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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., 3x^{2} -1 is the antiderivative of x^{3} -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) = $\int_{c}^{x} f(t)dt$.

In mathematics, the logarithm is the inverse function to exponentiation. We call the inverse of a^{x} **the logarithmic function with base a**, that is, log_{a}x=y ↔ a^{y}=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., log_{4}(64) = 3 ↭ 4^{3} = 64, log_{2}(16) = 4 ↭ 2^{4} = 16, log_{8}(512) = 3 ↭ 8^{3} = 512, but log_{2}(-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(a ^{r}) =** $ \int_{1}^{a^{r}} \frac{dt}{t} =$[Integration by substitution or change of variables, w = t

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:

- Dom(erf) = ℝ.
- $\lim_{x \to -∞} (\frac{2}{\sqrt{π}}\int_{0}^{x} e^{-t^{2}}dt) = \frac{2}{\sqrt{π}}\int_{0}^{-∞} e^{-t^{2}}dt = -\frac{2}{\sqrt{π}}\int_{-∞}^{0} e^{-t^{2}}dt =$[This integral requires a more advanced knowledge of Calculus, $\int_{-∞}^{0} e^{-t^{2}}dt = \frac{\sqrt{π}}{2}$] =$-\frac{2}{\sqrt{π}}·\frac{\sqrt{π}}{2} = -1$
- $\lim_{x \to ∞} (\frac{2}{\sqrt{π}}\int_{0}^{x} e^{-t^{2}}dt) = \frac{2}{\sqrt{π}}\int_{0}^{∞} e^{-t^{2}}dt =\frac{2}{\sqrt{π}}·\frac{\sqrt{π}}{2} = 1$, so there are two horizontal asymptotes y = -1 and y = 1.
- 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.
- 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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