nyuuzyou/wikis huggingface dataset:
Integral Ideal is Ideal of Ring

== Theorem ==
Let $J$ be a non-empty subset of the set of integers $\Z$.
Then:
:$J$ is an integral ideal
:
:$J$ is an ideal of the ring of integers $\struct {\Z, +, \times}$.
== Proof ==
Let $J \subseteq \Z$ fulfil the conditions of an integral ideal:
:$(1): \quad n, m \in J \implies m + n \in J, m - n \in J$
:$(2): \quad n \in J, r \in \Z \implies r n \in J$
First note that $J$ is non-empty by definition.
Then from $(1)$ we have in particular:
:$n, m \in J \implies m - n \in J$
Thus by the One-Step Subgroup Test:
:$\struct {J, +}$ is a subgroup of the additive group of integers $\struct {\Z, +}$.
Then from $(2)$ we have:
:$n \in J, r \in \Z \implies r n \in J$
and by Integer Multiplication is Commutative it follows that:
:$n \in J, r \in \Z \implies n r \in J$
Thus it is seen that the conditions are fulfilled for $J$ to be an ideal of the ring of integers $\struct {\Z, +, \times}$.
Suppose $J \subseteq \Z$ such that:
:$(3): \quad \struct {J, +}$ is a subgroup of $\struct {\Z, +}$
:$(4): \quad \forall n \in J: \forall r \in R: n r \in J \land r n \in J$
From $(4)$, $(2)$ immediately follows.
From $(3)$ it follows from the One-Step Subgroup Test that:
:$\forall n, m \in J: m - n \in J$
which is a restatement of $(1)$.
Thus the conditions are fulfilled for $J$ to be an integral ideal.
== Sources ==
* : $\S 1.8$ Algebraic properties of $p$-adic integers
Category:Integral Ideals
Category:Ideal Theory
Integral Ideal is Set of Integer Multiples

== Theorem ==
Let $J$ be an integral ideal.
Then $J$ is in the form of a set of integer multiples $m \Z$ for some $m \in \Z$.
== Proof ==
By definition, $J$ satisfies the following conditions:
:$(1): \quad n, m \in J \implies m + n \in J, m - n \in J$
:$(2): \quad n \in J, r \in \Z \implies r n \in J$
First note that the null ideal $\set 0$ is an integral ideal.
This is of the form $0 \Z$.
Let $J \ne \set 0$.
Then $\exists a \in J: a \ne 0$.
As $0 \in \Z$ it follows from $(2)$ that:
:$0 \times a = 0 \in J$
Thus from $(1)$ we have that:
:$0 - a \in J$
so both $a \in J$ and $-a \in J$
Thus there exists $a \in J: a > 0$.
Consider the set $S = \set {x \in J: x > 0}$.
$S$ is bounded below by $0$, for example.
$S$ is not empty, because it has been established that $a \in S$.
By Set of Integers Bounded Below has Smallest Element, $S$ has a smallest element, $m$, say.
Consider $y \in m \Z$.
By definition of $m \Z$, $y = r m$ for some $r \in \Z$.
Thus, by $(2)$, $y \in J$.
So by definition of subset, $m \Z \subseteq J$.
It remains to be demonstrated that $J \subseteq m \Z$.
 $k \in J: k \notin m \Z$.
As $k \in J$, it follows from $(1)$ that $0 - k = -k \in J$ also.
Either $k > 0$ or $-k > 0$.
, let $k > 0$.
If $k < m$, then that contradicts the statement that $m \in J$ is the smallest element of $J$ that is (strictly) positive.
So $k > m$.
Consider the set $T = \set {x \in m \Z: x > k}$.
$T$ is not empty, as $m \paren {k + 1}$, for example, is bound to be in $T$ (even if $m = 1$, the smallest it can be).
By Set of Integers Bounded Below has Smallest Element, $T$ has a smallest element, $m s$, say.
Then:
:$m \paren {s - 1} < k < m s$
As $k > m$ it follows also that $m \paren {s - 1} > 0$.
Equality cannot happen because $k \notin m \Z$.
But we have that $k \in J$ and so:
:$m s - k \in J$
and:
:$k - m \paren {s - 1} \in J$
But $m s - k < m$, otherwise $m \paren {s - 1} > k$ which contradicts the statement that $m s \in J$ is the smallest element of $T$
Thus $m s - k \in J$ and $0 < m s - k < m$.
This contradicts the statement that $m$ is the smallest element of $J$ that is (strictly) positive.
Hence by Proof by Contradiction there is no such $k$.
Hence $J \subseteq m \Z$ and the result follows.
== Sources ==
* : $\text {2-1}$ Euclid's Division Lemma: Exercise $5$
Category:Integral Ideals
Category:Sets of Integer Multiples
Integral Multiple Distributes over Ring Addition

