nyuuzyou/wikis huggingface dataset:
Integral with respect to Dirac Measure/Proof 2

== Theorem ==
== Proof ==
We first prove the result for positive simple functions.
Let $g : X \to \R$ be a positive simple function.
From Simple Function has Standard Representation, there exists:
:a finite sequence $a_1, \ldots, a_n$ of real numbers
:a partition $E_0, E_1, \ldots, E_n$ of $X$ into $\Sigma$-measurable sets
such that: 
:$\ds g = \sum_{i \mathop = 0}^n a_i \chi_{E_i}$
Then, from the definition of the $\delta_x$-integral of a positive simple function:
:$\ds \int g \rd \delta_x = \sum_{i \mathop = 1}^n a_i \map {\delta_x} {E_i}$
Since $E_0, E_1, \ldots, E_n$ is a partition of $X$, precisely one contains $x$. 
That is, $x \in E_i$ for precisely one $i$. 
Then:
:$\map {\delta_x} {E_j} = 0$ for $i \ne j$
leaving:
:$\ds \int g \rd \delta_x = a_i$
Note that: 
:$\ds \map g x = \sum_{j \mathop = 0}^n a_j \map {\chi_{E_j} } x$
and $\map {\chi_{E_j} } x = 1$ $j = i$, so we have: 
:$\map g x = a_i$
So we have: 
:$\ds \int g \rd \delta_x = \map g x$ 
for positive simple functions $g$.
Now consider a positive $\Sigma$-measurable function $f : X \to \overline \R_{\ge 0}$. 
From Measurable Function is Pointwise Limit of Simple Functions, there exists an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ of positive simple functions such that: 
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
for each $x \in X$.
For each $n \in \N$, we have: 
:$\ds \int f_n \rd \delta_x = \map {f_n} x$
from our previous work. 
From the Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions, we have: 
:$\ds \int f \rd \delta_x = \lim_{n \mathop \to \infty} \int f_n \rd \delta_x$
That is: 
:$\ds \int f \rd \delta_x = \lim_{n \mathop \to \infty} \map {f_n} x = \map f x$
Now consider an arbitrary $\Sigma$-measurable function $f : X \to \overline \R$. 
From Function Measurable iff Positive and Negative Parts Measurable, we have: 
:$f^+$ and $f^-$ are $\Sigma$-measurable.
Now, if $\map f x \ge 0$, we have:
:$\map {f^+} x = \map f x$
and:
:$\map {f^-} x = 0$
So:
:$\ds \int f^+ \rd \delta_x = \map f x$
and:
:$\ds \int f^- \rd \delta_x = 0$
giving:
If $\map f x < 0$, then we have: 
:$\map {f^+} x = 0$
and:
:$\map {f^-} x = -\map f x$
so that: 
:$\ds \int f^+ \rd \delta_x = 0$
and:
:$\ds \int f^- \rd \delta_x = -\map f x$
Then we have:
Hence the demand.
Category:Integral with respect to Dirac Measure
Integral with respect to Discrete Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$. 
Let $\ds \mu = \sum_{n \mathop \in \N} \lambda_n \delta_{x_n}$ be a discrete measure on $\struct {X, \Sigma}$.
Let $f \in \MM_{\overline \R}^+, f: X \to \overline \R$ be a positive measurable function.
Then:
:$\ds \int f \rd \mu = \sum_{n \mathop \in \N} \lambda_n \map f {x_n}$
where the integral sign denotes $\mu$-integration.
== Proof ==
We have: 
== Sources ==
* : $9.10 \ \text{(ii)}$
Category:Integral of Positive Measurable Function
Category:Discrete Measures
Integral with respect to Kernel Transformation of Measure

