Antidifferentiation

Antiderivatives

Definition: Antiderivative

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ and $F:{\mathcal{D}}_F\subseteq \mathbb{R}\to \mathbb{R}$ be real functions and let $S\subseteq {\mathcal{D}}_f\cap {\mathcal{D}}_F$.

We say that $F$ is an antiderivative of $f$ on $S$ if the derivative of $F$ is equal to $f$ for all $x\in S$:

$$F^{\prime }(x)=f(x)\forall x\in S$$

Theorem: Indefinite Integrals

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ be a real function, let $S\subseteq {\mathcal{D}}_f$ and let $F$ be an antiderivative of $f$ on $S$.

Another real function $\mathcal{F}$ is also an antiderivative of $f$ on $S$ if and only if there exists some real number $C\in \mathbb{R}$ such that

$$\mathcal{F}(x)=F(x)+C\forall x\in S$$

PROOF

Proof of (1):

If $\mathcal{F}$ is an antiderivative of $f$, then ${\mathcal{F}}^{\prime }(x)=f(x)$ for all $x\in S$. We thus have

$$(\mathcal{F}(x)-F(x){)}^{\prime }={\mathcal{F}}^{\prime }(x)-F^{\prime }(x)=f(x)-f(x)=0$$

for all $x\in S$. However, this is only possible if $\mathcal{F}(x)-F(x)$ is a constant.

Definition: Indefinite Integral

The set of all antiderivatives of $f$ is known as $f$‘s indefinite integral.

$$\{F\mid F^{\prime }=f\}$$

NOTATION

Most commonly, we use the following notation:

$$\int f(x)\mathrm{dx}=F(x)+C$$

This notation and the name “indefinite integral” are unfortunate remnants of the historical development of analysis and one should be very careful not to confuse them with Riemannn integrals.

Theorem: Linearity of Antidifferentiation

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ and $g:{\mathcal{D}}_g\subseteq \mathbb{R}\to \mathbb{R}$ be real functions and let $S\subseteq {\mathcal{D}}_f\cap {\mathcal{D}}_g$. Let $F$ and $G$ be antiderivatives on $S$ of $f$ and $g$, respectively.

If $f$ and $g$ are continuous on $S$, then for all $\alpha ,\beta \in \mathbb{R}$, the antiderivatives of $\alpha f+\beta g$ are given by

$$\int \alpha f(x)+\beta g(x)\mathrm{dx}=\alpha F(x)+\beta G(x)+C$$

PROOF

We differentiate $\alpha F(x)+\beta G(x)+C$:

$$\begin{array}{rl}(\alpha F(x)+\beta G(x)+C{)}^{\prime }& =(\alpha F(x){)}^{\prime }+(\beta G(x){)}^{\prime }+C^{\prime }\\ & =\alpha F^{\prime }(x)+\beta G^{\prime }(x)+0=\alpha F^{\prime }(x)+\beta G^{\prime }(x)\end{array}$$

Since $F$ is an antiderivative of $f$, we have $F^{\prime }(x)=f$. Similarly, $G^{\prime }(x)=g$. Therefore,

$$(\alpha F(x)+\beta G(x)+C{)}^{\prime }=\alpha f(x)+\beta g(x).$$

Theorem: Integration by Parts

Let $u:{\mathcal{D}}_u\subseteq \mathbb{R}\to \mathbb{R}$ and $v:{\mathcal{D}}_v\subseteq \mathbb{R}\to \mathbb{R}$ be real functions and let $S\subseteq {\mathcal{D}}_u\cap {\mathcal{D}}_v$.

If $u$ and $v$ are continuously differentiable on $S$, then the antiderivatives of $u{v}^{\prime }$ are related to the antiderivatives of $u^{\prime }v$ as follows:

$$\int u(x)v^{\prime }(x)\mathrm{dx}=u(x)v(x)-\int u^{\prime }(x)v(x)\mathrm{dx}$$

PROOF

We begin using the product rule for differentiation:

$$(u(x)v(x){)}^{\prime }=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$$

Now, the antiderivatives of the left-hand side must be equal to antiderivatives of the right-hand side, i.e.

$$\int (u(x)v(x){)}^{\prime }\mathrm{dx}=\int u^{\prime }(x)v(x)+u(x)v^{\prime }(x)\mathrm{dx}$$

The antiderivatives of $(u(x)v(x){)}^{\prime }$ are by definition $u(x)v(x)+C$. Therefore, we have

$$(u(x)v(x))+C=\int u^{\prime }(x)v(x)+u(x)v^{\prime }(x)\mathrm{dx}.$$

Next, we apply linearity of antidifferentiation to the right-hand side.

$$(u(x)v(x))+C=\int u^{\prime }(x)v(x)\mathrm{dx}+\int u(x)v^{\prime }(x)\mathrm{dx}.$$

Finally, we just rearrange the terms:

$$\int u(x)v^{\prime }(x)\mathrm{dx}=(u(x)v(x))-\int u^{\prime }(x)v(x)\mathrm{dx}+C$$

Theorem: Integration by Substitution

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ and $g:{\mathcal{D}}_g\subseteq \mathbb{R}\to \mathbb{R}$ be real functions such that the image of $g$ is contained in the domain of $f$ be and let $F$ be an antiderivative of $f$ on $g({\mathcal{D}}_g)$.

