nyuuzyou/wikis huggingface dataset:
Integral of Positive Function with respect to Counting Measure on Natural Numbers

== Theorem ==
Consider the measure space $\struct {\N, \map \PP \N, \mu}$ where $\mu$ is the counting measure on $\struct {\N, \map \PP \N}$.
Let $f : \N \to \R$ be a function. 
Then: 
:$\ds \int f \rd \mu = \sum_{n \mathop = 1}^\infty \map f n$
== Proof ==
Clearly we have: 
:$\set {x \in \N : \map f x \le \alpha} \in \map \PP \N$
for each $\alpha \in \R$, so any function $f : \N \to \R$ is $\map \PP \N$-measurable.
Similarly, an arbitrary subset of $\N$ is clearly $\map \PP \N$-measurable.
For each $n \in \N$, define $f_n : \N \to \R$ by: 
:$\ds \map {f_n} k = \begin{cases}\map f k & k \le n \\ 0 & k > n\end{cases}$
for $k \in \N$.
We show that $\sequence {f_n}_{n \mathop \in \N}$ is increasing.
Let $k \in \N$. 
We want to show that: 
:$\map {f_n} k \le \map {f_{n + 1} } k$ 
for each $n \in \N$. 
If $k \le n$, then we have: 
:$\map {f_n} k = \map f k$
and:
:$\map {f_{n + 1} } k = \map f k$
so we have: 
:$\map {f_n} k = \map {f_{n + 1} } k$
in this case.
If $k = n + 1$, then: 
:$\map {f_n} k = 0$
and:
:$\map {f_{n + 1} } k = \map f k$
so we have: 
:$\map {f_n} k \le \map {f_{n + 1} } k$
If $k > n + 1$, then: 
:$\map {f_n} k = \map f k$
and:
:$\map {f_{n + 1} } k = \map f k$
so that: 
:$\map {f_n} k = \map {f_{n + 1} } k$
So $\sequence {f_n}_{n \mathop \in \N}$ is increasing
We show that: 
:$\ds \map f k = \lim_{n \mathop \to \infty} \map {f_n} k$
Note that for all $n \ge k$, we have: 
:$\map {f_n} k = \map f k$
so we obtain the above from Tail of Convergent Sequence.
So $\sequence {f_n}_{n \mathop \in \N}$ is a increasing sequence of $\map \PP \N$-measurable functions converging to $f$. 
So, from the monotone convergence theorem, we obtain: 
:$\ds \int f \rd \mu = \lim_{n \mathop \to \infty} \int f_n \rd \mu$ 
It remains to compute: 
:$\ds \int f_n \rd \mu$
for each $n \in \N$.
We can rewrite: 
:$\ds f_n = \sum_{i \mathop = 1}^n \map f i \chi_{\set i}$
Clearly:
:$\set {\set 1, \set 2, \ldots, \set n}$ is pairwise disjoint
so, from the definition of the $\mu$-integral of a positive simple function, we have: 
:$\ds \map {I_\mu} {f_n} = \sum_{i \mathop = 1}^n \map f i \map \mu {\set i}$
From the definition of the counting measure, we have: 
:$\map \mu {\set i} = 1$ 
for each $i$, so:
:$\ds \map {I_\mu} {f_n} = \sum_{i \mathop = 1}^n \map f i$
From Integral of Positive Measurable Function Extends Integral of Positive Simple Function, we therefore have: 
:$\ds \int f_n \rd \mu = \sum_{i \mathop = 1}^n \map f i$
From the definition of infinite series, we then have: 
:$\ds \int f \rd \mu = \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \map f k = \sum_{k \mathop = 1}^\infty \map f k$
as required.
Category:Counting Measure
Integral of Positive Measurable Function Extends Integral of Positive Simple Function

