Definitions and Basics

In the previous chapter (Derivatives, Definition and Rules) we introduced the notation $\frac{d f}{d x}$ for the derivative of the function

f (x)

with respect to x . This notation was obtained from the limit of the two increments' ratio:

\lim_{Δ x \to 0} \frac{Δ f}{Δ x}

. Since this notation originates from the limit of two numeric entities, it behaves similarly to the ratio of two numbers. As an example the derivative of an inverse function can be expressed by the derivative of the function as known from (1.3.1.10):

$\frac{d y}{d x} = \frac{1}{\frac{d x}{d y}} where {\begin{cases} y = inv f (x) \\ x = f (y) \end{cases}}$

Although the notations df and dx do not have any quantitative meaning, they are called the differentials of f and of x accordingly. We consider them as abstract entities which have quantitative meaning in paricular circumstances as in the case of a ratio of two differentials. The notation and use of differentials is very helpful if the rules of differentiation are obeyed.

Below is a list of the rules, that can be easily verified by dividing the differentials by dx ( x being the independent variable of the functions).

$d f = \frac{d f}{d x} d x$	(2.1.1.1)

$\begin{array}{l} d (\sin x) = \cos x d x \\ d (\cosh x) = \sinh x d x \end{array}}$	(2.1.1.2)

The known chain rule for differentiation of a function of another function (1.3.1.11)

$\frac{d y}{d x} = \frac{d f}{d u} \frac{d u}{d x} where {\begin{cases} y = f (u) \\ u = u (x) \end{cases}}$

$\begin{array}{l} if y = f [u (x)] then \\ d y = d f = \frac{d f}{d u} d u = \frac{d f}{d u} \frac{d u}{d x} d x \end{array}}$	(2.1.1.3)

$\begin{array}{l} d (u + v) = d u + d v \\ d (u v) = v d u + u d v \\ d (\frac{u}{v}) = \frac{v d u - u d v}{v^{2}} \end{array}}$	(2.1.1.4)

Indefinite integral

One can look at the letter d from a differential of a variable as a kind of operator applied to the variable. By the same token one can define an inverse operator that cancels this operator, which will be called integral and denoted by $\int$ :

$I (x) = \int d I = \int \frac{d I}{d x} d x$	(2.1.1.5)

${\begin{cases} f (x) = \frac{d I}{d x} \\ I (x) = \int f d x \end{cases}}$	(2.1.1.6)

The function I(x) from (2.1.1.6) is called the indefinite integral of fdx and its meaning is that its derivative is the function f(x) . In other words the integration is the operation opposite to the differentiation. One should notice that the notations used for differentials and integrals stress the physical dimensions of the expressions.

As we know, if f(x) is the derivative of the function I(x) , then f(x) will remain unchanged if a constant is added to I(x) . This is the reason for the adjective "indefinite".

In order to clarify this point let's look at (2.1.1.2). If we apply the integral operator we obtain:

$\begin{array}{l} \begin{array}{l} \int \cos x d x = \int d (\sin x) = \sin x + C \\ \int \sinh x d x = \int d (\cosh x) = \cosh x + C \end{array}} \\ which means that: \\ \begin{array}{l} \frac{d}{d x} (\sin x + C) = \cos x \\ \frac{d}{d x} (\cosh x + C) = \sinh x \end{array}} \end{array}}$	(2.1.1.7)

and C is an arbitrary additive constant. We use the notation C , for any additive constant obtained from an indefinite integral.

Elementary rules

In the previous chapter we saw that there are clear rules for differentiation. The rules for integration consist mostly of methods for converting an integral to a familiar form which we know how to deal with. Even so, there are cases that cannot be solved analytically, and one has to use numerical evaluations.

Tip. Since the differentiation has clear and sound rules, it is a very good custom to check the integrals you've calculated by taking their first derivatives.

We will list only the obvious rules, and the known integrals. The methods mentioned earlier will be studied in the next section.