== Theorem ==
Let $\struct {R, +, \times}$ be a ring, or a field.
Let $a, b \in R$ and $m, n \in \Z$.
Then:
:$(1): \quad \paren {m + n} \cdot a = \paren {m \cdot a} + \paren {n \cdot a}$
:$(2): \quad m \cdot \paren {a + b} = \paren {m \cdot a} + \paren {m \cdot b}$
where $m \cdot a$ is as defined in integral multiple.
== Proof ==
We have that the additive group $\struct {R, +}$ is an abelian group.
$(1): \quad \paren {m + n} \cdot a = \paren {m \cdot a} + \paren {n \cdot a}$:
This is an instance of Powers of Group Elements: Sum of Indices when expressed in additive notation:
:$\forall n, m \in \Z: \forall a \in R: m a + n a = \paren {m + n} a$
$(2): \quad m \cdot \paren {a + b} = \paren {m \cdot a} + \paren {m \cdot b}$:
This is an instance of Power of Product in Abelian Group when expressed in additive notation:
:$\forall n \in \Z: \forall a, b \in R: n \paren {a + b} = n a + n b$
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
Category:Ring Theory
Category:Field Theory
Category:Examples of Distributive Operations
Integral Multiple of Integral Multiple

== Theorem ==
Let $\struct {F, +, \times}$ be a field.
Let $a \in F$ and $m, n \in \Z$.
Then:
:$\paren {m n} \cdot a = m \cdot \paren {n \cdot a}$
where $n \cdot a$ is as defined in integral multiple.
== Proof ==
We have that $\struct {F^*, \times}$ is the multiplicative group of $\struct {F, +, \times}$.
Let $a \in F^*$, that is, $a \in F: a \ne 0_F$, where $0_F$ is the zero of $F$.
This is an instance of Powers of Group Elements when expressed in additive notation:
:$\forall m, n \in \Z: \paren {m n} a = m \paren {n a}$
Now suppose $a = 0_F$.
Then by definition of the zero element of $F$, we have that:
:$\paren {m n} \cdot a = 0_F = m \cdot \paren {n \cdot a}$
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
Category:Field Theory
Integral Multiple of Ring Element

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $n \cdot x$ be an integral multiple of $x$:
:$n \cdot x = \begin {cases}
0_R & : n = 0 \\
x & : n = 1 \\
\paren {n - 1} \cdot x + x & : n > 1
\end {cases}$
that is:
:$n \cdot x = \underbrace {x + x + \cdots + x}_{\text {$n$ times} }$
For $n < 0$ we use:
:$n \cdot x = \paren{-n} \cdot \paren {-x}$
Then:
:$\forall n \in \Z: \forall x \in R: \paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$
=== General Result ===
== Proof ==
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$
First we verify $\map P 0$.
When $n = 0$, we have:
So $\map P 0$ holds.
=== Basis for the Induction ===
Now we verify $\map P 1$:
So $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\paren {k \cdot x} \circ x = k \cdot \paren {x \circ x} = x \circ \paren {k \cdot x}$
Then we need to show:
:$\paren {\paren {k + 1} \cdot x} \circ x = \paren {k + 1} \cdot \paren {x \circ x} = x \circ \paren {\paren {k + 1} \cdot x}$
=== Induction Step ===
This is our induction step:
A similar construction shows that $\paren {k + 1} \cdot \paren {x \circ x} = x \circ \paren {\paren {k + 1} \cdot x}$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: \paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$
The result for $n < 0$ follows directly from Powers of Group Elements.
Category:Ring Theory
Category:Proofs by Induction
Integral Multiple of Ring Element/General Result