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $N: X \times \Sigma \to \overline \R_{\ge 0}$ be a kernel.
Let $f: X \to \overline \R$ be a positive measurable function.
Then:
:$\ds \int f \map \rd {\mu N} = \int N f \rd \mu$
where:
:The integral sign denotes integration with respect to a measure
:$\mu N$ is the transformation of $\mu$ by $N$
:$N f$ is the transformation of $f$ by $N$
Writing $\map \mu f$ in place of $\ds \int f \rd \mu$, the theorem statement can be conveniently expressed as:
:$\map {\mu N} f = \map \mu {N f}$
== Proof ==
== Sources ==
* : $\S 9$: Problem $11 \ \text{(iii)}$
Category:Kernel Transformation of Measure
Integral with respect to Pushforward Measure

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\struct {X', \Sigma'}$ be a measurable space.
Let $T: X \to X'$ be a $\Sigma \, / \, \Sigma'$-measurable mapping.
Let $f: X' \to \overline \R$ be a positive $\Sigma'$-measurable function.
Let $\map T \mu$ be the pushforward measure of $\mu$ under $T$.
Then $f \circ T: X \to \overline \R$ is positive and $\Sigma$-measurable with: 
:$\ds \int_{X'} f \rd \map T \mu = \int_X f \circ T \rd \mu$
=== Corollary ===
== Proof ==
From Composition of Measurable Mappings is Measurable:
:$f \circ T$ is $\Sigma$-measurable.
Clearly $f \circ T \ge 0$, so $f \circ T$ is a positive $\Sigma$-measurable function.
We first show the proposition for characteristic functions $f$.
Suppose $f = \chi_A$ for $A \in \Sigma'$.
Then, we have: 
So, we have: 
:$\ds \int_{X'} \chi_A \rd \map T \mu = \int_X \chi_A \circ T \rd \mu$
whenever $A \in \Sigma'$. 
Now suppose that $f$ is a positive simple function.
From Simple Function has Standard Representation, there exists:
:a finite sequence $a_0, \ldots, a_n$ of non-negative real numbers
:a partition $E_0, E_1, \ldots, E_n$ of $X$ into $\Sigma'$-measurable sets
such that: 
:$\ds f = \sum_{i \mathop = 0}^n a_i \chi_{E_i}$
Then: 
Now suppose that $f$ is a general positive $\Sigma'$-measurable function.
From Measurable Function is Pointwise Limit of Simple Functions, there exists an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ of positive simple functions such that: 
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
for each $x \in X$.
For each $n \in \N$, we have:
:$\ds \int_{X'} f_n \rd \map T \mu = \int_X f_n \circ T \rd \mu$
Then:
== Sources ==
* : $14.1$
Category:Integral of Integrable Function
Category:Pushforward Measures
Category:Integral with respect to Pushforward Measure
Integral with respect to Pushforward Measure/Corollary

== Corollary ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\struct {X', \Sigma'}$ be a measurable space.
Let $T: X \to X'$ be a $\Sigma \, / \, \Sigma'$-measurable mapping.
Let $f: X' \to \overline \R$ be a positive $\Sigma'$-measurable function.
Let $\map T \mu$ be the pushforward measure of $\mu$ under $T$.
Then $f \circ T : X \to \overline \R$ is $\mu$-integrable $f : X' \to \overline \R$ is $\map T \mu$-integrable.
In this case, we have: 
:$\ds \int_{X'} f \rd \map T \mu = \int_X f \circ T \rd \mu$
== Proof ==
From Function Measurable iff Positive and Negative Parts Measurable, we have: 
:$f^+$ and $f^-$ are $\Sigma'$-measurable.
From Composition of Measurable Mappings is Measurable, we have: 
:$f \circ T$ is $\Sigma$-measurable.
Then from Function Measurable iff Positive and Negative Parts Measurable, we have:
:$\paren {f \circ T}^+$ and $\paren {f \circ T}^-$ are $\Sigma$-measurable.
Then, we have: 
and:
So:
:$\ds \int_X \paren {f \circ T}^+ \rd \mu < \infty$ and $\ds \int_X \paren {f \circ T}^- \rd \mu <\infty$
:
:$\ds \int_{X'} f^+ \rd \map T \mu < \infty$ and $\ds \int_{X'} f^- \rd \map T \mu < \infty$. 
So $f \circ T : X \to \overline \R$ is $\mu$-integrable $f : X' \to \overline \R$ is $\map T \mu$-integrable
In this case: 
== Sources ==
* : $14.2$
Category:Integral with respect to Pushforward Measure
Integral with respect to Restriction of Measure to Trace Sigma-Algebra of Measurable Set

