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Antidifferentiability (Real Functions)#

Definition: Antiderivative

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

We say that \(f\) is antidifferentiable on \(S \subseteq \mathcal{D}_f\) if there exists some real function \(F: \mathcal{D}_F \subseteq \mathbb{R} \to \mathbb{R}\) whose derivative on \(S\) is \(f\):

\[F'(x) = f(x) \qquad \forall x \in S\]

Any such \(F\) is known as an antiderivative of \(f\) on \(S\).

Example

Consider the function \(f: \mathbb{R} \to \mathbb{R}\) defined as

\[f(x) \overset{\text{def}}{=} 3x\]

for all \(x \in \mathbb{R}\).

The functions \(F_1, F_2, F_3: \mathbb{R} \to \mathbb{R}\) defined as

\[\begin{aligned}F_1(x) & \overset{\text{def}}{=} \frac{3}{2}x^2 \\ F_2(x) & \overset{\text{def}}{=} \frac{3}{2}x^2 + 7 \\ F_1(x) & \overset{\text{def}}{=} \frac{3}{2}x^2 - \pi \end{aligned}\]

for all \(x \in \mathbb{R}\) are antiderivatives of \(f\) on \(\mathbb{R}\).

Definition: Indefinite Integral

The set of \(f\)'s antiderivatives on \(S\) is known as \(f\)'s indefinite integral on \(S\):

Notation

The indefinite integral is denoted in the following way:

\[ \int f(x) \,\mathrm{d}x \]

This notation and the name "indefinite integral" are historical remnants and one should be very careful not to confuse them with actual integrals.

Theorem: Antiderivatives on Intervals

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) be a real function, let \(I \subseteq \mathcal{D}\) be an interval and let \(F: \mathcal{D}_F \subseteq \mathbb{R} \to \mathbb{R}\) be an antiderivative of \(f\) on \(I\).

A real function \(G: \mathcal{D}_G \subseteq \mathbb{R} \to \mathbb{R}\) is also an antiderivative of \(f\) on \(I\) if and only if there exists some real number \(C \in \mathbb{R}\) such that

\[F(x) = G(x) + C\]

for all \(x \in I\).

Notation

This fact allows us to introduce simplified notation for the indefinite integral. Instead of writing

\[\int f(x) \,\mathrm{d}x = \{F\mid F \text{ is an antiderivative of } f \text{ on } I\},\]

we write

\[\int f(x) \,\mathrm{d}x = F(x) + C,\]

where \(F\) is any particular antiderivative.

Proof

TODO

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

If \(f\) and \(g\) are antidifferentiable on \(S \subseteq \mathcal{D}_f \cap \mathcal{D}_g\), then so is \(\alpha f + \beta g\) for all \(\alpha, \beta \in \mathbb{R}\) with

\[\int \alpha f(x) + \beta g(x) \,\mathrm{d}x = \alpha \int f(x) \,\mathrm{d}x + \beta \int g(x) \,\mathrm{d}x.\]
Proof

TODO

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 \(uv'\) are related to the antiderivatives of \(u'v\) as follows:

\[\int u(x) v'(x) \mathop{\mathrm{d}x} = u(x)v(x) - \int u'(x)v(x) \mathop{\mathrm{d}x}\]
Proof

We begin using the product rule for differentiation:

\[(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)\]

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

\[\int (u(x)v(x))' \mathop{\mathrm{d}x} = \int u'(x)v(x) + u(x)v'(x) \mathop{\mathrm{d}x}\]

The antiderivatives of \((u(x)v(x))'\) are by definition \(u(x)v(x) + C\). Therefore, we have

\[(u(x)v(x)) + C = \int u'(x)v(x) + u(x)v'(x) \mathop{\mathrm{d}x}.\]

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

\[(u(x)v(x)) + C = \int u'(x)v(x) \mathop{\mathrm{d}x} + \int u(x)v'(x) \mathop{\mathrm{d}x}.\]