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $n \cdot x$ be an integral multiple of $x$:
:$n \cdot x = \begin {cases}
0_R & : n = 0 \\
x & : n = 1 \\
\paren {n - 1} \cdot x + x & : n > 1
\end {cases}$
that is:
:$n \cdot x = x + x + \cdots \paren n \cdots x$
For $n < 0$ we use $-n \cdot x = n \cdot \paren {-x}$.
Then:
:$\forall m, n \in \Z: \forall x \in R: \paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$.
== Proof ==
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$
In what follows, we make extensive use of Integral Multiple of Ring Element:
:$\forall n \in \Z: \forall x \in R: \paren {m \cdot x} \circ x = m \cdot \paren {x \circ x} = x \circ \paren {m \cdot x}$
First we verify $\map P 0$.
When $n = 0$, we have:
So $\map P 0$ holds.
=== Basis for the Induction ===
Next we verify $\map P 1$.
When $n = 1$, we have:
So $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\paren {m \cdot x} \circ \paren {k \cdot x} = \paren {m k} \cdot \paren {x \circ x}$
Then we need to show:
:$\paren {m \cdot x} \circ \paren {\paren {k + 1} \cdot x} = \paren {m \paren {k + 1} } \cdot \paren {x \circ x}$
=== Induction Step ===
This is our induction step:
So $\map P K \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall m \in \Z: \forall n \in \N: \paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$
The result for $n < 0$ follows directly from Powers of Group Elements.
Category:Ring Theory
Category:Proofs by Induction
Integral Multiple of an Algebraic Number

== Theorem ==
Let $K$ be a number field and $\alpha \in K$.
Then there exists a positive $n \in \Z$ such that $n \alpha \in \OO_K$.
In this context, $ \OO_K$ denotes the algebraic integers in $K$.
== Proof ==
If $\alpha = 0$ then any integer works and the proof is finished.
Let $\alpha \ne 0$.
Let $\map f x = x^d + a_{d - 1} x^{d - 1} + \dotsb + a_0$ be the minimal polynomial of $\alpha$ over $\Q$. 
Suppose that $a_i = \dfrac {b_i} {c_i}$ is a reduced fraction for each $i$ such that $a_i \ne 0$.
Let $n$ be the least common multiple of the $c_i$, of which there must be at least one by our assumptions. 
Consider the polynomial:
Note that $g$ is a monic polynomial with coefficients in $\Z$ by our choice of $n$.
Furthermore, by construction, we see that $n \alpha$ is a root of $g$ and is therefore an algebraic integer.
== Sources ==
* : Chapter $2$ Ring of Integers: Integral elements: Proposition $2.6$
Category:Number Theory
Category:Algebraic Number Theory
Integral Operator is Linear

== Theorem ==
Let $T$ be an integral operator.
Let $f$ and $g$ be integrable real functions on a domain appropriate to $T$.
Then $T$ is a linear operator:
:$\forall \alpha, \beta \in \R: \map T {\alpha f + \beta g} = \alpha \map T f + \beta \map T g$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Let $T$ be expressed in its full form as an integral fransform:
:$\map T f := \ds \int_a^b \map f x \map K {p, x} \rd x$
for some integrable function $\map K {p, x}$.
Then:
Category:Integral Transforms
Category:Integral Operator is Linear
Integral Operator is Linear/Corollary 1

== Theorem ==
Let $T$ be an integral operator.
Let $f$ and $g$ be integrable real functions on a domain appropriate to $T$.
Then:
:$\forall \alpha, \beta \in \R: \map T {f + g} = \map T f + \map T g$
== Proof ==
From Integral Operator is Linear:
:$\forall \alpha, \beta \in \R: \map T {\alpha f + \beta g} = \alpha \map T f + \beta \map T g$
The result follows by setting $\alpha = \beta = 1$.
== Sources ==
* : $\S 1.1$. The Idea of an Integral Transform
Category:Integral Operator is Linear
Integral Operator is Linear/Corollary 2

== Theorem ==
Let $T$ be an integral operator.
Let $f$ be an integrable real function on a domain appropriate to $T$.
Then:
:$\forall \alpha \in \R: \map T {\alpha f} = \alpha \map T f$
== Proof ==
From Integral Operator is Linear:
:$\forall \alpha, \beta \in \R: \map T {\alpha f + \beta g} = \alpha \map T f + \beta \map T g$
The result follows by setting $\beta = 0$.
== Sources ==
* : $\S 1.1$. The Idea of an Integral Transform
Category:Integral Operator is Linear
Integral Points of Elliptic Curve y^2 = x^3+3x