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \R, f \in \EE^+$ be a positive simple function.
Then:
:$\ds \int f \rd \mu = \map {I_\mu} f$
where:
:$\ds \int \cdot \rd \mu$ denotes the $\mu$-integral of positive measurable functions
:$I_\mu$ denotes the $\mu$-integral of positive simple functions.
That is:
:$\ds \int \cdot \rd \mu \restriction_{\EE^+} = I_\mu$
using the notion of restriction, $\restriction$.
== Proof ==
From the definition of the integral of a positive measure function, we have: 
:$\ds \int f \rd \mu = \sup \set {\map {I_\mu} g: g \le f, g \in \EE^+}$
Let $g \in \EE^+$ be such that $g \le f$. 
Then, from Integral of Positive Simple Function is Increasing, we have: 
:$\map {I_\mu} g \le \map {I_\mu} f$ 
So $\map {I_\mu} f$ is an upper bound of $\set {\map {I_\mu} g: g \le f, g \in \EE^+}$. 
Since $f \in \EE^+$, and $f \le f$, we have: 
:$\map {I_\mu} f \in \set {\map {I_\mu} g: g \le f, g \in \EE^+}$
So:
:$\map {I_\mu} f$ is the greatest element of $\set {\map {I_\mu} g: g \le f, g \in \EE^+}$.
so, from Greatest Element is Supremum:
:$\map {I_\mu} f = \sup \set {\map {I_\mu} g: g \le f, g \in \EE^+}$
So:
:$\ds \int f \rd \mu = \map {I_\mu} f$
== Sources ==
* : $9.5$
Category:Integral of Positive Measurable Function
Integral of Positive Measurable Function is Additive

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ and $g : X \to \overline \R$ be positive $\Sigma$-measurable functions.
Then:
:$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$
where:
:$f + g$ is the pointwise sum of $f$ and $g$
:the integral sign denotes $\mu$-integration
This can be summarized by saying that $\ds \int \cdot \rd \mu$ is additive.
=== Corollary ===
== Proof ==
We are given that $f : X \to \overline \R$ and $g : X \to \overline \R$ is a positive $\Sigma$-measurable functions, which is a measurable function, so we can apply Measurable Function is Pointwise Limit of Simple Functions.
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 $x \in X$. 
Similarly, there exists an increasing sequence $\sequence {g_n}_{n \mathop \in \N}$ of positive simple functions such that: 
:$\ds \map g x = \lim_{n \mathop \to \infty} \map {g_n} x$
for each $x \in X$. 
From the Sum Rule for Real Sequences, we have: 
:$\ds \map f x + \map g x = \lim_{n \mathop \to \infty} \paren {\map {f_n} x + \map {g_n} x}$
for each $x \in X$.
We then have: 
== Sources ==
* : $9.8 \ \text{(iii)}$, $\S 9$: Problem $2$
Category:Integral of Positive Measurable Function
Category:Integral of Positive Measurable Function is Additive
Integral of Positive Measurable Function is Additive/Corollary

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ and $g : X \to \overline \R$ be positive $\Sigma$-measurable functions.
Let $A \in \Sigma$. 
Then: 
:$\ds \int_A \paren {f + g} \rd \mu = \int_A f \rd \mu + \int_A g \rd \mu$
where:
:$f + g$ is the pointwise sum of $f$ and $g$
:the integral sign denotes $\mu$-integration over $A$.
This can be summarized by saying that $\ds \int_A \cdot \rd \mu$ is additive.
== Proof ==
We have: 
Category:Integral of Positive Measurable Function is Additive
Category:Integral of Positive Measurable Function over Measurable Set‎
Integral of Positive Measurable Function is Monotone