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $A \in \Sigma$.
Let $\Sigma_A$ be the trace $\sigma$-algebra of $A$ in $\Sigma$.
Let $\mu \restriction_{\Sigma_A}$ be the restriction of $\mu$ to $\Sigma_A$. 
Let $f : X \to \overline \R$ be a positive measurable function.
Let $f \restriction_A$ be the restriction of $f$ to $A$. 
Define $f^\ast : X \to \overline \R$ by: 
:$\ds \map {f^\ast} x = \begin{cases}\map f x & x \in A \\ 0 & x \not \in A\end{cases}$
for each $x \in X$. 
Then, we have: 
:$\ds \int \paren {f \restriction_A} \rd \paren {\mu \restriction_{\Sigma_A} } = \int f^\ast \rd \mu$
== Proof ==
From Restriction of Measurable Function is Measurable on Trace Sigma-Algebra, $f \restriction_A$ is $\Sigma_A$-measurable.
Consider the case that $f$ is a positive simple function.
From Simple Function has Standard Representation, there exists: 
:a finite sequence $a_0, \ldots, a_n$ of non-negative real numbers
:a partition $E_0, E_1, \ldots, E_n$ of $\Sigma_A$-measurable sets
such that: 
:$\ds f \restriction_A \, = \sum_{j \mathop = 0}^n a_j \chi_{E_j}$
From Trace Sigma-Algebra of Measurable Set: 
:$E_0, E_1, \ldots, E_n$ are in fact $\Sigma$-measurable subsets of $A$. 
Then, since $\map {f^\ast} x = 0$ for $x \in X \setminus A$, we have: 
:$\ds f^\ast = \sum_{j \mathop = 0}^n a_j \chi_{E_j}$
Then $f^\ast$ is a positive simple function. 
Then, we have: 
Now suppose that $f$ is a general positive $\Sigma$-measurable function.
From Measurable Function is Pointwise Limit of Simple Functions, there exists an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ of positive simple functions such that: 
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
for each $x \in X$.
In particular: 
:$\ds \map {f \restriction_A} x = \lim_{n \mathop \to \infty} \map {f_n \restriction_A} x$
for each $x \in A$. 
We have established that $f^\ast$ (using the notation introduced in the hypotheses) is a positive simple function for each positive simple function $f$.
In particular, $f_n^\ast$ is a positive simple function for each $n \in \N$.
We now show that $f_n^\ast \to f^\ast$ pointwise.
Then we have, for $x \in A$: 
For $x \in X \setminus A$, we have: 
From Pointwise Limit of Measurable Functions is Measurable, we therefore have: 
:$f^\ast$ is $\Sigma$-measurable.
From our previous computation with positive simple functions, we have: 
:$\ds \int \paren {f_n \restriction_A} \rd \paren {\mu \restriction_{\Sigma_A} } = \int f_n^\ast \rd \mu$
Then, we have:
as required.
Category:Trace Sigma-Algebras
Integral with respect to Series of Measures

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\ds \mu := \sum_{n \mathop \in \N} \lambda_n \mu_n$ be a series of measures on $\struct {X, \Sigma}$.
Then for all positive measurable functions $f: X \to \overline \R, f \in \MM_{\overline \R}^+$:
:$\ds \int f \rd \mu = \sum_{n \mathop \in \N} \lambda_n \int f \rd \mu_n$
where the integral signs denote integration with respect to a measure.
== Proof ==
== Sources ==
* : $\S 9$: Problem $7$
Category:Integral of Positive Measurable Function
Category:Series of Measures
Integrally Closed is Local Property

