4. Integral

Antiderivative

Definition

In calculus, an antiderivative, inverse derivative, primitive function, primitive integral or indefinite integral of a function $f$ is a differentiable function $F$ whose derivative is equal to the original function $f$.

Suppose $F'(x)=f(x)$, the notation is

$$\int f(x)\mathrm dx=F(x)$$

So all the antiderivative of $f$ become a family set $\{F(x)+C|C\in\mathbb R\}$.

also the equation below is obviously.

$$\mathrm d \int f(x) \mathrm{d} x=f(x) \mathrm{d} x, \quad \int F^{\prime}(x) \mathrm{d} x=F(x)+c$$

Theorem: Integration by parts

$$\int u(x)v'(x)\,\mathrm dx=u(x)v(x)-\int u'(x)v(x)\,\mathrm dx$$

Example: Wallis product

the Wallis product for $\pi$, published in 1656 by John Wallis states that

$$\begin{aligned}{\frac {\pi }{2}}&=\prod _{n=1}^{\infty }{\frac {4n^{2}}{4n^{2}-1}}=\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)\\[6pt]&={\Big (}{\frac {2}{1}}\cdot {\frac {2}{3}}{\Big )}\cdot {\Big (}{\frac {4}{3}}\cdot {\frac {4}{5}}{\Big )}\cdot {\Big (}{\frac {6}{5}}\cdot {\frac {6}{7}}{\Big )}\cdot {\Big (}{\frac {8}{7}}\cdot {\frac {8}{9}}{\Big )}\cdot \;\cdots \\\end{aligned}$$

Prove:

$$\begin{align} I(n) &= \int_0^\pi \sin^nx\,\mathrm dx=\int_0^\pi -u \,\mathrm dv = -uv |_{x=0}^{x=\pi}-\int_0^\pi -v \,\mathrm du \\ {} &= -\sin^{n-1}x\cos x |_{x=0}^{x=\pi} - \int_0^\pi - \cos x(n-1) \sin^{n-2}x \cos x \,\mathrm dx \\ {} &= 0 - (n-1) \int_0^\pi -\cos^2x \sin^{n-2}x\,\mathrm dx, n > 1 \\ {} &= (n - 1) \int_0^\pi (1-\sin^2 x) \sin^{n-2}x \,\mathrm dx \\ {} &= (n - 1) \int_0^\pi \sin^{n-2}x\,\mathrm dx - (n - 1) \int_0^\pi \sin^{n}x \,\mathrm dx \\ {} &= (n - 1) I(n-2)-(n-1) I(n) \\ {} &= \frac{n-1}{n} I(n-2) \\ \Rightarrow \frac{I(n)}{I(n-2)} &= \frac{n-1}{n} \\ \Rightarrow \frac{I(2n-1)}{I(2n+1)} &=\frac{2n+1}{2n} \end{align}$$

so that:

$$\int_0^{\frac{\pi} {2}} \sin^nx\,\mathrm dx = \left\{\begin{matrix}\frac{\pi}{2}\frac{(2n-1)!!}{(2n)!!}& \text{n is even} \\ \\ \frac{(2n-1)!!}{(2n)!!}&\text{n is odd} \end{matrix} \right.$$

so that:

$$\begin{aligned} \Rightarrow& \lim _{n \rightarrow \infty} \frac{I(2 n)}{I(2 n+1)}=1 \\ &\lim _{n \rightarrow \infty} \frac{I(2 n)}{I(2 n+1)}=\frac{\pi}{2} \lim _{n \rightarrow \infty} \prod_{k=1}^{n}\left(\frac{2 k-1}{2 k} \cdot \frac{2 k+1}{2 k}\right)=1 \\ \Rightarrow& \frac{\pi}{2}=\prod_{k=1}^{\infty}\left(\frac{2 k}{2 k-1} \cdot \frac{2 k}{2 k+1}\right)=\frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots \end{aligned}$$

Simplify the Polynomial and Integral

If $Q(z)=\left(z-z{1}\right)^{k{1}} \cdots\left(z-z{p}\right)^{k{p}}$ and $\frac{P(z)}{Q(z)}$ is a proper fraction, there exists a unique representation of the fraction $\frac{P(z)}{Q(z)}$ in the form

$$\frac{P(z)}{Q(z)}=\sum_{j=1}^{p}\left(\sum_{k=1}^{k_{j}} \frac{a_{j k}}{\left(z-z_{j}\right)^{k}}\right)$$

and if $P(x)$ and $Q(x)$ are polynomials with real coefficients and

$$Q(x)=\left(x-x_{1}\right)^{k_{1}} \cdots\left(x-x_{l}\right)^{k_{l}}\left(x^{2}+p_{1} x+q_{1}\right)^{m_{1}} \cdots\left(x^{2}+p_{n} x+q_{n}\right)^{m_{n}}$$

there exists a unique representation of the proper fraction $\frac{P(x)}{Q(x)}$ in the form

$$\frac{P(x)}{Q(x)}=\sum_{j=1}^{l}\left(\sum_{k=1}^{k_{j}} \frac{a_{j k}}{\left(x-x_{j}\right)^{k}}\right)+\sum_{j=1}^{n}\left(\sum_{k=1}^{m_{j}} \frac{b_{j k} x+c_{j k}}{\left(x^{2}+p_{j} x+q_{j}\right)^{k}}\right)$$

where $a_{jk}, b_{jk},$ and $c_{j k}$ are real numbers.

