avewright/proofwiki-math huggingface dataset:
Theorem: Cauchy's Residue Theorem

Let $U$ be a simply connected open subset of the complex plane $\C$.
Let $a_1, a_2, \dots, a_n$ be finitely many points of $U$.
Let $f: U \to \C$ be analytic in $U \setminus \set {a_1, a_2, \dots, a_n}$. 
Let $L$ be a contour in $\C$ oriented anticlockwise.
Let $\partial U_k$ denote the closed contour bounding $U_k$.
Then:
:$\ds \oint_L \map f z \rd z = 2 \pi i \sum_{k \mathop = 1}^n \Res f {a_k}$
where $\Res f {a_k}$ denotes the residue at $a_k$ of $f$.

Proof:
Let $\set {U_1, \dotsc, U_n}$ be a set of open subsets of $U$ such that $a_i \in U_i$, and $a_i \notin U_j$ for $i \ne j$.
Let $U_i \cap U_j = \O$ for all $i \ne j$.
By Existence of Laurent Series, around each $a_k$ there is an expansion:
:$\ds \map f z = \sum_{j \mathop = -\infty}^\infty c_j \paren {z - a_k}^j$ 
convergent in $U_k$.
Write: 
:$\ds X = \bigcup_{i \mathop = 1}^n U_i$
Then, by Contour Integral of Concatenation of Contours: 
:$\ds \oint_L \map f z \rd z = \oint_{\partial \paren {U \setminus X} } \map f z \rd z + \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z$ 
As all poles of $f$ in $U$ are contained in $X$, $f$ is holomorphic on $U \setminus X$.
So by the Cauchy-Goursat Theorem: 
:$\ds \oint_{\partial \paren {U \setminus X} } \map f z \rd z = 0$
giving: 
:$\ds \oint_L \map f z \rd z = \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z$
we have: 
{{begin-eqn}}
{{eqn | l = \oint_{\partial U_k} \map f z \rd z
      | r = \oint_{\partial U_k} \sum_{j \mathop = -\infty}^\infty c_j \paren {z - a_k}^j \rd z
}}
{{eqn | r = \oint_{\partial U_k} \paren {\sum_{j \mathop = -\infty}^{-2} c_j \paren {z - a_k}^j + \frac {c_{-1} } {z - a_k} + \sum_{j \mathop = 0}^\infty c_j \paren {z - a_k}^j} \rd z
}}
{{eqn | r = \sum_{j \mathop = -\infty}^{-2} \oint_{\partial U_k} c_j \paren {z - a_k}^j \rd z + \oint_{\partial U_k} \paren {\frac {c_{-1} } {z - a_k} } \rd z + \sum_{j \mathop = 0}^\infty \oint_{\partial U_k} c_j \paren {z - a_k}^j \rd z
}}
{{eqn | r = \oint_{\partial U_k} \paren {\frac {c_{-1} } {z - a_k} } \rd z 
      | c = by Contour Integral of Power, terms where $j \ge 0$ and $j < -1$ vanish
}}
{{eqn | r = 2 \pi i c_{-1}
      | c = Contour Integral of Power 
}}
{{eqn | r = 2 \pi i \Res f {a_k}
      | c = {{Defof|Residue (Complex Analysis)}}
}}
{{end-eqn}}
So:
:$\ds \oint_L \map f z \rd z = \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z = 2 \pi i \sum_{k \mathop = 1}^n \Res f {a_k}$
{{qed}}
Theorem: Cauchy's Integral Formula/General Result

Let $n \in \N$ be a natural number.
Then for each $a$ in the interior of $D$, the $n$-th derivative of $f$ at $a$ exists and can be written as:
:$\ds \map {f^\paren n} a = \frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$
where $\partial D$ is the boundary of $D$, and is traversed anticlockwise.

Proof:
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \map {f^\paren n} a = \frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$
Theorem: Characteristic of Subfield of Complex Numbers is Zero

The characteristic of any subfield of the field of complex numbers is $0$.