== Theorem ==
:
The elliptic curve:
:$y^2 = x^3 + 3 x$
has exactly $7$ lattice points:
:$\tuple {0, 0}, \tuple {1, \pm 2}, \tuple {3, \pm 6}, \tuple {12, \pm 42}$
== Proof ==
== Sources ==
* 
Category:Examples of Elliptic Curves
Integral Representation of Bernoulli Number

== Theorem ==
Bernoulli numbers can be expressed in integral form as follows:
:$\ds \size {B_{2 n} } = 4 n \int_0^\infty \frac {t^{2 n - 1} } {e^{2 \pi t} - 1} \rd t$
where:
:$B_n$ are the Bernoulli numbers
:$n$ is a positive integer.
== Proof ==
== Sources ==
* : $7.2$: The Bernoullian numbers and the Bernoullian polynomials
Category:Bernoulli Numbers
Category:Definite Integrals involving Exponential Function
Category:Gamma Function
Category:Riemann Zeta Function
Category:Analytic Number Theory
Integral Representation of Bessel Function of the First Kind/Integer Order

== Theorem ==
Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.
Let $n \in \Z$ be an integer.
Then:
:$\ds \map {J_n} x = \dfrac 1 \pi \int_0^\pi \map \cos {n \theta - x \sin \theta} \rd \theta$
== Proof ==
Take $C$ to be the unit circle.
Write $t = e^{-i \theta}$ and let $\theta$ be decreasing from $\pi$ to $-\pi$, so that we integrate along $C$ anticlockwise.
Then:
== Sources ==
* : $\S 24$: Bessel Functions: Integral Representations for Bessel Functions: $24.99$
* 
* 
Category:Bessel Functions
Integral Representation of Bessel Function of the First Kind/Non-Integer Order

== Theorem ==
Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.
Let $n \in \Z$ be an integer.
Then:
:$\ds \map {J_n} x = \dfrac {x^n} {2^n \sqrt \pi \map \Gamma {n + \frac 1 2} } \int_0^\pi \map \cos {x \sin \theta} \cos^{2 n} \theta \rd \theta$
== Proof ==
== Sources ==
* : $\S 24$: Bessel Functions: Integral Representations for Bessel Functions: $24.100$
Category:Bessel Functions
Integral Representation of Dirichlet Beta Function in terms of Gamma Function

== Theorem ==
:$\ds \map \beta s = \frac 1 {\map \Gamma s} \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 + e^{-2 x} } \rd x$
where: 
:$\beta$ denotes the Dirichlet beta function
:$\Gamma$ denotes the gamma function
:$s$ is a complex number with $\map \Re s > 0$.
== Proof ==
We have, by Laplace Transform of Power: 
:$\ds \frac {\paren {-1}^n \map \Gamma s} {\paren {2 n + 1}^s} = \paren {-1}^n \int_0^\infty x^{s - 1} e^{-\paren {2 n + 1} x} \rd x$
for $\map \Re s > 0$.
Summing, we have: 
We have: 
Therefore: 
giving: 
:$\ds \map \beta s = \frac 1 {\map \Gamma s} \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 + e^{-2 x} } \rd x$
Category:Dirichlet Beta Function
Category:Gamma Function
Integral Representation of Dirichlet Eta Function in terms of Gamma Function

== Theorem ==
:$\ds \map \eta s = \frac 1 {\map \Gamma s} \int_0^\infty \frac {x^{s - 1} } {e^x + 1} \rd x$
where: 
:$s$ is a complex number with $\map \Re s > 0$
:$\eta$ denotes the Dirichlet eta function
:$\Gamma$ denotes the Gamma function.
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.82$
Category:Definite Integrals involving Exponential Function
Category:Dirichlet Eta Function
Category:Gamma Function
Integral Representation of Riemann Zeta Function in terms of Fractional Part