From basic principles (2.1.1.6) one obtains the following rules for integration:

$\int c f (x) d x = c \int f (x) d x (where c = constant)$	(2.1.1.8)

$\int [f_{1} (x) + f_{2} (x)] d x = \int f_{1} (x) d x + \int f_{2} (x) d x$	(2.1.1.9)

A power series can be integrated term by term within the interval of absolute convergence without affecting the convergence. Since the integration is opposite to the differentiation and this statement is correct for differentiation, it should hold also for integration. The only difference is the additional arbitrary constant that could not affect the convergence.

The integral of the power function x^p could be obtained from the differential

$d (x^{p + 1}) = (p + 1) x^{p} d x$

$\int x^{p} d x = \frac{x^{p + 1}}{p + 1} + C$	(2.1.1.10)

$\frac{d}{d x} (\ln \| x \|) = {\begin{cases} \frac{d}{d x} (\ln x) = \frac{1}{x} = x^{- 1} for x > 0 \\ \frac{d}{d x} (\ln (- x)) = \frac{1}{x} = x^{- 1} for x < 0 \end{cases}}$	(2.1.1.11)

$\int \frac{d x}{x} = \ln \| x \| + C$	(2.1.1.12)

The notation of the dx under the integral (e.g. (2.1.1.6)) is very important not only for the physical dimensions, but it also shows what the variable of the function is. This is important in particular when the integrand (the function to be integrated) can be integrated with respect to different variables. A word of caution: New students often make the mistake of omitting the differential, which leaves the integral entirely senseless.

This point is emphasized in an example from mechanics. A point with mass m moves along the x axis, its velocity is defined by $v = \frac{d x}{d t}$ , its acceleration is $a = \frac{d v}{d t}$ and t is the time. The same function will be integrated, once with respect to t and once with respect to x :

For those who still struggle with the notation of differentials, the second row of (2.1.1.13) might look a bit odd. An easier way to understanding it is:

$\begin{array}{l} m \int a d t = m \int \frac{d v}{d t} d t = m \int d v = m v + C (momentum) \\ m \int a d x = m \int \frac{d v}{d t} d x = m \int v d v = m \frac{v^{2}}{2} + C (energy) \end{array}}$	(2.1.1.13)

$\frac{d v}{d t} d x = \frac{d v}{d t} \frac{d x}{d t} d t = \frac{d x}{d t} \frac{d v}{d t} d t = v \frac{d v}{d t} d t = v d v$

The following list of known integrals is obtained by inverting the differentiation of the elementary functions from the previous chapter, section 3.

${\begin{cases} \int \cos x d x = \sin x + C \\ \int \sin x d x = - \cos x + C \\ \int (1 + \tan^{2} x) d x = \int \frac{d x}{\cos^{2} x} = \tan x + C \\ \int (1 + \cot^{2} x) d x = \int \frac{d x}{\sin^{2} x} = - \cot x + C \end{cases}}$	(2.1.1.14)

From the derivatives of the inverse-trigonometric functions (1.3.3.7) one obtains:

${\begin{cases} \int \frac{d x}{\sqrt{1 - x^{2}}} = (or) {\begin{cases} \pm asin x + C \\ \mp acos x + C \end{cases} \\ \int \frac{d x}{1 + x^{2}} = (or) {\begin{cases} + atan x + C \\ - acot x + C \end{cases} \end{cases}}$	(2.1.1.15)

As already known from the previous chapter, the possibility of two signs of asin and acos depend on the branch of the functions used. In this case the branches of increasing functions should be used with the positive sign and - decreasing with the negative sign.

From the derivative of lnx we already obtained (2.1.1.12) above, and from the derivative of expx we have:

$\int \exp x d x = \exp x + C$	(2.1.1.16)

${\begin{cases} \int \cosh x d x = \sinh x + C \\ \int \sinh x d x = \cosh x + C \\ \int \frac{d x}{\cosh^{2} x} = \int (1 - \tanh^{2} x) d x = \tanh x + C \\ \int \frac{d x}{\sinh^{2} x} = \int (\coth^{2} x - 1) d x = - \coth x + C \end{cases}}$	(2.1.1.17)

From the derivatives of the inverse hyperbolic functions (1.3.3.15) one obtains:

${\begin{cases} \int \frac{d x}{\sqrt{1 + x^{2}}} = asinh x + C \\ \int \frac{d x}{\sqrt{x^{2} - 1}} = \mp acosh x + C \\ \int \frac{d x}{1 - x^{2}} = atanh x + C (\| x \| < 1) \\ \int \frac{d x}{1 - x^{2}} = acoth x + C (\| x \| > 1) \end{cases}}$	(2.1.1.18)

The two possible signs of acosh depend on the branch of acosh that is used, as known from the previous chapter.

