2.2 The Natural Logarithm

We want to find a mathematical function \(f(x)\) that can be evaluated in a straightforward way, and that has the properties of a logarithm that were discussed in the previous chapter. For instance, the function would need to have the following properties:

• The logarithm of 1 must zero for any choice of base. Hence, \(f(1) = 0.\)
  
• The function should be positive for values of \(x\) greater than one, and negative for values less than one: \(f(x<1) < 0; ~~ f(x > 1) >0.\)
  
• As \(x\) approaches zero from the positive side, the function should approach negatively infinite numbers: \(f(x)\rightarrow -\infty\) as \(x\rightarrow 0.\)
  
• And, as \(x\) gets more and more positive, the function should tend toward infinite numbers: \(f(x)\rightarrow +\infty\) as \(x\rightarrow +\infty .\)
  
• Lastly, but perhaps most importantly, since the logarithm of a product of two numbers must be the sum of the logarithms of the two numbers, we must have: \(f(a\times b) = f(a) + f(b).\)

It turns out that the area under a specific geometrical figure, which can be accurately computed through the use of calculus, has the above properties and hence can be used to define a natural logarithm. This may seem like a very strange approach toward defining a logarithm which is, after all, an exponent applied to a base number, a seemingly unrelated mathematical exercise. But we’ll see that for one particular curve, these two operations have a remarkable connection and the power of calculus allows us to compute the answers in a straightforward way.

The curve in question is that of a hyperbola. In terms of the coordinates in a plane, \(u\) and \(v\), the equation of a hyperbola can be written as \(u\cdot v = 1\) or, as a function of \(u\), \(v(u) = 1/u\).

If we take the branch of the hyperbola for positive values of \(u\) we find, after a bit of inspection, that our properties of a logarithm can be met by defining a new function that is evaluated by finding the \(area\) between the hyperbolic curve and the horizontal axis. The area in question is shown below:

The shaded area always starts at \(u=1\) and stops at an arbitrary value \(x\), and so yields a function of \(x\), call it \(f(x)\). This function, therefore, can be computed using calculus by performing the following integration: \[ {\cal Area} = f(x) = \int_1^x \frac{1}{u}\;du. \] This integral has the properties of a logarithm as listed above. First of all, it obeys the rule of a logarithm in that the logarithm of \(a\times b\) is equal to the sum of the individual logarithms of \(a\) and of \(b\): \[\begin{eqnarray*} f(a\times b) &=& \int_1^{ab} \frac{1}{u}\;du \\ &=& \int_1^{a} \frac{1}{u}\;du + \int_a^{ab} \frac{1}{u}\;du \\ &=& \int_1^{a} \frac{1}{u}\;du + \int_1^{b} \frac{1}{w}\;dw ~~~~~~~(w \equiv u/a)\\ &~& ~~ \\ &=& f(a) + f(b). \end{eqnarray*}\] Secondly, for \(x\) = 1 the integration range will have zero length and hence \(f(1)\) will be zero. For values of \(x<1\) the integral will be negative, just as logarithms are negative for \(x<1\). In addition, the integral goes to \(-\infty\) as \(x\) approaches 0 and to \(+\infty\) as \(x\) approaches \(+\infty\). The function \(f(x)\) given by the above integral obeys all the basic properties of a logarithm.

For this reason, this integral represents a natural logarithm and is given the designation \(\ln x\) with the formal definition, \[ \ln x \equiv \int_1^x \frac{1}{u}\;du. \]