User:Physis/Substitution theorem for indefinite integral

Motivation
I have read a variant of the theorem of integration by substitution which is formulated directly for indefinite integral. I write it only here, not to the article page, because my knowledge in the topic lacks both the overview and the details.

Indefinite integral
The indefinite integral of a function f is notated and conceived as the set of all its primitive functions:
 * $$\int f = \left\{\,F \in \dots \mid F^\prime = f\,\right\}$$.

Point-free style
Abuse of notation (with writing variable in stead of function) will be avoided: a point-free style notation will be used. Operations on real numbers will be "transferred" to real functions in a pointwise way. Composition notation will be explicit in most cases.

Extensional definition of equality
Equality of between two functions is meant extensionally. let f and g have the same domain X (if not said otherwise), then let us notate them equal iff for all x in X both functions result in coinciding values.
 * $$f = g \iff \forall x \in X \left(f\left(x\right) = g\left(x\right)\right)$$

Implicit notation of pointwise transition of operations
The pointwise transition of operation symbols will be meant implicitly

\begin{align} - f &= x \mapsto -f(x) \\ f + g &= x \mapsto f(x) + g(x) \\ f - g &= x \mapsto f(x) - g(x) \\ f \cdot g &= x \mapsto f(x) \cdot g(x) \\ \frac fg &= x \mapsto \frac{f(x)}{g(x)} \end{align} $$ (appropriate restriction applies for g in the last case.)

Implicit notation for pointwise transition of powers
Here, powers of real functions will be defined as an implicite notation of pointwise transition of powers among real numbers:
 * $$f^n = x \mapsto \left(f\left(x\right)\right)^n$$

But in further examples, need for inverse function or iteration of-composition may cause a problem, possibly changing the convention for power notation.

Constant functions
The "pointwise" transition of a real number to a real function is done its corresponding constant functions

\begin{align} \mathrm{const}_{\mathbb R, c_0} &: \mathbb R \to \left\{\,c_0\,\right\} \\ \mathrm{const_{\mathbb R, c_0}}(x) &= c_0 \\ \mathrm{Const}_{\mathbb R,\mathbb R} &= \left\{ \mathrm{const}_{\mathbb R, c_0} \mid c_0 \in \mathbb R\right\} \end{align} $$

Explicit notation of composition
Every other pointwise transition will be notated explicity by use of composition sign:
 * $$\ln \circ \;\mathrm{abs} = x \mapsto \ln \left| x \right|$$

Algebrae for function composition
As mentioned above, powers have been defined here as here as an implicite notation of pointwise transition:
 * $$f^n = x \mapsto \left(f\left(x\right)\right)^n$$

But in further examples, need for inverse function or iteration of-composition may cause a problem, possibly changing the convention for power notation.

Iteration
Functions of a common domain, where the range of each is subset of the [common] domain, form a semigroup with composition. Each of its member can be composed with each other member, including itself, and this can be iterated any finite times.



\begin{align} f^0 &= \mathrm{id} \\ f^1 &= f \\ f^2 &= f \circ f \\ \vdots \\ f^{n +1} &= f \circ f^n \end{align} $$

Motivation
In some differential equation examples, we may get the integral curve in a "reversed" way. We may fail to be able to express the "dependent variable" y explicitly in terms of the independent variable x:



\begin{align} \dot y &= \dots x \dots y \dots \\ &\vdots \\ y &= \dots x \dots y \dots \end{align} $$ but by changing the role of variables x and y, we can express x explicitly in terms of y
 * $$x = \dots y$$

and this may lead to a correct graph of the desired integral curve, the solution of the differential equation. Although it seems for me yet as an abuse of notation, but its correctness makes me think why it is correct at all and how its idea can be formulated in a more transparent way.

The notion of this "transposition" of the independent and dependent variables makes the graph of the function reflected to the "diagonal" line of the identity function. It is exactly the graph of the inverse function (if any). Thus, the idea can be rendered also in a point-free style treatment.

Symmetric group
The inverse of a function f (if any) can be notated as $$f^{-1}$$, supported by the below considerations. Permutations on a common set are not only composable without restriction, they are also invertible (the inverses are also member of the permutation set). They form a symmetry group, we can augment the above notation of powers with the inverses, allowing the exponents ranging over the whole $$\mathbb Z$$ (thus, including all negative integers in addition to the naturals).
 * $$(f^{-1})^n = (f^n)^{-1}$$

the coincidence can give rise to the notion of $$f^{-n}$$

Any kind of "inverse" notation needs care if f is not a permutation, for example only injective, but not surjective.