== Theorem ==
Let $\zeta$ be the Riemann zeta function.
Let $s \in \C$ be a complex number with real part $\sigma > 1$.
Then 
:$\ds \map \zeta s = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$
where $\fractpart x$ denotes the fractional part of $x$.
== Proof ==
First, we observe that:
Rearranging terms, we get:
Therefore, we have:
Next we note:
Therefore, we have:
== Also see ==
* Analytic Continuation of Riemann Zeta Function using Mellin Transform of Fractional Part
Category:Riemann Zeta Function
Integral Representation of Riemann Zeta Function in terms of Gamma Function

== Theorem ==
For $\Re \paren s > 1$, the Riemann Zeta function is given by:
:$\ds \map \zeta s = \frac 1 {\map \Gamma s} \int_0^\infty \frac {t^{s - 1}} {e^t - 1} \rd t$
where $\Gamma$ is the Gamma function.
=== Corollary ===
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.80$
* : $\S 35$: Riemann Zeta Function $\map \zeta x = \dfrac 1 {1^x} + \dfrac 1 {2^x} + \dfrac 1 {3^x} + \cdots$: $35.24$
Category:Integral Representation of Riemann Zeta Function in terms of Gamma Function
Category:Definite Integrals involving Exponential Function
Category:Gamma Function
Category:Riemann Zeta Function
Category:Analytic Number Theory
Integral Representation of Riemann Zeta Function in terms of Gamma Function/Corollary

== Corollary to Integral Representation of Riemann Zeta Function in terms of Gamma Function ==
For $\Re \paren s > 1$, the Riemann Zeta function is given by:
:$\ds \map \zeta s = \frac 1 {\map \Gamma s} \int_0^1 \frac {\paren {\map \ln {\dfrac 1 u} }^{s - 1} } {1 - u} \rd u$
where $\Gamma$ is the Gamma function.
== Proof ==
Category:Integral Representation of Riemann Zeta Function in terms of Gamma Function
Category:Primitives involving Logarithm Function
Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function

== Theorem ==
Let $s \in \C: \map \Re s > 1$.
Let $x \in \R_{>0}$.
Then:
:$\ds \pi^{-s / 2} \map \Gamma {\frac s 2} \map \zeta s = -\frac 1 {s \paren {1 - s} } + \dfrac 1 2 \int_1^\infty \paren {x^{s / 2 - 1} + x^{-\paren {s + 1} / 2} } \paren {\map {\vartheta_3} {0, e^{-\pi x} } - 1} \rd x$
where:
:$\map \Gamma s$ is the gamma function
:$\map \zeta s$ is the Riemann zeta function
:$\ds \map {\vartheta_3} {0, e^{-\pi x} }$ is the Jacobi theta function of the third type.
== Proof ==
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
== Also see ==
* Analytic Continuation of Riemann Zeta Function using Jacobi Theta Function
* Functional Equation for Riemann Zeta Function
Category:Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function
Category:Jacobi Theta Functions
Category:Riemann Zeta Function
Category:Gamma Function
Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function/Lemma 1

== Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function: Lemma 1 ==
:$\ds \pi^{-s / 2} \map \Gamma {\frac s 2} \map \zeta s = \int_0^1 x^{s / 2 - 1} \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x } \rd x + \int_1^\infty x^{s / 2 - 1} \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x } \rd x$
where:
:$\map \Gamma s$ is the gamma function
:$\map \zeta s$ is the Riemann zeta function
:$s \in \C$ is a complex number with real part $s > 1$
:$x \in \R_{>0}$.
== Proof ==
The gamma function $\Gamma: \C \to \C$ is defined, for the open right half-plane, as:
:$\ds \map \Gamma z = \int_0^{\infty} t^{z - 1} e^{-t} \rd t$
Setting $z = \dfrac s 2$:
:$\ds \map \Gamma {\dfrac s 2} = \int_0^{\infty} t^{s/2 - 1} e^{-t} \rd t$
Substituting $t = \pi n^2 x$ and $\rd t = \pi n^2 \rd x$:
Category:Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function
Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function/Lemma 2

== Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function: Lemma 2 ==
:$\ds \int_0^1 x^{s / 2 - 1} \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x} \rd x = -\frac 1 {s \paren {1 - s} } + \int_1^\infty x^{-\paren {s + 1} / 2} \sum_{n \mathop = 1}^\infty e^{-\paren {\pi n^2 x} } \rd x$
where:
:$s \in \C: \map \Re s > 1$
:$x \in \R_{>0}$.
== Proof ==
Let $u = \dfrac 1 x$.
Then:
Hence:
Recall the Fourier Transform of $e^{-t^2}$:
Therefore, by the Poisson Summation Formula:
Category:Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function
Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function/Lemma 3

== Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function: Lemma 3 ==
:$\ds \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x} = \dfrac 1 2 \paren {\map {\vartheta_3} {0, e^{-\pi x} } - 1}$
where:
:$\map {\vartheta_3} {0, e^{-\pi x} }$ is the Jacobi theta function of the third type
:$x \in \R_{>0}$.
== Proof ==
Recall the definition of the Jacobi theta function of the third type:
:
Category:Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function
Integral Resulting in Arcsecant

== Theorem ==
:$\ds \int \frac 1 {x \sqrt {x^2 - a^2} } \rd x = \begin{cases}
\dfrac 1 {\size a} \arcsec \dfrac x {\size a} + C & : x > \size a \\
-\dfrac 1 {\size a} \arcsec \dfrac x {\size a} + C & : x < -\size a
\end{cases}$
where $a$ is a constant.
== Proof ==
Substitute:
:$\sec \theta = \dfrac x {\size a} \iff \size a \sec \theta = x$
for $\theta \in \openint 0 {\dfrac \pi 2} \cup \openint {\dfrac \pi 2} \pi$.
This substitution is valid for all $\dfrac x {\size a} \in \R \setminus \openint {-1} 1$.
By hypothesis:
so this substitution will not change the domain of the integrand.
Thus:
and so:
By Shape of Tangent Function and the stipulated definition of $\theta$:
:$(A): \quad \dfrac x {\size a} > 1 \iff \theta \in \openint 0 {\dfrac \pi 2}$
and
:$(B): \quad \dfrac x {\size a} < -1 \iff \theta \in \openint {\dfrac \pi 2} \pi$
If $(A)$:
If $(B)$:
== Also see ==
* Derivative of Arcsecant Function
Category:Integral Calculus
Integral Transform is Mapping

== Theorem ==
Let $\map F p$ be an integral transform:
:$\map F p = \ds \int_a^b \map f x \map K {p, x} \rd x$
Let $T$ be the integral operator associated with $\map F p$.
Then $T$ is a mapping from the domain of $T$ to its image.
That is, for every $\map f x$ there exists a unique $\map F p$.
== Proof ==
Let $p$ be fixed.
In this context, $\map f x \map K {p, x}$ is the pointwise product of the functions $\map f x$ and $\map K {p, x}$.
From Pointwise Operation is Well-Defined, it follows that $\map f x \map K {p, x}$ is a real function on $x$.
We have that both $\map f x$ and $\map K {p, x}$ are integrable.
It follows from Pointwise Product of Integrable Real Functions is Integrable that $\map f x \map K {p, x}$ is an integrable function.
From Definite Integral is Unique, for a given $p$ there is exactly one real number $\map F p$ for which:
:$\map F p = \ds \int_a^b \map f x \map K {p, x} \rd x$
== Sources ==
* : $\S 1.1$. The Idea of an Integral Transform
Category:Integral Transforms
Integral between Limits is Independent of Direction

== Theorem ==
Let $f$ be a real function which is integrable on the interval $\openint a b$.
Then:
:$\ds \int_a^b \map f x \rd x = \int_a^b \map f {a + b - x} \rd x$
=== Corollary ===
== Proof ==
Let $z = a + b - x$.
Then:
:$\dfrac {\d z} {\d x} = -1$
and:
:$x = a \implies z = a + b - a = b$
:$x = b \implies z = a + b - b = a$
So:
Category:Definite Integrals
Integral from 0 to 1 of Complete Elliptic Integral of First Kind

== Theorem ==
Let $G$ denote Catalan's constant.
Then:
:$2 G = \ds \int_0^1 \map K k \rd k$
where $\map K k$ denotes the complete elliptic integral of the first kind:
:$\map K k = \ds \int \limits_0^{\pi / 2} \dfrac {\d \phi} {\sqrt {1 - k^2 \sin^2 \phi} }$
== Proof ==
== Sources ==
* : $0,91596 55941 77219 01505 \ldots$
Category:Catalan's Constant
Category:Complete Elliptic Integral of the First Kind