If $f$ is continuous and $g$ is differentiable, then the antiderivatives of $f(g(x))g^{\prime }(x)$ are given by $F(g(x))+C$.

NOTATION

We commonly use the following notation to express this:

$$\int f(\underset{u}{\underbrace{g(x)}})\underset{\mathrm{du}}{\underbrace{g^{\prime }(x)\mathrm{dx}}}=\int f(u)\mathrm{du}$$

PROOF

TODO

THEOREM

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ be a real function and let $S\subseteq {\mathcal{D}}_f$.

If $f$ is differentiable on $S$ and $f(x)>0$ for all $x\in S$, then the antiderivatives of the quotient $f^{\prime }/f$ are given by the composition of the real natural logarithm and $f$:

$$\int \frac{f^{\prime }(x)}{f(x)}\mathrm{dx}=\ln (f(x))+C$$

PROOF

TODO

Theorem: Antiderivatives of Polynomial Functions

$$\int x^{\alpha }\mathrm{dx}=\frac{1}{\alpha +1}{x}^{\alpha +1}+C\forall \alpha \ne -1\in \mathbb{R}$$ $$\int (ax+b{)}^{\gamma }\mathrm{dx}=\frac{(ax+b{)}^{\gamma +1}}{a(\gamma +1)}+C$$

PROOF

TODO

Theorem: Antiderivatives of Rational Functions

$$\int \frac{c}{ax+b}\mathrm{dx}=\frac{c}{a}\ln |ax+b|+C$$

For all $n\in \mathbb{N}$ and $a\in \mathbb{R}$:

$$\int \frac{\mathrm{dx}}{(x-a{)}^n}=\{\begin{array}{l}\ln |x-a|+C,\text{ if }n=1\\ -\frac{1}{(n-1)(x-a{)}^{n-1}}+C,\text{ if }n\ge 2\end{array}$$

For all $a,b\in \mathbb{R}$ and all polynomials $x^2+px+q$ with $p^2-4q<0$:

$$\int \frac{\mathrm{dx}}{x^2+px+q}=\frac{2}{\sqrt{4q-p^2}}\arctan \frac{2x+p}{\sqrt{4q-p^2}}+C$$ $$\int \frac{ax+b}{x^2+px+q}\mathrm{dx}=\frac{a}{2}\ln (x^2+px+q)-\left(b-\frac{ap}{2}\right)\int \frac{\mathrm{dx}}{x^2+px+q}$$

For all $n\ge 2\in \mathbb{N}$ and all polynomials $x^2+px+q$ with $p^2-4q<0$:

$$\int \frac{\mathrm{dx}}{(x^2+px+q{)}^n}=\frac{2x+p}{(n-1)(4q-p^2)(x^2+px+q{)}^{n-1}}+\frac{2(2n-3)}{(n-1)(4q-p^2)}\int \frac{\mathrm{dx}}{(x^2+px+q{)}^{n-1}}$$ $$\int \frac{ax+b}{(x^2+px+q{)}^n}\mathrm{dx}=-\frac{a}{2(n-1)(x^2+px+q{)}^{n-1}}+\left(b-\frac{ap}{2}\right)\int \frac{\mathrm{dx}}{(x^2+px+q{)}^{n-1}}$$

PROOF

TODO

Theorem: Antiderivatives of Exponential Functions

$$\int e^{\alpha x}\mathrm{dx}=\frac{1}{\alpha }{e}^{\alpha x}+C$$ $$\int f^{\prime }(x)e^{f(x)}\mathrm{dx}=e^{f(x)}+C$$ $$\int a^x\mathrm{dx}=\frac{a^x}{\ln a}+C$$ $$\int e^x(f(x)+f^{\prime }(x))\mathrm{dx}=e^{x}f(x)+C$$

PROOF

TODO

Theorem: Antiderivatives of Logarithmic Functions

$$\int \ln x\mathrm{dx}=x(\ln x-1)+C$$ $$\int {\log }_{a}x\mathrm{dx}=\frac{x}{\ln a}(\ln x-1)+C$$

PROOF

TODO

Theorem: Antiderivatives of Trigonometric Functions

$$\int \sin x\mathrm{dx}=-\cos x+C$$ $$\int \cos x\mathrm{dx}=\sin x+C$$ $$\int \tan x\mathrm{dx}=-\ln |\cos x|+C$$ $$\int \cot x\mathrm{dx}=\ln |\sin x|+C$$ $$\int \arctan x\mathrm{dx}=x\arctan x-\frac{1}{2}\ln (x^2+1)+C$$

PROOF

TODO