== Theorem ==
Let $A$ be an integral domain.
For a prime ideal $\mathfrak p$ of $A$, let $A_{\mathfrak p}$ denote the localization at $S = A \divides \mathfrak p$.
Then the following are equivalent:
:$(1): \quad A$ is integrally closed
:$(2): \quad A_{\mathfrak p}$ is integrally closed for all prime ideals $\mathfrak p$.
:$(3): \quad A_{\mathfrak m}$ is integrally closed for all maximal ideals $\mathfrak m$.
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $\map Q R$ denote the field of quotients of an integral domain $R$.
We have by Localization Preserves Integral Closure that:
:$\map Q {A_{\mathfrak p}} = \map Q A$
Hence $A_{\mathfrak p}$ is integrally closed for all prime ideals $\mathfrak p$.
=== $(2)$ implies $(3)$ ===
This is true because a Maximal Ideal of Commutative and Unitary Ring is Prime Ideal.
=== $(3)$ implies $(1)$ ===
Category:Algebraic Number Theory
Integrals Involving x squared minus a squared, x squared less than a squared/Mistake

== Source Work ==
:Chapter $17$: Tables of Special Indefinite Integrals
::$(8)$ Integrals Involving $x^2 - a^2$, $x^2 < a^2$
== Mistake ==
:$(8)$ Integrals Involving $x^2 - a^2$, $x^2 < a^2$
== Correction ==
This family of primitives actually involve $a^2 - x^2$, $x^2 < a^2$.
== Sources ==
* : $\S 17$: Tables of Special Indefinite Integrals: $(8)$ Integrals Involving $x^2 - a^2$, $x^2 < a^2$: $17.8.1.$
Category:Mistakes/Mathematical Handbook of Formulas and Tables
Integrated Linear Differential Mapping is Continuous

== Theorem ==
Let $C^1 \closedint a b := \map {C^1} {\closedint a b, \R}$ be the space of real functions of differentiability class $C^1$.
Let $S$ be the set of differentiable functions on closed real interval vanishing at their endpoints:
:$S := \set {\mathbf h \in C^1 \closedint a b : \map {\mathbf h} a = \map {\mathbf h} b = 0}$
Let $S \subseteq C^1 \closedint a b$ be equiped with the $C^1$ norm. 
Let $\mathbf A, \mathbf B \in C \closedint a b$ be continuous real functions.
Let $L : S \to \R$ be the integrated linear differential mapping:
:$\ds \map L {\mathbf h} = \int_a^b \paren {\map {\mathbf A} t \map {\mathbf h} t + \map {\mathbf B} t \map {\mathbf h'} t} \rd t$
where $\mathbf h \in S$.
Then $L$ is continuous.
== Proof ==
We have that the Integrated Linear Differential Mapping is Linear. 
For $\mathbf h \in S$ we have:
where: 
:$\ds M : = \int_a^b \paren {\size {\map {\mathbf A} t} + \size {\map {\mathbf B} t} } \rd t$
By Continuity of Linear Transformation between Normed Vector Spaces $L$ is continuous.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$
Category:Operator Theory
Category:Continuous Mappings
Category:Linear Transformations
Integrating Factor for First Order ODE