Proof:
{{AimForCont}} to the contrary.
Let $K$ be a subfield of $\C$ such that $\Char K = n$ where $n \in \N, n > 0$.
Then from Characteristic times Ring Element is Ring Zero:
:$\forall a \in K: n \cdot a = 0$
But as $K$ is a subfield of $\C$ it follows that $K \subseteq \C$ which means:
:$\exists a \in \C: n \cdot a = 0$
Thus, by definition of characteristic:
:$0 < \Char \C \le n$
But $\C$ is infinite and so $\Char \C = 0$.
{{questionable|$\Char \C {{=}} 0$ is not a result from the infiniteness of $\C$. See Field has Algebraic Closure and Algebraically Closed Field is Infinite.|Fake Proof}}
From that contradiction follows the result.
{{qed}}
{{expand|See Field has Characteristic of Zero iff exists Monomorphism from Rationals.}}
Theorem: Characteristic times Ring Element is Ring Zero

Let $\struct {R, +, \circ}$ be a ring with unity.
Let the zero of $R$ be $0_R$ and the unity of $R$ be $1_R$.
Let the characteristic of $R$ be $n$.
Then:
:$\forall a \in R: n \cdot a = 0_R$

Proof:
If $a = 0_R$ then $n \cdot a = 0_R$ is immediate.
So let $a \in R: a \ne 0_R$.
Then:
{{begin-eqn}}
{{eqn | l = n \cdot a
      | r = n \cdot \paren {1_R \circ a}
      | c = {{Defof|Unity of Ring}}
}}
{{eqn | r = \paren {n \cdot 1_R} \circ a
      | c = Integral Multiple of Ring Element
}}
{{eqn | r = 0_R \circ a
      | c = {{Defof|Characteristic of Ring}}
}}
{{eqn | r = 0_R
      | c = {{Defof|Ring Zero}}
}}
{{end-eqn}}
{{qed}}
Theorem: Characteristic of Finite Ring is Non-Zero

Let $\struct {R, +, \circ}$ be a finite ring with unity.
Then the characteristic of $R$ is not zero.

Proof:
We have that $\struct {R, +, \circ}$ is finite, so its additive group $\struct {R, +}$ is likewise finite.
The result follows by Element of Finite Group is of Finite Order and the definition of characteristic.
{{qed}}
Theorem: Field of Rational Functions is Field

Let $K$ be a field.
Let $K \sqbrk x$ be the integral domain of polynomial forms on $K$.
Let $\map K x$ be the field of rational functions on $K$.
Then $\map K x$ forms a field.
If the characteristic of $K$ is $p$, then the characteristic of $\map K x$ is non-zero.

Proof:
{{ProofWanted|Reverted this to the version 9th May 2011 as it's mutated into something that makes less sense.}}
Theorem: Hadamard Factorization Theorem

Let $f: \C \to \C$ be an entire function of finite order $\omega$.
Let $0$ be a zero of $f$ of multiplicity $m \ge 0$.
Let $\sequence {a_n}$ be the sequence of non-zero zeroes of $f$, repeated according to multiplicity.
Then:
:$f$ has finite rank $p \le \omega$
and:
:there exists a polynomial $g$ of degree at most $\omega$ such that:
::$\ds \map f z = z^m e^{\map g z} \prod_{n \mathop = 1}^\infty E_p \paren {\frac z {a_n} }$
where $E_p$ denotes the $p$th Weierstrass elementary factor.

Proof:
By Convergence Exponent is Less Than Order, $f$ has finite exponent of convergence $\tau \le \omega$.
By Relation Between Rank and Exponent of Convergence, $f$ has finite rank $p \leq \omega$.
{{ProofWanted}}
Theorem: Approximation to Stirling's Formula for Gamma Function

Let $\Gamma: \R \to \R$ be the gamma function restricted to the real numbers.
Let $x \in \R_{>0}$.
Then:
:$\map \Gamma x \sim \sqrt {\dfrac {2 \pi} x} \paren {\dfrac x e}^x$
where $\sim$ denotes asymptotic equivalence.