and with these formulas below:

$$\begin{aligned} \int \frac{\mathrm{d} x}{(x-a)^{m}}&=\left\{\begin{array}{ll} -\frac{1}{(m-1)(x-a)^{m-1}}+c, & m \neq 1 \\ \\ \log |x-a|+c, & m=1 \end{array}\right. \\ \\ \int \frac{2 x \mathrm{d} x}{\left(x^{2}+a^{2}\right)^{m}}&=\int \frac{\mathrm{d}\left(x^{2}+a^{2}\right)}{\left(x^{2}+a^{2}\right)^{m}}=\left\{\begin{array}{c} -\frac{1}{(m-1)\left(x^{2}+a^{2}\right)^{m-1}}+c, \quad &m \neq 1 \\ \\ \log \left(x^{2}+a^{2}\right)+c, \quad &m=1 \end{array}\right. \\\\ \int \frac{\mathrm{d} x}{\left(x^{2}+a^{2}\right)^{m}}&=\frac{x}{\left(x^{2}+a^{2}\right)^{m}}+\int \frac{2 m x^{2} \mathrm{d} x}{\left(x^{2}+a^{2}\right)^{m+1}} \\&=\frac{x}{\left(x^{2}+a^{2}\right)^{m}}+\int \frac{2 m \mathrm{d} x}{\left(x^{2}+a^{2}\right)^{m}}-2 m a^{2} \int \frac{\mathrm{d} x}{\left(x^{2}+a^{2}\right)^{m+1}} \end{aligned}$$

And from that we get the recursion:

$$\int \frac{\mathrm{d} x}{\left(x^{2}+a^{2}\right)^{m+1}}=\frac{1}{2 m a^{2}} \frac{x}{\left(x^{2}+a^{2}\right)^{m}}+\frac{2 m-1}{2 m a^{2}} \int \frac{\mathrm{d} x}{\left(x^{2}+a^{2}\right)^{m}}$$

Primitives of the Form $\int R(\cos x, \sin x)\mathrm dx$

We make the change of variable $t = \tan \frac{x} {2}$ . Since:

$$\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}, \qquad \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$$

so that

$$\mathrm{d} t=\frac{\mathrm{d} x}{2 \cos ^{2} \frac{x}{2}} \quad \Rightarrow\quad \mathrm{d} x=\frac{2 \mathrm{d} t}{1+\tan ^{2} \frac{x}{2}}=\frac{2\,\mathrm{d} t}{1+t^{2}}$$

It follows that

$$\int R(\cos x, \sin x) \mathrm{d} x=\int R\left(\frac{1-t^{2}}{1+t^{2}}, \frac{2 t}{1+t^{2}}\right) \frac{2}{1+t^{2}} \mathrm{d} t$$

not only $\sin ,\cos$can to do this, but here are a lot of formula:

$\tan a=\frac{2 \tan \frac{a}{2}}{1-\tan ^{2} \frac{a}{2}}$, $\cot \alpha=\frac{1-\tan ^{2} \frac{\alpha}{2}}{2 \tan \frac{\alpha}{2}}$, $\sec \alpha=\frac{1+\tan ^{2} \frac{\alpha}{2}}{1-\tan ^{2} \frac{\alpha}{2}}$,$\csc \alpha=\frac{1+\tan ^{2} \frac{\alpha}{2}}{2 \tan \frac{\alpha}{2}}$

Integration

Riemann Sums

partition

A partition P of a closed interval $[a,b]$, $a < b$, is a finite system of points $x_0,\cdots,x_n$ of the interval such that $a = x_0 < x_1 <\cdots < x_n = b$.

If a function $f$ is defined on the closed interval $[a, b]$ and $(P, \xi)$ is a partition with distinguished points on this closed interval, the sum

$$\sigma(f ; P, \xi):=\sum_{i=1}^{n} f\left(\xi_{i}\right) \Delta x_{i}$$

where $\Delta x_i = x_i − x_{i−1}$, is the Riemann sum of the function $f$ corresponding to the partition $(P, \xi)$ with distinguished points on $[a,b]$.

The largest of the lengths of the intervals of the partition $P$ , denoted $\lambda(P)$, is called the mesh of the partition.

we define:

$$\int_{a}^{b} f(x) \mathrm{d} x:=\lim _{\lambda(P) \rightarrow 0} \sum_{i=1}^{n} f\left(\xi_{i}\right) \Delta x_{i}$$

Integral mean value theorem

If $f$ is a continuous function on the closed, bounded interval $[a,b]$, then there is at least one number $\xi$ in $(a , b )$ for which

$$\int_{a}^{b} f(x) \mathrm{d} x=f(\xi)(b-a)$$

The second Integral mean value theorem

If $f , g$ are continuous functions on the closed, bounded interval $[a,b]$,$g$ is monotonous on $[a, b]$, then there is at least one number $\xi$ in $(a , b )$ for which

$$\int_{a}^{b}(f \cdot g)(x) \mathrm{d} x=g(a) \int_{a}^{\xi} f(x) \mathrm{d} x+g(b) \int_{\xi}^{b} f(x) \mathrm{d} x$$

Newton-Leibniz formula

Let $f$ be a continuous real-valued function defined on a closed interval $[a, b]$. Let $F$ be the function defined, s.t.

$$\frac{d}{\mathrm{d} x} \int_{a}^{x} f(t) \mathrm{d} t=f(x), \quad \forall x \in[a, b]$$

Substitution Rule For Definite Integrals

Suppose $f \in C[a, b], \varphi:[\alpha, \beta] \rightarrow[a, b]$and $\varphi^{\prime} \in \mathcal{R}[\alpha, \beta],\varphi(\alpha)=a, \varphi(\beta)=b$, s.t.

$$\int_{a}^{b} f(x) \mathrm{d} x=\int_{\alpha}^{\beta} f(\varphi(t)) \varphi^{\prime}(t) \mathrm{d} t$$

$\blacksquare$