== Theorem ==
Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be non-homogeneous and not exact.
By Existence of Integrating Factor, if $(1)$ has a general solution, there exists an integrating factor $\map \mu {x, y}$ such that:
:$\ds \map \mu {x, y} \paren {\map M {x, y} + \map N {x, y} \frac {\d y} {\d x} } = 0$
is an exact differential equation.
Unfortunately, there is no systematic method of finding such a $\map \mu {x, y}$ for all such equations $(1)$."In general this is quite difficult."
::: -- : $\S 2.9$: Integrating Factors
However, there are certain types of first order ODE for which an integrating factor can be found procedurally.
=== Function of One Variable: $x$ or $y$ only ===
=== Function of $x + y$ ===
=== Function of $x y$ ===
== Proof ==
We have one of these:
: Integrating Factor for First Order ODE: Function of One Variable: $x$ or $y$ only
: Integrating Factor for First Order ODE: Function of $x + y$
: Integrating Factor for First Order ODE: Function of $x y$
=== Conclusion of Proof ===
== Technique for finding an Integrating Factor ==
== Examples ==
== References ==
== Sources ==
* : $\S 2.9$: Integrating Factors
Category:First Order ODEs
Category:Integrating Factors
Integrating Factor for First Order ODE/Conclusion

== Theorem ==
Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be non-homogeneous and not exact.
Let $\map \mu {x, y}$be an integrating factor for $(1)$.
If one of these is the case:
:$\mu$ is a function of $x$ only
:$\mu$ is a function of $y$ only
:$\mu$ is a function of $x + y$
:$\mu$ is a function of $x y$
then:
:$\mu = e^{\int \map f w \rd w}$
where $w$ depends on the nature of $\mu$.
== Proof ==
We have one of these:
:Integrating Factor for First Order ODE: Function of One Variable: $x$ or $y$ only
:Integrating Factor for First Order ODE: Function of $x + y$
:Integrating Factor for First Order ODE: Function of $x y$
We have an equation of the form:
:$\dfrac 1 \mu \dfrac {\d \mu} {\d w} = \map f w$
which is what you get when you apply the Chain Rule for Derivatives and Derivative of Logarithm Function to:
:$\dfrac {\map \d {\ln \mu} } {\d w} = \map f w$
Thus:
:$\ds \ln \mu = \int \map f w \rd w$
and so:
:$\mu = e^{\int \map f w \rd w}$
Hence the results as stated.
== Sources ==
* : $\S 2.9$: Integrating Factors
Category:Integrating Factors
Integrating Factor for First Order ODE/Examples

== Examples of Integrating Factor for First Order ODE ==
This page gathers examples of first order ODEs solved using the technique of Solution of First Order ODE by Integrating Factor.
=== $y \rd x + \paren {x^2 y - x} \rd y = 0$ ===
=== $\paren {3 x^2 - y^2} \rd y - 2 x y \rd x = 0$ ===
=== $\paren {x y - 1} \rd x + \paren {x^2 - x y} \rd y = 0$ ===
=== $y \rd x + x \rd y + 3 x^3 y^4 \rd y = 0$ ===
Category:Integrating Factors
Integrating Factor for First Order ODE/Function of One Variable

== Theorem ==
Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be non-homogeneous and not exact.
Suppose that:
:$\map g x = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } {\map N {x, y} }$
is a function of $x$ only.
Then:
:$\map \mu x = e^{\int \map g x \rd x}$
is an integrating factor for $(1)$.
Similarly, suppose that:
:$\map h y = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } {\map M {x, y} }$
is a function of $y$ only.
Then:
:$\map \mu y = e^{\int -\map h y \rd y}$
is an integrating factor for $(1)$.
== Proof ==
=== Preliminary Work ===
=== Proof for Function of $x$ or $y$ only ===
Suppose that $\mu$ is a function of $x$ only.
Then:
:$\dfrac {\partial \mu} {\partial x} = \dfrac {d \mu} {d x}, \dfrac {\partial \mu} {\partial y} = 0$
which, when substituting in $(3)$, leads us to:
:$\dfrac 1 \mu \dfrac {\d \mu} {\d x} = \dfrac {\map P {x, y} } {\map N {x, y} } = \map g x$
where $\map g x$ is the function of $x$ that we posited.
Similarly, if $\mu$ is a function of $y$ only, we find that:
:$\dfrac 1 \mu \dfrac {\d \mu} {\d y} = \dfrac {\map P {x, y} } {-\map M {x, y} } = \map h y$
where $\map h y$ is the function of $y$ that we posited.
=== Conclusion of Proof ===
== Sources ==
* : $\S 2.9$: Integrating Factors
Category:Integrating Factors
Integrating Factor for First Order ODE/Function of Product of Variables