Proof:
{{Explain|How do these inequalities follow from Bohr-Mollerup Theorem, or perhaps Log of Gamma Function is Convex on Positive Reals?}}
From Bohr-Mollerup Theorem:
{{begin-eqn}}
{{eqn | n = 1
      | l = \paren {y + n}^y n!
      | r = \paren {y + n}^y \map \Gamma {n + 1}
      | c = Gamma Function Extends Factorial
}}
{{eqn | o = \le
      | r = \map \Gamma {y + n + 1}
      | c = 
}}
{{eqn | o = \le
      | r = \paren {n + 1}^y \map \Gamma {n + 1}
      | c = 
}}
{{eqn | r = \paren {n + 1}^y n!
      | c = 
}}
{{end-eqn}}
for $0 < y \le 1$ and $n \in \N$.
Let $x$ be given.
Let $n + 1$ be the largest natural number such that $n + 1 \le x$.
Let $x = y + n + 1$, and thus $0 < y \le 1$.
Then:
{{begin-eqn}}
{{eqn | l = \frac {\map \Gamma x} {\sqrt {2 \pi} x^x x^{-1/2} e^{-x} }
      | r = \frac {\map \Gamma {y + n + 1} } {\sqrt {2 \pi} x^x x^{-1/2} e^{-x} }
      | c = $x = y + n + 1$
}}
{{eqn | o = \le
      | r = \frac {\paren {n + 1}^y n!} {\sqrt {2 \pi} x^x x^{-1/2} e^{-x} }
      | c = from $(1)$
}}
{{eqn | o = \sim
      | r = \frac {\paren {n + 1}^y \paren {\sqrt {2 \pi} n^n n^{1/2} e^{-n} } } {\sqrt {2 \pi} x^x x^{-1/2} e^{-x} }
      | c = Stirling's Formula
}}
{{eqn | r = \frac {\paren {n + 1}^y n^n n^{1/2} e^{-n} } {\paren {y + n + 1}^{y + n} \paren {y + n + 1}^{1/2} e^{-y - n - 1} }
      | c = dividing top and bottom by $\sqrt {2 \pi}$ and substituting $x = y + n + 1$
}}
{{eqn | r = \paren {\frac {n + 1} {y + n + 1} }^y \paren {\frac n {y + n + 1} }^{1/2} \paren {\frac n {y + n + 1} }^n e^{y + 1}
      | c =  dividing top and bottom by $e^{-n}$
}}
{{eqn | r = \paren {\frac {n + 1} {y + n + 1} }^y \paren {\frac n {y + n + 1} }^{1/2} \paren {\frac {y + n + 1} n }^{-n} e^{y + 1}
      | c = Exponent Combination Laws Negative Power
}}
{{eqn | r = \paren {\frac {n + 1} {y + n + 1} }^y \paren {\frac n {y + n + 1} }^{1/2} \paren {1 + \frac {y + 1} n}^{-n} e^{y + 1}
      | c =  
}}
{{eqn | o = \to
      | r = 1 \cdot 1 \cdot \frac 1 {e^{y + 1} } e^{y + 1} \text { as } n \to \infty
      | c = Exponential Function as Limit of Sequence and Properties of Limit at Infinity of Real Function/Product Rule
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
Therefore, we have:
:$\dfrac {\map \Gamma x} {\sqrt {\dfrac {2 \pi} x} \paren {\dfrac x e}^x} \to 1$ as $x \to +\infty$.
which by the Definition of Asymptotic Equality is equivalent to:
:$\map \Gamma x \sim \sqrt {\dfrac {2 \pi} x} \paren {\dfrac x e}^x$
{{qed}}
Theorem: Poisson Summation Formula