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \overline \R$ be positive $\Sigma$-measurable functions.
Suppose that $f \le g$, where $\le$ denotes pointwise inequality.
Then:
:$\ds \int f \rd \mu \le \int g \rd \mu$
where the integral sign denotes $\mu$-integration.
This can be summarized by saying that $\ds \int \cdot \rd \mu$ is monotone.
=== Corollary ===
== Proof ==
By the definition of $\mu$-integration, we have: 
:$\ds \int f \rd \mu = \sup \set {\map {I_\mu} h: h \le f, h \in \EE^+}$
and:
:$\ds \int g \rd \mu = \sup \set {\map {I_\mu} h : h \le g, h \in \EE^+}$
where:
:$\EE^+$ denotes the space of positive simple functions
:$\map {I_\mu} g$ denotes the $\mu$-integral of the positive simple function $g$.
Since $f \le g$:
:if $h \in \EE^+$ satisfies $h \le f$ then it satisfies $h \le g$.
So:
:$\set {\map {I_\mu} h: h \le f, h \in \EE^+} \subseteq \set {\map {I_\mu} h : h \le g, h \in \EE^+}$
From Supremum of Subset, we therefore have: 
:$\sup \set {\map {I_\mu} h: h \le f, h \in \EE^+} \le \sup \set {\map {I_\mu} h : h \le g, h \in \EE^+}$
That is: 
:$\ds \int f \rd \mu \le \int g \rd \mu$
== Sources ==
* : $9.8 \ \text{(iv)}$, $\S 9$: Problem $2$
Category:Integral of Positive Measurable Function
Category:Integral of Positive Measurable Function is Monotone
Integral of Positive Measurable Function is Monotone/Corollary

== Corollary to Integral of Positive Measurable Function is Monotone ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \overline \R$ be positive $\Sigma$-measurable functions.
Let $A \in \Sigma$.
Suppose that $f \le g$, where $\le$ denotes pointwise inequality.
Then:
:$\ds \int_A f \rd \mu \le \int_A g \rd \mu$
where the integral sign denotes $\mu$-integration over $A$.
This can be summarized by saying that $\ds \int_A \cdot \rd \mu$ is monotone.
== Proof ==
From the definition of $\mu$-integration over $A$, we have: 
:$\ds \int_A f \rd \mu = \int \paren {\chi_A \times f} \rd \mu$
and:
:$\ds \int_A g \rd \mu = \int \paren {\chi_A \times g} \rd \mu$ 
We show that: 
:$f \times \chi_A \le g \times \chi_A$
If $x \in A$, we have: 
and:
So:
:$\map {\paren {f \times \chi_A} } x \le \map {\paren {g \times \chi_A} } x$ for all $x \in A$. 
Now take $x \in X \setminus A$. 
We have: 
and:
So:
:$\map {\paren {f \times \chi_A} } x \le \map {\paren {g \times \chi_A} } x$ for all $x \in X \setminus A$
giving:
:$f \times \chi_A \le g \times \chi_A$
From Integral of Positive Measurable Function is Monotone, we therefore have: 
:$\ds \int \paren {f \times \chi_A} \rd \mu \le \int \paren {g \times \chi_A} \rd \mu$
From the definition of $\mu$-integration over $A$, we have: 
:$\ds \int_A f \rd \mu \le \int_A g \rd \mu$
Category:Integral of Positive Measurable Function is Monotone
Category:Integral of Positive Measurable Function over Measurable Set
Integral of Positive Measurable Function is Positive Homogeneous