== Theorem ==
Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be non-homogeneous and not exact.
Let $(1)$ be such that:
:$\map g z = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}} {N y - M x}$
is a function of $z$, where $z = x y$.
Then:
:$\map \mu {x y} = \map \mu z = e^{\int \map g z \d z}$
is an integrating factor for $(1)$.
== Proof ==
=== Preliminary Work ===
=== Proof for Function of x y ===
Suppose that $\mu$ is a function of $z = x y$.
Then:
:$\dfrac {\partial z} {\partial x} = y$
:$\dfrac {\partial z} {\partial y} = x$
Thus:
:$\dfrac {\partial \mu} {\partial x} = \dfrac {\d \mu} {\d z} \dfrac {\partial z} {\partial x} = y \dfrac {\d \mu} {\d z}$
:$\dfrac {\partial \mu} {\partial y} = \dfrac {\d \mu} {\d z} \dfrac {\partial z} {\partial y} = x \dfrac {\d \mu} {\d z}$
which, when substituting in $(3)$, leads us to:
:$\dfrac 1 \mu \dfrac {\d \mu} {\d z} = \dfrac {\map P {x, y} } {N y - M x} = \map g z$
where $\map g z$ is the function of $z$ that we posited.
=== Conclusion of Proof ===
== Sources ==
* : $\S 2.9$: Integrating Factors: Problem $1$
Category:Integrating Factors
Integrating Factor for First Order ODE/Function of Sum of Variables

== Theorem ==
Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be non-homogeneous and not exact.
Suppose that:
:$\map g z = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } {\map N {x, y} - \map M {x, y} }$
is a function of $z$, where $z = x + y$.
Then:
:$\map \mu {x + y} = \map \mu z = e^{\int \map g z \rd z}$
is an integrating factor for $(1)$.
== Proof ==
=== Preliminary Work ===
=== Proof for Function of $x + y$ ===
Suppose that $\mu$ is a function of $z = x + y$.
Then:
:$\dfrac {\partial z} {\partial x} = 1 = \dfrac {\partial z} {\partial y}$
Thus:
:$\dfrac {\partial \mu} {\partial x} = \dfrac {d \mu} {d z} \dfrac {\partial z} {\partial x} = \dfrac {\d \mu} {\d z} = \dfrac {\d \mu} {\d z} \dfrac {\partial z} {\partial y} = \dfrac {\partial \mu} {\partial y}$
which, when substituting in $(3)$, leads us to:
:$\dfrac 1 \mu \dfrac {\d \mu} {\d z} = \dfrac {\map P {x, y} } {\map N {x, y} - \map M {x, y} } = \map g z$
where $\map g z$ is the function of $z$ that we posited.
=== Conclusion of Proof ===
== Sources ==
* : $\S 2.9$: Integrating Factors: Problem $3$
Category:Integrating Factors
Integrating Factor for First Order ODE/Preliminary Work