Finally, we just rearrange the terms:

\[\int u(x)v'(x) \mathop{\mathrm{d}x} = (u(x)v(x)) - \int u'(x)v(x) \mathop{\mathrm{d}x} + C\]
Example: \(\int x \mathrm{e}^x \,\mathrm{d}x\)

We want to find the antiderivatives of \(f(x) = x \mathrm{e}^x\). We have

\[f(x) = u(x) v'(x)\]

with \(u(x) = x\), \(u'(x) = 1\), \(v'(x) = \mathrm{e}^x\) and \(v(x) = \mathrm{e}^x\). Since both \(u\) and \(v\) are continuously differentiable on \(\mathbb{R}\), we have:

\[\begin{aligned}\int x \mathrm{e}^x \,\mathrm{d}x & = u(x)v(x) - \int u'(x) v(x) \,\mathrm{d}x \\ & = x \mathrm{e}^x - \int \mathrm{e}^x \,\mathrm{d}x \\ & = x \mathrm{e}^x - \mathrm{e}^x + C \\ & = \mathrm{e}^x (x - 1) + C \end{aligned}\]
Example: \(\int \mathrm{e}^x \sin x \,\mathrm{d}x\)

We want to find the antiderivatives of \(f(x) = \mathrm{e}^x \sin x\). We have

\[f(x) = u(x) v'(x)\]

with \(u(x) = \sin x\), \(u'(x) = \cos x\), \(v'(x) = \mathrm{e}^x\) and \(v(x) = \mathrm{e}^x\). Since both \(u\) and \(v\) are continuously differentiable on \(\mathbb{R}\), we have:

\[\begin{aligned}\int \mathrm{e}^{x}\sin x\,\mathrm{d}x & = u(x)v(x) - \int u'(x) v(x)\,\mathrm{d}x \\ & = \mathrm{e}^x \sin x - \int \mathrm{e}^x \cos x \, \mathrm{d}x \end{aligned}\]

We can use the same procedure for \(\mathrm{e}^x \cos x\), since

\[\mathrm{e}^x \cos x = u(x)v'(x)\]

with \(u(x) = \cos x\), \(u'(x) = -\sin x\), \(v'(x) = \mathrm{e}^x\) and \(v(x) = \mathrm{e}^x\).

Since both \(u\) and \(v\) are continuously differentiable on \(\mathbb{R}\), we have:

\[\begin{aligned}\int \mathrm{e}^x \cos x \,\mathrm{d}x & = u(x)v(x) - \int u'(x)v(x) \,\mathrm{d}x \\ & = \mathrm{e}^x \cos x - \int \mathrm{e}^x (-\sin x) \,\mathrm{d}x \\ & = \mathrm{e}^x \cos x + \int \mathrm{e}^x \sin x \,\mathrm{d}x \end{aligned}\]

We substitute this result into the previous expression:

\[\begin{aligned}\int \mathrm{e}^{x}\sin x\,\mathrm{d}x & = \mathrm{e}^x \sin x - \int \mathrm{e}^x \cos x \, \mathrm{d}x \\ & = \mathrm{e}^x \sin x - \left(\mathrm{e}^x \cos x + \int \mathrm{e}^x \sin x \,\mathrm{d}x\right) \\ & = \mathrm{e}^x \sin x - \mathrm{e}^x \cos x - \int \mathrm{e}^x \sin x \,\mathrm{d}x \end{aligned}\]

At first, it might seem that we have achieved nothing because we still have \(\int \mathrm{e}^x \sin x \,\mathrm{d}x\). However, on the left side we have it with a plus and on the right side we have it with a minus. Thus, we can move the one on the right to the left and solve for it:

\[2\int \mathrm{e}^x \sin x \,\mathrm{d}x = \mathrm{e}^x \sin x - \mathrm{e}^x \cos x\]
\[\int \mathrm{e}^x \sin x \,\mathrm{d}x = \frac{1}{2} \mathrm{e}^x (\sin x - \cos x) + C\]
Example: \(\int \ln x \,\mathrm{d}x\)

We want to find the antiderivatives of the real natural logarithm.

