbola: $$x^{2} – y^{2} = a^{2}$$, where $$a = OA$$. We note that the point P has coodinates $$(x,y)$$ where

$$x = a \sec\theta = a \cosh u$$
$$y = a \tan\theta = a \sinh u$$.

Thus we have:

$$\sec\theta = \cosh u$$
$$\tan\theta = \sinh u$$

from which we can use the basic definitions of the trig and hyperbolic functions to derive all of the relations between them.

At this point, we must define $$u$$. For a circle of radius $$a$$ that makes an angle $$\theta$$ with the positive x-axis, the area of a circular sector is $$\int_{0}^{a}\int_{0}^{\theta} r dr d\phi = \frac{1}{2}a^{2}\theta$$ = area of OTA along the arc of the red circle.

As point P moves along the hyperbola, a hyperbolic area sector is swept out. This area is that of OAP along the arc of the hyperbola in dark blue. From the graph, we note that this area plus that of ANP, also along the arc of the hyperbola, is equal to the area of the triangle ONP.

Let us compute the area of ANP. With $$a$$ equal to OA and $$x$$ equal to ON, the area we want is the integral of the hyperbola between these two points: $$\int_{a}^{x} y \, dt$$. Using the equations for $$x$$ and $$y$$ in terms of hyperbolic functions, we have area of ANP =

\begin{align}
& = \int_{a}^{x} y \, dt \\
& = \int_{0}^{u} a^{2} \sinh^{2} v \, dv \\
& = \frac{1}{2}a^{2} \int_{0}^{u} (\cosh 2v – 1) \, dv \\
& = \frac{1}{2}a^{2} (\frac{1}{2} \sinh 2v – v)|_{0}^{u} \\
& = \frac{1}{4}a^{2}\sinh 2u – \frac{1}{2}a^{2}u \\
\end{align}

The area of the triangle ONP = $$\frac{1}{2}xy = \frac{1}{2}(a \cosh u)(a \sinh u) = \frac{1}{4}a^{2}\sinh 2u$$. We now have

area of OAP = area of the triangle ONP – area of ANP = $$\frac{1}{2}a^{2}u$$.

So for a hyperbolic sector, $$u$$ plays the role of $$\theta$$ for a circular sector.

Edwards concludes, “It is this connection with the hyperbola from which these transcendental functions are termed hyperbolic functions.”

We then have $$\theta = \mathrm{gd} u$$ and $$u = \mathrm{gd}^{-1}\theta$$.

Now we derive the integral expression of the Gudermannian function.

This is done by working backwards. As mentioned above, we can relate trigonometric functions of $$\theta$$ to hyperbolic functions of $$u$$. These relations result in multiple expressions of $$u = f(\theta)$$. 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 our desired result $$\mathrm{gd} u = \int_{0}^{u} \frac{1}{\cosh t} dt$$.

Note that the Gudermannian function was named in honor of Christoph Gudermann.

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