== Theorem ==
Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be non-homogeneous and not exact.
Let $\map \mu x$ be an integrating factor for $(1)$.
Let:
:$\map P {x, y} := \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
Then:
:$\dfrac 1 \mu = \dfrac {\map P {x, y} } {N \dfrac {\partial \mu} {\partial x} - M \dfrac {\partial \mu} {\partial y} }$
== Proof ==
Let us for ease of manipulation express $(1)$ in the form of differentials:
:$(2): \quad \map M {x, y} \rd x + \map N {x, y} \rd y = 0$
Let $\mu$ be an integrating factor for $(2)$.
Then, by definition:
:$\mu \, \map M {x, y} \rd x + \mu \, \map N {x, y} \rd y = 0$
is an exact differential equation.
By Solution to Exact Differential Equation:
:$\dfrac {\map \partial {\mu M} } {\partial y} = \dfrac {\map \partial {\mu N} } {\partial x}$
Evaluating this, using the Product Rule for Derivatives:
:$\mu \dfrac {\partial M} {\partial y} + M \dfrac {\partial \mu} {\partial y} = \mu \dfrac {\partial N} {\partial x} + N \dfrac {\partial \mu} {\partial x}$
which leads us to:
:$\dfrac 1 \mu \paren {N \dfrac {\partial \mu} {\partial x} - M \dfrac {\partial \mu} {\partial y} } = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
Let us use $\map P {x, y}$ for $\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$.
Then:
:$(3): \quad \dfrac 1 \mu = \dfrac {\map P {x, y} } {N \dfrac {\partial \mu} {\partial x} - M \dfrac {\partial \mu} {\partial y}}$
== Sources ==
* : $\S 2.9$: Integrating Factors
Category:Integrating Factors
Integrating Factor for First Order ODE/Technique for finding Integrating Factor

== Proof Technique ==
Suppose you have a first order ODE which is in (or can be manipulated into) the form:
:$\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
and it is not homogeneous, exact or linear.
Then what you can do is evaluate:
:$\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
and see what you get when you divide it by each of $N$, $M$, $N - M$ and $N y - M x$ in turn.
Then examine what you get to see if you have a function in $x$ only, $y$ only, $x + y$ or $x y$ respectively.
Suppose this has been achieved.
Then from Integrating Factor for First Order ODE, you have found an integrating factor and can solve the equation by using the technique defined in Solution to Exact Differential Equation.
Category:Integrating Factors
Category:Proof Techniques
Integration by Partial Fractions

== Theorem ==
Let $\map R x = \dfrac {\map P x} {\map Q x}$ be a rational function over $\R$ such that the degree of the polynomial $P$ is strictly smaller than the degree of the polynomial $Q$.
Consider the primitive:
:$\ds \int \map R x \rd x$
Let $\map R x$ be expressible by the partial fractions expansion:
:$\map R x = \ds \sum_{k \mathop = 0}^n \dfrac {\map {p_k} x} {\map {q_k} x}$
where:
:each of the $\map {p_k} x$ are polynomial functions
:the degree of $\map {p_k} x$ is strictly less than the degree of $\map {q_k} x$ for all $k$.
Then:
:$\ds \int \map R x \rd x = \sum_{k \mathop = 0}^n \int \dfrac {\map {p_k} x} {\map {q_k} x} \rd x$
This technique is known as Integration by Partial Fractions.
== Proof ==
== Examples ==
== Also see ==
* Integration by Substitution
* Integration by Parts
== Sources ==
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* 
Category:Integration by Partial Fractions
Category:Integral Calculus
Category:Partial Fractions Expansions
Category:Named Theorems
Category:Proof Techniques
Integration by Partial Fractions/Examples/Arbitrary Example 1

== Example of Use of Integration by Partial Fractions ==
Let $\map R x = \dfrac {\map P x} {\map Q x}$ be a rational function over $\R$ such that the degree of the polynomial $P$ is strictly smaller than the degree of the polynomial $Q$.
Let $\map Q x$ be expressible as:
:$\map Q x = \paren {x - a} \paren {x - b}^2 \paren {x^2 + c x + d}$
Then:
where $A$, $B_1$, $B_2$, $C$ and $D$ are constants dependent upon $a$, $b$, $c$ and $d$.
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 8$. Change of Variable
* : $6$. Integral Calculus: Appendix: Rules and Techniques of Integration: $2.3$ Integration by partial fractions
Category:Examples of Use of Integration by Partial Fractions