We have

\[ \ln x = u(x) v'(x) \]

with \(u(x) = \ln x\), \(u'(x) = \frac{1}{x}\), \(v'(x) = 1\) and \(v(x) = x\) for all \(x \in (0, \infty)\). Since \(u\) and \(v\) are continuously differentiable on \((0,\infty)\), we have the following:

\[ \begin{aligned}\int \ln x \,\mathrm{d}x & = u(x) v(x) - \int u'(x) v(x) \,\mathrm{d}x \\ & = x \ln x - \int \frac{1}{x}x\,\mathrm{d}x \\ & = x \ln x - x + C\end{aligned} \]

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.

If \(g\) is differentiable on \(S \subseteq \mathcal{D}_g\) and \(F\) is an antiderivative of \(f\) on \(g(S)\), then the composition \(F \circ g\) is an antiderivative of \((f\circ g) g'\) on \(S\).

Example: \(\int x \mathrm{e}^{x^2} \,\mathrm{d}x\)

We want to determine the antiderivatives of \(x \mathrm{e}^{x^2}\):

\[\int x \mathrm{e}^{x^2} \,\mathrm{d}x\]

We notice that

\[x\mathrm{e}^{x^2} = \frac{1}{2}f(g(x))g'(x)\]

with \(g(x) = x^2\), \(g'(x) = 2x\) and \(f(x) = \mathrm{e}^x\). One antiderivative \(F(x)\) of \(f(x)\) is \(\mathrm{e}^x\). We can verify that \(f\) and \(g\) satisfy the requirements of the theorem and so \((F \circ g)(x) = \mathrm{e}^{x^2}\) is an antiderivative of \(2x \mathrm{e}^{x^2}\). Therefore, \(\frac{1}{2}\mathrm{e}^{x^2}\) is an antiderivative of \(x \mathrm{e}^{x^2}\).

Using the integral notation, this would be written in the following way with \(u = g(x) = x^2\):

\[\begin{aligned}\int x \mathrm{e}^{x^2} \,\mathrm{d}x & = \int \frac{1}{2}f(g(x))g'(x)\,\mathrm{d}x \\ & = \frac{1}{2} \int f(g(x))g'(x) \,\mathrm{d}x \\ & = \frac{1}{2}\int f(u)\,\mathrm{d}u \\ & = \frac{1}{2}\mathrm{e}^u + C \\ & = \frac{1}{2}\mathrm{e}^{g(x)} + C \\ & = \frac{1}{2}\mathrm{e}^{x^2} + C\end{aligned}\]

Algorithm: Integration by Substitution

Integration by substitution can often be used to find the antiderivatives of some real function \(h\), when we have an analytic expression for \(h\):

\[\int h(x) \,\mathrm{d}x\]

The goal is to find \(f\) and \(g\) such that \(h(x) = f(g(x))g'(x)\).

  1. Pick some part \(u\) of the expression for \(h\) such that \(h\) can be expressed only in terms of \(u\) without \(x\).
  • Here, \(u\) plays the role of our guess for \(g\).
  • The expression for \(h\) in terms of \(u\) is our guess for \(f\).
  1. Compute the derivative of \(u\) with respect to \(x\) to obtain some expression for \(g'\):
\[\frac{\mathrm{d}u}{\mathrm{d}x} = \text{expression}\]
  • If the resulting expression can be written only in terms of \(u\) without \(x\), then proceed. If not, go back to step 1.
  1. Write \(\mathrm{d}x\) as the quotient of \(\mathrm{d}u\) and this expression:
\[\mathrm{d}x = \frac{\mathrm{d}u}{\text{expression}}\]

