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The Logarithm Defined as an Integral. The error function

Science is a differential equation, religion is a boundary condition, Alan Turing.

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Antiderivatives are fundamental concepts in calculus. They are the inverse operation of derivatives.

Given a function f(x), an antiderivative, also known as indefinite integral, F, is the function that can be differentiated to obtain the original function, that is, F’ = f, e.g., 3x2 -1 is the antiderivative of x3 -x +7 because $\frac{d}{dx} (x^3-x+7) = 3x^2 -1$. Symbolically, we write F(x) = $\int f(x)dx$.

The process of finding antiderivatives is called integration.

The Second Fundamental Theorem of Calculus. If f is a continuous function and c is a constant, then f has a unique antiderivative A that satisfies A(c) = 0, and that antiderivative is given by the rule A(x) = $\int_{c}^{x} f(t)dt$.

The Logarithm Defined as an Integral.

In mathematics, the logarithm is the inverse function to exponentiation. We call the inverse of ax the logarithmic function with base a, that is, logax=y ↔ ay=x, that means that the logarithm of a number x to the base a is the exponent to which a must be raised to produce x, e.g., log4(64) = 3 ↭ 43 = 64, log2(16) = 4 ↭ 24 = 16, log8(512) = 3 ↭ 83 = 512, but log2(-3) is undefined.

The logarithm’s domain consists of all real positive numbers. Its range is ℝ (figure i and ii). x-intercept: (1, 0), y-intercept: none. It is one-to-one and has a vertical asymptote along the y-axis at x = 0.

Recall, $\int x^n = \frac{x^{n+1}}{n+1}+C$, n ≠ 1.

We already know that L’(x) = 1/x, L(1) = 0.

By the The Second Fundamental Theorem of Calculus, the antiderivative of 1/x is L(x) = $\int_{1}^{x} \frac{dt}{t}$, L’(x) = 1/x and c = 1 (L(1) = 0). Notice that we could take L(x) as the definition of the logarithm. In other words, the natural logarithm can be defined as the definite integral L(x) = $\int_{1}^{x} \frac{dt}{t}$. Besides, L(x) equals the area between the x-axis and the graph of the function 1/t, ranging from t = 1 to t = x.

Futhermore, L’’(x) = $\frac{-1}{x^2}$, so it’s concave down. Besides, L’(x) > 0 (x > 0), so L is constantly increasing. Euler’s number e is defined as the constant e such that L(e) = 1 -Figure 1.a-.


Let’s prove the well-known logarithm’s property L(ab) = L(a) + L(b). $L(ab) = \int_{1}^{ab} \frac{dt}{t}$ =[Definite Integrals on Adjacent Intervals] $\int_{1}^{a} \frac{dt}{t} + \int_{a}^{ab} \frac{dt}{t}$ =[Integration by substitution or change of variables in the second integral, t = au, dt = adu where a is a constant, t = a ↭ a = au ↭ u = 1, t = ab ↭ u = b] = $\int_{1}^{a} \frac{dt}{t} + \int_{1}^{b} \frac{adu}{au} = \int_{1}^{a} \frac{dt}{t} + \int_{1}^{b} \frac{du}{u}$ = L(a) + L(b).

Similarly, L(ar) = $ \int_{1}^{a^{r}} \frac{dt}{t} =$[Integration by substitution or change of variables, w = t1/r ⇒ t = wr, dt = rwr-1dw. Besides, t = 1 ↭ w = 1, t = ar ↭ w = a] = $\int_{1}^{a} \frac{rw^{r-1}dw}{w^{r}} = r\int_{1}^{a} \frac{dw}{w} =$ rL(a).

The error function

By the The Second Fundamental Theorem of Calculus, ⇒[If f is a continuous function and c is a constant, then f has a unique antiderivative A that satisfies A(c) = 0, and that antiderivative is given by the rule A(x) = $\int_{c}^{x} f(t)dt$.] the equation y’ = $e^{-x^{2}}, y(0) = 0$ has a solution F(x) = $\int_{0}^{x} e^{-t^{2}}dt$ and c = 0 (F(0) = 0).

Therefore, F’(x) = $e^{-x^{2}} > 0$, hence F is always increasing, F’(0) = $e^{-0^{2}} = 1$ ⇒ the slope of the tangent line is 1 at x = 0. Besides, F(0) = 0. F’’(x) = $-2xe^{-x^{2}}$.

$F’’(x) = \begin{cases} F’’ < 0, &x > 0 \\ F’’ > 0, &x < 0 \end{cases}$

Therefore, F is concave down to the right of the vertical axis, F is concave up to the left of the vertical axis, with an inflection point at the origin of coordinates (0, F(0)) = (0, 0). Futhermore, F is an odd function ↭ F(-x) = -F(x) -Figure 1.b-. The plots shown below correspond to F and F'.


Futhermore, $\lim_{x \to ∞} \int_{0}^{x} e^{-t^{2}}dt = \frac{\sqrt{π}}{2}, \lim_{x \to -∞} \int_{0}^{x} e^{-t^{2}}dt = -\frac{\sqrt{π}}{2}$ and this is by no means easy to demonstrate.

The error function (erf) (also called the Gauss error function), often denoted by erf (Figure 1.c), is a special function defined as: $erf(x) =\frac{2}{\sqrt{π}}\int_{0}^{x} e^{-t^{2}}dt = \frac{2}{\sqrt{π}}F(x)$, and it has the following properties:

  1. Dom(erf) = ℝ.
  2. $\lim_{x \to -∞} (\frac{2}{\sqrt{π}}\int_{0}^{x} e^{-t^{2}}dt) = \frac{2}{\sqrt{π}}\int_{0}^{-∞} e^{-t^{2}}dt = -\frac{2}{\sqrt{π}}\int_{-∞}^{0} e^{-t^{2}}dt =$[This integral requires a more advanced knowledge of Calculus, $\int_{-∞}^{0} e^{-t^{2}}dt = \frac{\sqrt{π}}{2}$] =$-\frac{2}{\sqrt{π}}·\frac{\sqrt{π}}{2} = -1$
  3. $\lim_{x \to ∞} (\frac{2}{\sqrt{π}}\int_{0}^{x} e^{-t^{2}}dt) = \frac{2}{\sqrt{π}}\int_{0}^{∞} e^{-t^{2}}dt =\frac{2}{\sqrt{π}}·\frac{\sqrt{π}}{2} = 1$, so there are two horizontal asymptotes y = -1 and y = 1.
  4. erf’(x) =[By the The Second Fundamental Theorem of Calculus] $\frac{2}{\sqrt{π}}e^{-x^2}$, erf’(x) > 0, hence erf is always increasing.
  5. It is odd, that is, erf (-x) = -erf (x).

This non-elementary integral is a sigmoid (a function whose graph has a characteristic S-shaped or sigmoid curve) function that occurs often in probability, statistics, and partial differential equations.


This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
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  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
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