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ be a positive $\Sigma$-measurable function.
Let $\lambda \in \overline \R$ be an extended real number with $\lambda \ge 0$. 
Then:
:$\ds \int \lambda f \rd \mu = \lambda \int f \rd \mu$
where:
:$\lambda f$ is the pointwise $\lambda$-multiple of $f$
:The integral sign denotes $\mu$-integration
This can be summarized by saying that $\ds \int \cdot \rd \mu$ is positive homogeneous.
=== Corollary ===
== Proof ==
Suppose that $\lambda < \infty$. 
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: 
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
From the Multiple Rule for Real Sequences, we have: 
:$\ds \lambda \map f x = \lim_{n \mathop \to \infty} \paren {\lambda \map {f_n} x}$
for each $x \in X$. 
We then have: 
so we get the demand in the case $\lambda < \infty$.
Now suppose that $\lambda = \infty$. 
We can write: 
:$\ds \lambda f = \lim_{k \mathop \to \infty} k f$
Since $f \ge 0$, the sequence $\sequence {k f}_{k \mathop \in \N}$ is increasing, we have: 
:$\ds \int \lambda f \rd \mu = \lim_{k \mathop \to \infty} \int k f \rd \mu$ 
from the monotone convergence theorem.
From our earlier work, we have: 
:$\ds \int k f \rd \mu = k \int f \rd \mu$
so that: 
giving the demand in the case $\lambda = \infty$.
== Also see ==
* Integral of Positive Simple Function is Positive Homogeneous, a similar result for positive simple functions.
== Sources ==
* : $9.8 \ \text{(ii)}$, $\S 9$: Problem $2$
Category:Integral of Positive Measurable Function
Category:Integral of Positive Measurable Function is Positive Homogeneous
Integral of Positive Measurable Function is Positive Homogeneous/Corollary

== Corollary to Integral of Positive Measurable Function is Positive Homogeneous ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ be a positive $\Sigma$-measurable function.
Let $\lambda \in \overline \R$ be an extended real number with $\lambda \ge 0$. 
Let $A \in \Sigma$. 
Then: 
:$\ds \int_A \lambda f \rd \mu = \lambda \int_A f \rd \mu$
where:
:$\lambda f$ is the pointwise $\lambda$-multiple of $f$
:the integral sign denotes $\mu$-integration over $A$.
This can be summarized by saying that $\ds \int_A \cdot \rd \mu$ is positive homogeneous.
== Proof ==
We have: 
Category:Integral of Positive Measurable Function is Positive Homogeneous
Category:Integral of Positive Measurable Function over Measurable Set‎
Integral of Positive Measurable Function over Disjoint Union

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ be a positive $\Sigma$-measurable function.
Let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint $\Sigma$-measurable sets.
Let: 
:$\ds D = \bigcup_{n \mathop = 1}^\infty D_n$
Then: 
:$\ds \int_D f \rd \mu = \sum_{n \mathop = 1}^\infty \int_{D_n} f \rd \mu$
== Proof ==
We have: 
Category:Integral of Positive Measurable Function over Measurable Set‎
Integral of Positive Measurable Function over Measurable Set is Well-Defined

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $A \in \Sigma$. 
Let $f : X \to \overline \R$ be a positive $\Sigma$-measurable function.
Then the $\mu$-integral of $f$ over $A$ defined by: 
:$\ds \int_A f \rd \mu = \int \paren {\chi_A \cdot f} \rd \mu$
is well-defined. 
== Proof ==
We simply need to show that: 
:$\chi_A \cdot f$ is a positive $\Sigma$-measurable function.
For $x \in A$, we have: 
For $x \in X \setminus A$, we have:
so:
:$\chi_A \cdot f$ is non-negative. 
We now show that $\chi_A \cdot f$ is $\Sigma$-measurable.
From Characteristic Function Measurable iff Set Measurable, we have: 
:$\chi_A$ is $\Sigma$-measurable.
From Pointwise Product of Measurable Functions is Measurable, we then have: 
:$\chi_A \cdot f$ is $\Sigma$-measurable.
So:
:$\chi_A \cdot f$ is a positive $\Sigma$-measurable function.
So its $\mu$-integral is well-defined.
Category:Integral of Positive Measurable Function over Measurable Set‎
Integral of Positive Measurable Function with respect to Restricted Measure