  1. Verify that \(f\), \(g\) and \(g'\) satisfy the requirements of the theorem.

  2. Substitute \(\frac{\mathrm{d}u}{\text{expression}}\) and \(f(u)\) back into the original integral:

\[\int h(x) \,\mathrm{d}x = \int f(u) \frac{\mathrm{d}u}{\text{expression}}\]
  • Since \(f\) and \(\text{expression}\) contain only \(u\)'s and no \(x\)'s, compute the integral as you would normally.
  1. Substitute the expression (in terms of \(x\)) for \(u\) back into the result from step 5.
Example: \(\int \frac{1}{\mathrm{e}^x + 1}\,\mathrm{d}x\)

We want to determine the antiderivatives of \(\frac{1}{\mathrm{e}^x + 1}\):

\[\int \frac{1}{\mathrm{e}^x + 1} \, \mathrm{d}x\]

Let's try to substitute \(\mathrm{e}^x = u\). We have

\[\frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^x = u \implies \mathrm{d}x = \frac{\mathrm{d}u}{u}\]

Therefore:

\[\begin{aligned}\int \frac{1}{\mathrm{e}^x + 1} \, \mathrm{d}x & = \int \frac{1}{u + 1}\frac{1}{u}\,\mathrm{d}u \\ & = \int \frac{1}{u} - \frac{1}{u + 1}\,\mathrm{d}u \\ & = \ln |u| - \ln |u + 1| + C \\ & = \ln \left\vert\frac{u}{u+1}\right\vert + C \\ & = \ln \frac{\mathrm{e}^x}{\mathrm{e}^x + 1} + C\end{aligned}\]
Proof

TODO

Theorem: Antidifferentiation of Logarithmic Derivatives

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

If \(f\) is differentiable on an interval \(I \subseteq \mathcal{D}\) and \(f(x) \ne 0\) for all \(x \in I\), then the antiderivatives of the quotient \(f'/f\) are given by the composition of the real natural logarithm and \(|f|\):

\[ \int \frac{f'(x)}{f(x)} \mathop{\mathrm{d}x} = \ln |f(x)| + C \]
Proof

TODO

Theorem: Antiderivatives of Polynomial Functions

\[ \int x^\alpha \mathop{\mathrm{d}x} = \frac{1}{\alpha + 1} x^{\alpha + 1} + C \qquad \forall \alpha \ne -1 \in \mathbb{R} \]
\[ \int (ax + b)^\gamma \mathop{\mathrm{d}x} = \frac{(ax+b)^{\gamma+1}}{a(\gamma +1)} + C \]
Proof

TODO

Theorem: Antiderivatives of Rational Functions

\[ \int \frac{c}{ax+b} \mathop{\mathrm{d}x} = \frac{c}{a} \ln |ax+b| + C \]

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

\[ \int \frac{\mathop{\mathrm{d}x}}{(x-a)^n} = \begin{cases}\displaystyle \ln |x-a| + C, \text{ if } n = 1 \\\displaystyle -\frac{1}{(n-1)(x-a)^{n-1}} + C, \text{ if } n\ge 2 \end{cases} \]

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

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

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

\[ \int\frac{\mathop{\mathrm{d}x}}{(x^2+px+q)^n} = \frac{2x+p}{(n-1)(4q-p^2)(x^2+px+q)^{n-1}}+\frac{2(2n-3)}{(n-1)(4q-p^2)}\int\frac{\mathop{\mathrm{d}x}}{(x^2+px+q)^{n-1}} \]
\[ \int\frac{ax+b}{(x^2+px+q)^n}\mathop{\mathrm{d}x} = -\frac{a}{2(n-1)(x^2+px+q)^{n-1}}+\left(b-\frac{ap}{2}\right)\int\frac{\mathop{\mathrm{d}x}}{(x^2+px+q)^{n-1}} \]
Proof

TODO