Theorem
.

If
 * $$F \in \int f$$

then
 * $$F \circ g \in \int \left(f \circ g\right) \cdot g^\prime$$

Application
An application can be found in solving separable differential equations. Let us see first a general scheme lacking concrete details. Let f and g be known functions. It is y the equation must be solved for.
 * $$y^\prime = f \cdot \left(g \circ y\right)$$

is a form characteristic of what we mean by "separable" differential equation (if we use this point-free style notation).
 * $$\frac{y^\prime}{g \circ y} = f$$

we try to avoid abuse of notation.
 * $$\frac1{g \circ y} \cdot y^\prime = f$$

let use "factor out" a little more
 * $$\left(\frac1g \circ y\right) \cdot y^\prime = f$$

Let us introduce the notation a for 1/g
 * $$\left(\underbrace{\frac1g}_a \circ \;y\right) \cdot y^\prime = f$$

and notate one of a's primitive functions as A. Then, by the substitution theorem of indefinite integrals
 * $$A \circ y \in \int \left(a \circ y\right) \cdot y^\prime$$

which can be applied very well for the next step in solving the separable differential equation:
 * $$A \circ y \in \int f$$

Now, if we are lucky, and in the concrete differential equation the part A (and F, a primitive function of f) are of simple form, then we can make a good specification for y, or even cover it with explicit formulae.

Example
Let us see a concrete example.
 * $$- x \cdot y^\prime = y^2$$

Let us use again a point-free notation
 * $$- \mathrm{id} \cdot y^\prime = y^2$$

Let us write it in an explicit form:
 * $$y^\prime = -\frac{y^2}{\mathrm{id}}$$

Singular solution
We can see at once a singular solution
 * $$y = \mathrm{const}_{\mathbb R, 0}$$

General solution
Let us avoid again abuse of notation and use the theorems directly in a transparent form. Then, separation of variables begins with step
 * $$\frac{y^\prime}{y^2} = -\frac1{\mathrm{id}}$$

some "factoring out" steps helps to find the form where the substitution theorem will fit in
 * $$\frac1{y^2} \cdot y^\prime = -\frac1{\mathrm{id}}$$

"splitting" $$y^2$$ into a composition
 * $$\frac1{\mathrm{id}^2 \circ y} \cdot y^\prime = -\frac1{\mathrm{id}}$$

factoring "pre-composition with y" even more outward:
 * $$\left(\frac1{\mathrm{id}^2} \circ y\right) \cdot y^\prime = -\frac1{\mathrm{id}}$$

Let f denote the reciprocal of squared identity:
 * $$\left(\underbrace{\frac1{\mathrm{id}^2}}_f \circ \;y\right) \cdot y^\prime = -\frac1{\mathrm{id}}$$

and choose F as one of its primitive functions. Then by substitution theorem for indefinite integral,
 * $$F \circ y \in \int \left(f \circ y\right) \cdot y^\prime$$,

which means in the subsequent step in solving the differential equation:
 * $$F \circ y \in - \int \frac1{\mathrm{id}}$$

Let us choose for F the simplest primitive function of f: let F be $$-\frac1{\mathrm{id}}$$
 * $$-\frac1{\mathrm{id}} \circ y \in - \int \frac1{\mathrm{id}}$$

expanding the right-hand side
 * $$-\frac1{\mathrm{id}} \circ y \in \left\{\,- \ln \circ \;\mathrm{abs} + c \mid c\in \mathrm{Const}_{\mathbb R,\mathbb R}\,\right\}$$

An equation (for functions) can be given in parametric form, with parameter c running over all possible constant functions $$\mathrm{Const}_{\mathbb R,\mathbb R} = \left\{\mathrm{const}_{\mathbb R, c_0} : \mathbb R \to \left\{\,c_0\,\right\} \mid c_0 \in \mathbb R\right\}$$
 * $$\frac1y = \ln \circ \;\mathrm{abs} + c$$

The solution can be expressed in explicit form
 * $$y = \frac1{\left(\ln \circ \;\mathrm{abs}\right) + c}$$