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space. 
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra of $\Sigma$.
Let $\mu \restriction_\GG$ be the restriction of $\mu$ to $\GG$.
Let $f : X \to \overline \R$ be a positive $\GG$-measurable function.
Then: 
:$\ds \int f \rd \mu = \int f \rd \mu \restriction_\GG$
== Proof ==
We first prove the result for positive simple functions.
Let $f : X \to \R$ be a positive simple function that is $\GG$-measurable.
From Simple Function has Standard Representation there exists:
:a finite sequence $a_1, \ldots, a_n$ of real numbers
:a partition $E_1, \ldots, E_n$ of $\GG$-measurable sets
such that: 
:$f = \ds \sum_{j \mathop = 1}^n a_j \chi_{E_j}$
Then we have: 
So we get the theorem for positive simple functions.
Now let $f$ be a general positive $\GG$-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 $x \in X$.
Then for each $n \in \N$ we have: 
:$\ds \int f_n \rd \mu = \int f_n \rd \mu \restriction_\GG$
From Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions, taking $n \to \infty$ we obtain: 
:$\ds \int f \rd \mu = \int f \rd \mu \restriction_\GG$
Category:Restricted Measures
Category:Integral of Positive Measurable Function
Integral of Positive Simple Function is Additive

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \R$, $f, g \in \EE^+$ be positive simple functions.
Then $\map {I_\mu} {f + g} = \map {I_\mu} f + \map {I_\mu} g$, where:
:$f + g$ is the pointwise sum of $f$ and $g$
:$I_\mu$ denotes $\mu$-integration
This can be summarized by saying that $I_\mu$ is additive.
== Proof ==
== Sources ==
* : $9.3 \ \text{(iii)}$
Category:Integral of Positive Simple Function
Integral of Positive Simple Function is Increasing

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \R$, $f, g \in \EE^+$ be positive simple functions.
Suppose that:
: $f \le g$
where $\le$ denotes pointwise inequality.
Then:
:$\map {I_\mu} f \le \map {I_\mu} g$
where $I_\mu$ denotes $\mu$-integration
This can be summarized by saying that $I_\mu$ is an increasing mapping.
== Proof ==
Note that: 
:$g - f \ge 0$ 
From Scalar Multiple of Simple Function is Simple Function and Pointwise Sum of Simple Functions is Simple Function, we then have that: 
:$g - f \in \EE^+$ 
Write: 
:$g = f + \paren {g - f}$ 
From Integral of Positive Simple Function is Additive, we then have: 
:$\map {I_\mu} g = \map {I_\mu} f + \map {I_\mu} {g - f}$ 
Since: 
:$\map {I_\mu} {g - f} \ge 0$ 
we obtain: 
:$\map {I_\mu} f \le \map {I_\mu} g$
== Sources ==
* : $9.3 \ \text{(iv)}$
Category:Integral of Positive Simple Function
Integral of Positive Simple Function is Positive Homogeneous

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \R, f \in \EE^+$ be a positive simple function.
Let $\lambda \in \R_{\ge 0}$ be a positive real number.
Then:
:$\map {I_\mu} {\lambda \cdot f} = \map {\lambda \cdot I_\mu} f$
where:
:$\lambda \cdot f$ is the pointwise $\lambda$-multiple of $f$
:$I_\mu$ denotes $\mu$-integration.
This can be summarized by saying that $I_\mu$ is positive homogeneous.
== Proof ==
Remark that $\lambda \cdot f$ is a positive simple function by Scalar Multiple of Simple Function is Simple Function.
Let:
:$f = \ds \sum_{i \mathop = 0}^n a_i \chi_{E_i}$
be a standard representation for $f$.
Then we also have, for all $x \in X$:
and it is immediate from the definition that this yields a standard representation for $\lambda \cdot f$.
Therefore, we have:
Hence the result.
== Also see ==
* Integral of Positive Measurable Function is Positive Homogeneous, an extension of this result to positive measurable functions
== Sources ==
* : $9.3 \ \text{(ii)}$
Category:Integral of Positive Simple Function
Integral of Positive Simple Function is Well-Defined

