It is very convenient to define the logarithm as an integral

The exponential and logarithmic functions are generally introduced in the form of the "laws of exponents" and their intuitive exensions. These laws are: a^{x}a^{y} = a^{x + y}, a^{x}/a^{y} = a^{x - y}, 1/a^{x} = a^{-x}, 1/a^{-x} = a^{x}, a^{0} = 1, where a, x and y are real numbers (the extension to complex numbers is also valid). These rules are introduced for x and y integral, from which they are obvious by definition, then extended to rationals and finally to reals. In this way the exponential function y = a^{x} is defined. It is continuous, positive and monotonic.

The logarithmic function x = log_{a} y is defined as the inverse of the exponential function. The subscript a is the *base* of the logarithms. Then, a^{loga y} = y, and log_{a}a^{x} = x. Since a^{0} = 1, then log_{a} 1 = 0. Also, log_{a}a = 1. The familiar properties of logarithms follow from the laws of exponents: log (xy) = log x + log y, log (x/y) = log x - log y, log x^{n} = n log x, log ^{n}√x = (1/n)log x. The subscript giving the base has been dropped here.

The number e = 2.71828182846... as the base of exponential and logarithmic functions is generally pulled out of the blue as the sum of the series 1 + 1/1! + 1/2! + ..., or as the limit lim(z→0)(1 + z)^{1/z}. This is very unsatisfactory. One then finds the derivative of log x by the delta method, in which the above limit appears, and one finds (d/dx)log_{a}x = log_{a}e/x. If a is chosen so that log_{a}e = 1, or a = e, then we have simply d(log_{e} x)/dx = 1/x. For logarithms to base e, we write ln x here. Then, log x means an arbitrary understood base, or base 10, the Briggsian logarithms useful for decimal calculations. Natural logarithms are sometimes called Naperian to honor the inventor of logarithms, but Napier did not use natural logarithms himself.

A much more satisfactory way to introduce logarithms is to define the logarithmic function y = ln x by the integral y = ∫(1,x)du/u. This function is clearly continuous and has a continuous derivative 1/x for 0 < x < ∞. Since 1/x is always positive, ln x is monotonic. Obviously, ln 1 = 0. For x < 1, ln x is negative and approaches -∞ as x → 0, while for x > 1 it is positive and approaches +∞. ln x = 1 corresponds to some positive number e such that 1 = ∫(1,e)du/u [the limits are in parentheses following the integral sign in our notation]. That is, the area below the curve 1/x between x = 1 and x = e is 1. This defines e in a very clear and nonmysterious way. This is perhaps the simplest case of defining a new function by an integral. Though all algebraic functions can be differentiated, not all can be integrated, of which 1/x is the prime example. We use the integral of 1/x to define a new function, the logarithm.

The integrand, du/u has the property that if we substitute u' = au, where a is any constant, then the integrand remains du'/u'. This means that ln e^{2} = ∫(1,e^{2})du/u = ∫(1,e)du/u + ∫(1,e)du'/u' = 2. In the second integral we simply substituted u' = eu and adjusted the limits accordingly. This can be done generally by writing ln e^{m/n} = ∫(1,e^{m/n})du/u = ∫(1,e^{1/n})du/u + ∫(e^{1/n},e^{2/n}du/u + ... ∫(e^{(m-1)/n},e^{m/n}du/u. Making the substitutions as before, we have ln e^{m/n} = m ∫(1,e^{1/n})du/u = m/n. From this result, it is easy to establish all the rules of logarithms. The role played by e is very clear; it simply gives unit area. The exponential function y = a^{x} now becomes y = (e^{ln a})^{x}, from which all its properties follow.

It is easy to prove the addition law for logarithms. We have ln (ab) = ∫(1,ab)du/u = ∫(1,a)du/u + ∫(a,ab)du/u. The substitution u' = au in the second integral immediately gives ln (ab) = ln a + ln b. In terms of the E functions, we have E(α)E(β) = E(α + β), where α = ln a, and β = ln b. Note that if α = ln a, then a = E(α).

If x = log a, then a = 10^{x} = e^{x ln 10}, or ln a = (log a)(ln 10), where log means log to base 10. ln 10 = 2.302585093. 1/ln 10 = 0.4342944819 = log e. Also, log a = (ln a)(log e). Note that if e^{ln 10} = 10 and 10^{log e} = e, then e^{(ln 10)(log e)} = e, so that (ln 10)(log e) = 1. The same relation holds between any two bases.

Since y = ln x is continuous and monotonic, it has a continuous and monotonic inverse x = E(y), and dx/dy = 1/(dy/dx) = x. This is the usual differential equation for the inverse function, from which we can get the power series that shows that the inverse function is indeed x = e^{y}. It expresses the fact that the derivative of E(y) is E(y) itself. We use the E(y) notation to show that this function is independent of the earlier definitions and the laws of exponents, arising purely from the integral.

The slope of the curve y = ln x is 1/x, so it decreases steadily towards zero as x increases. The curve cannot approach a horizontal asymptote, however, because it increases beyond all bounds. If you propose some large number A for an asymptote y = A, then I can always find an x for which ln x is greater than A. For, ln 2^{n} = n ln 2, so if I take any n > A/ln 2, ln 2^{n} will be greater than A. It is not usually noticed that the curves y = ln x and y = e^{x} have the same shape; one is transformed into the other upon reflection in the line y = x. The integral ∫(1,x)du/u does not converge as x → ∞. This recalls the nonconvergence of the harmonic series 1 + 1/2 + 1/3 + ..., which diverges logarithmically. We may also note that although ∫(0,∞)dx/x = ∞, ∫(0,∞)(sin x)/x = π/2, just as 1 - 1/2 + 1/3 - ... = ln 2.

The familiar limit formula for e is obtained as follows. 1/x = d(ln x)/dx = lim_{h→0}[ln(x + h) - ln x]/h = lim_{h→0}ln(1 + h/x)/h. Then, e^{1/x} = lim_{h→0} e^{ln(1+h/x)/h} = lim_{h→0}(1 + h/x)^{1/h}. If h takes the values 1, 1/2, 1/3, ..., 1/n this gives e^{1/x} = lim_{n→∞}(1 + 1/xn)^{n}, or e^{x} = lim_{n→∞}(1 + x/n)^{n}. Letting x = 1, we find e = lim_{n→∞}(1 + 1/n)^{n}. This is, of course, the same expression used in the elementary definition of logarithms. If the binomial is expanded by the binomial theorem, we find the infinite series for e, or the power series for e^{x}.

Trigonometric functions can also be expressed as integrals of algebraic functions. For example, tan^{-1}y = ∫(0,y)du/(1 + u^{2}) = y, from which y = tan x can be found. This integral shows that for y << 1, tan^{-1}y = y, or y = tan y (further terms are easily found by expanding the integrand with the binomial theorem. However, the integrand does not have the very convenient properties of 1/u. We can find π by numerical integration, from π = 4∫(0,1)du/(1 + u^{2}), for example.

The word *logarithm* is derived from the Greek *logos*, here implying arithmetic, and *arithmos*, number.

R. Courant, *Differential and Integral Calculus*, Vol. I (London: Blackie & Son, 1937). pp. 167-183.

H. J. Gay, *Analytic Geometry and Calculus* (New York: McGraw-Hill, 1950). Chapter 13. An example of the way the exponential and logarithmic functions are introduced in elementary mathematics. Any similar text will serve as well.

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Composed by J. B. Calvert

Created 27 November 2004

Last revised