Parity

We saw in the previous chapter that the differentiation inverts the parity of a function. Since the integration is a process in the opposite direction we could in principle expect a similar conclusion. Can we? The derivative is defined uniquely, but not the integral. Does the addition of a constant to a function change its parity? NO for an even function, but YES for an odd one, in which case the parity property could be lost (Do not forget that a constant is an even function). Conclusion: if the function to be integrated is odd, then the integral is even, but in the case of an even function the obtained integral could be odd or without a defined parity.

In (2.1.1.7) sinx + C is odd for C = 0 but without defined parity otherwise.

In (2.1.1.7) coshx + C is even for any C.

In (2.1.1.10) the integral is even if the power p = odd.

In (2.1.1.10) if the power is even, the integral is odd (for C = 0) or without parity otherwise.

Exercises

Note: The additive constants are essential for the solution of integrals!

Exercise 1. Use (2.1.1.10) and D'Alambert's test to prove that:

A power series can be integrated term by term within the interval of absolute convergence, without affecting the convergence in the following cases of power series:

$\sum_{n = 0}^{\infty} b_{n} x^{n}$
$\sum_{n = 0}^{\infty} b_{n} x^{2 n + 1}$

Exercise 2. The Gaussian function

$f (x) = \exp (- \frac{x^{2}}{2})$

is well known in statistics and is extensively used in science for error calculations. Its integral is of great importance, but unfortunately it cannot be expressed in terms of the elementary functions. One way to deal with it is to use power series.

Express the Gaussian in power series (Maclaurin)!
Integrate the Gaussian $I (x) = \int f d x$ !
What is necessary for I(x) to be a function with defined parity and what is the parity?
For I(x) write a general expression of the n-th derivative at x=0 !

Exercise 3. Calculate the following integrals:

$\int \tan^{2} x d x$
$\int \frac{d x}{\tan^{2} x}$
$\int \frac{d x}{1 - x^{4}}$
Hint for the last case: $(1 - x^{4}) = (1 + x^{2}) (1 - x^{2})$

Exercise 4. According to Newton's mechanics for one dimension along the x axis, a point with mass m has an acceleration a: $\begin{array}{l} a = \frac{F}{m} where F = force \\ a = \frac{d v}{d t} where {\begin{cases} v = \frac{d x}{d t} = velocity \\ t = time \end{cases} \end{array}$

In this particular case the force is: $F = C \sin (ω t)$ where C and ω are known constants.

The following rule is necessary for the solution of the exercise: $\begin{array}{l} If \int f (x) d x = I (x) \\ then \int f (b x) d x = \frac{I (b x)}{b} \end{array}$ where b is a constant. Prove it!
Obtain the velocity as a function of the time!
Obtain the position x as a function of the time!
In the expression of x as a function of time that you found, prove that all the terms have the correct physical dimensions!

Previous topic: chapter 1 Differentiation, section 3 Derivatives, page 5 Taylor series

Next topic: page 2 Methods

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 2: Integration; Section 1: Indefinite Integrals; page 1

Home | Table of Contents | A-Z Index | Help

Definitions and Basics

Differential

Indefinite integral

Elementary rules

Parity

Exercises

Home | Table of Contents | A-Z Index | Help

Math Animated™ Mathematical Introduction for Physics and Engineering by Samuel Dagan (Copyright © 2007-2020)

Chapter 2: Integration; Section 1: Indefinite Integrals; page 1

Home | Table of Contents | A-Z Index | Help

Definitions and Basics

Differential

Indefinite integral

Elementary rules

Parity

Exercises

Home | Table of Contents | A-Z Index | Help

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)