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \R, f \in \EE^+$ be a positive simple function.
Then the $\mu$-integral of $f$, $\map {I_\mu} f$, is well-defined.
That is, for any two standard representations for $f$, say:
:$\ds f = \sum_{i \mathop = 0}^n a_i \chi_{E_i} = \sum_{j \mathop = 0}^m b_j \chi_{F_j}$
it holds that:
:$\ds \sum_{i \mathop = 0}^n a_i \map \mu {E_i} = \sum_{j \mathop = 0}^m b_j \map \mu {F_j}$
== Proof ==
The sets $F_0, \ldots, F_m$ are pairwise disjoint, and:
:$X = \ds \bigcup_{j \mathop = 0}^m F_j$
From Characteristic Function of Disjoint Union, we have:
:$\chi_X = \ds \sum_{j \mathop = 0}^m \chi_{F_j}$
Remark that $\map {\chi_X} x = 1$ for all $x \in X$, so that we have:
Repeating the argument with the $E_i$ and $F_j$ interchanged also yields:
:$f = \ds \sum_{j \mathop = 0}^m \sum_{i \mathop = 0}^n b_j \chi_{F_j \cap E_i}$
Now whenever $x \in E_i \cap F_j$, for some $i, j$, then since the $E_i$, $F_j$ are disjoint, we find:
:$x \in E_{i'} \cap F_{j'}$ implies $i = i'$ and $j = j'$
Thus, evaluating both expressions for $\map f x$ we find:
:$a_i = \map f x = b_j$
In conclusion, we have:
:$a_i = b_j$
if $E_i \cap F_j \ne \O$.
Furthermore, we have for all $i$ that:
:$\ds E_i = E_i \cap X = E_i \cap \paren {\bigcup_{j \mathop = 0}^m F_j} = \bigcup_{j \mathop = 0}^m \paren {E_i \cap F_j}$
by Intersection Distributes over Union: General Result.
Similarly, we obtain for all $j$:
:$\ds F_j = F_j \cap X = F_j \cap \paren {\bigcup_{i \mathop = 0}^n E_i} = \bigcup_{i \mathop = 0}^n \paren {F_j \cap E_i}$
With this knowledge, we compute:
Hence the result.
== Sources ==
* : $9.1$
Category:Integral of Positive Simple Function
Integral of Power

== Theorem ==
:$\ds \forall n \in \R_{\ne -1}: \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$
== Conventional Proof ==
== Fermat's Proof ==
This proof is is valid only for positive rational numbers, that is, it proves that:
:$\ds \forall n \in \Q_{>0}: \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.13$: Fermat ()
Category:Integral Calculus
Integral of Power/Conventional Proof

== Theorem ==
== Proof ==
From the Fundamental Theorem of Calculus:
:$(1): \quad \ds \int_0^b x^n \rd x = \bigintlimits {\map F x} 0 b = \map F b - \map F 0$
where $\map F x$ is a primitive of $x^n$.
By Primitive of Power, $\dfrac {x^{n + 1} } {n + 1}$ is a primitive of $x^n$.
Then:
Category:Integral Calculus
Integral of Power/Fermat's Proof

== Theorem ==
:$\ds \forall n \in \Q_{>0}: \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$
== Proof ==
First let $n$ be a positive integer.
Take a real number $r \in \R$ such that $0 < r < 1$ but reasonably close to $1$.
Consider a subdivision $S$ of the closed interval $\closedint 0 b$ defined as:
:$S = \set {0, \ldots, r^2 b, r b, b}$
that is, by taking as the points of subdivision successive powers of $r$.
Now we take the upper Darboux sum $\map U S$ over $S$ (starting from the right):
Now we let $r \to 1$ and see that each of the terms on the bottom also approach $1$.
Thus:
:$\ds \lim_{r \mathop \to 1} S = \frac {b^{n + 1} } {n + 1}$
That is:
:$\ds \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$
for every positive integer $n$.
Now assume $n = \dfrac p q$ be a strictly positive rational number.
We set $s = r^{1/q}$ and proceed:
As $r \to 1$ we have $s \to 1$ and so that last expression shows:
So the expression for the main result still holds for rational $n$.
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.5$: Fermat's Calculation of $\int_0^b x^n \rd x$ for Positive Rational $n$
Category:Integral Calculus
Integral of Power/Fermat's Proof/Historical Note

== Historical Note on proof of Integral of Power ==
The integral of a power for rational power was used by , and predated the work done by and by a considerable period.
== Sources ==
* : Chapter $\text {B}.5$: Fermat's Calculation of $\int_0^b x^n \rd x$ for Positive Rational $n$
Category:Historical Notes
Integral of Power/Historical Note

== Theorem ==
The conventional proof of Integral of Power of course holds for all real $n \ne -1$, not just where $n$ is a strictly positive rational.
However, the real point of this page is Fermat's proof, which demonstrates how integration was achieved before the full machinery of calculus had been thoroughly constructed.
 had previously made progress with this problem, proving it for integral $1 \le n \le 9$ but the algebra for the proof of each power was more difficult than the previous one, and he found $10$ too much hard work. The clear beauty of 's approach was that it works for all $n$, rational as well as integral.
== Sources ==
* : Chapter $\text {A}.13$: Fermat ()
* : Chapter $\text {A}.14$: Cavalieri ()
Category:Historical Notes
Integral of Probability Density Function over the Reals is Equal to One

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an absolutely continuous random variable. 
Let $f_X$ be a probability density function for $X$. 
Let $\map \BB \R$ be the Borel $\sigma$-algebra on $\R$. 
Let $\lambda$ be the Lebesgue measure on $\struct {\R, \map \BB \R}$. 
Then: 
:$\ds \int_\R f_X \rd \lambda = 1$
where $\ds \int_\R f_X \rd \lambda$ denotes the Lebesgue integral of $f_X$. 
== Proof ==
Let $P_X$ be the probability distribution of $X$. 
From the definition of a probability density function, $f_X$ is a Radon-Nikodym derivative of $P_X$ with respect to $\lambda$.
We then have:
Category:Probability Density Functions
Integral of Reciprocal is Divergent

== Theorem ==
=== Unbounded Above ===
=== To Zero ===
Thus the improper integrals $\ds \int_1^{\to +\infty} \frac {\d x} x$ and $\ds \int_{\to 0^+}^1 \frac {\d x} x$ do not exist.
In particular:
$\ds \int_{\to 0^+}^{\to +\infty} \frac {\d x} x$
certainly does not exist.
== Also see ==
* Logarithm Tends to Infinity
== Sources ==
* : $\S 13.33$
Category:Integral of Reciprocal is Divergent
Category:Divergent Improper Integrals
Category:Reciprocals
Integral of Reciprocal is Divergent/To Zero

== Theorem ==
:$\ds \int_\gamma^1 \frac {\d x} x \to -\infty$ as $\gamma \to 0^+$
Thus the improper integral $\ds \int_{\to 0^+}^1 \frac {\d x} x$ does not exist.
== Proof ==
Put $x = \dfrac 1 z$.
Then:
But as $\gamma \to 0+$, we have that $\dfrac 1 \gamma \to +\infty$.
Hence, from Integral of Reciprocal Unbounded Above is Divergent:
:$\ds \int_1^{1 / \gamma} \frac {\d z} z \to +\infty$
as $\dfrac 1 \gamma \to +\infty$.
The result follows.
== Sources ==
* : $\S 13.33$
Category:Integral of Reciprocal is Divergent
Integral of Reciprocal is Divergent/Unbounded Above

== Theorem ==
:$\ds \int_1^n \frac {\d x} x \to +\infty$ as $n \to + \infty$
Thus the improper integral $\ds \int_1^{\to +\infty} \frac {\d x} x$ does not exist.
== Proof 1 ==
== Proof 2 ==
Category:Integral of Reciprocal is Divergent