Let $f: \R \to \C$ be a Schwarz function.
Let $\hat f$ be its Fourier transform.
Then:
:$\ds \sum_{n \mathop \in \Z} \map f n = \sum_{m \mathop \in \Z} \map {\hat f} m$

Proof:
Let:
:$\ds \map F x = \sum_{n \mathop \in \Z} \map f {x + n}$
Then $\map F x$ is $1$-periodic (because of absolute convergence), and has Fourier coefficients:
{{begin-eqn}}
{{eqn | l = \hat F_k
      | r = \int_0^1 \sum_{n \mathop \in \Z} \map f {x + n} e^{-2\pi i k x} \rd x
      | c = 
}}
{{eqn | r = \sum_{n \mathop \in \Z} \int_0^1 \map f {x + n} e^{-2\pi i k x} \rd x
      | c = because $f$ is Schwarz, so convergence is uniform
}}
{{eqn | r = \sum_{n \mathop \in \Z} \int_n^{n + 1} \map f x e^{-2\pi i k x} \rd x
      | c = 
}}
{{eqn | r = \int_\R \map f x e^{-2\pi i k x} \rd x
      | c = 
}}
{{eqn | r = \map {\hat f} k
      | c = where $\hat f$ is the Fourier transform of $f$
}}
{{end-eqn}}
Therefore by the definition of the Fourier series of $f$:
:$\ds \map F x = \sum_{k \mathop \in \Z} \map {\hat f} k e^{2 \pi i k x}$
Choosing $x = 0$ in this formula:
:$\ds \sum_{n \mathop \in \Z} \map f n = \sum_{k \mathop \in \Z} \map {\hat f} k$
as required.
{{qed}}
{{Namedfor|Siméon-Denis Poisson|cat = Poisson}}
Theorem: Harmonic Properties of Schwarz Functions

Let $f, g : \R \to \C$ be Schwarz functions.
Let $\hat f$, $\hat g$ be the Fourier transforms of $f$ and $g$ respectively.
Then:
:$(1): \quad \hat f$, $\hat g$ are Schwarz functions.
:$(2): \quad \map {\widehat {\paren {\hat f} } } x = \map f {-x}$ for all $x \in \R$.
:$(3): \quad$ If $f * g$ is the convolution of $f$ and $g$, then:
{{disambiguate|Definition:Convolution}}
::::$\widehat {f * g} = \hat f \hat g$

Proof:
{{ProofWanted}}
Category:Harmonic Analysis
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Theorem: Euler's Reflection Formula

:$\forall z \notin \Z: \map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$

Proof:
We have the Weierstrass products:
:$\ds \map \sin {\pi z} = \pi z \prod_{n \mathop \ne 0} \paren {1 - \frac z n} \map \exp {\frac z n}$
From the Weierstrass form of the Gamma function:
:$\ds \frac 1 {\map \Gamma z} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \paren {1 + \frac z n} \map \exp {-\frac z n}$
from which:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {-z \, \map \Gamma z \map \Gamma {-z} }
      | r = \frac {-z^2 \map \exp {\gamma z} \map \exp {-\gamma z} } {-z} \prod_{n \mathop = 1}^\infty \paren {1 + \frac z n} \paren {1 - \frac z n} \map \exp {\frac z n} \map \exp {-\frac z n}
}}
{{eqn | r = z \prod_{n \mathop = 1}^\infty \paren {1 - \frac {z^2} {n^2} }
}}
{{eqn | r = \dfrac {\map \sin {\pi z} } \pi 
      | c = Euler Formula for Sine Function
}}
{{end-eqn}}
whence:
{{begin-eqn}}
{{eqn | l = \map \Gamma z \map \Gamma {1 - z}
      | r = -z \, \map \Gamma z \map \Gamma {-z}
      | c = Gamma Difference Equation
}}
{{eqn | r = \frac \pi {\map \sin {\pi z} }
}}
{{end-eqn}}
{{qed}}
Theorem: Legendre's Duplication Formula

:$\forall z \notin \set {-\dfrac n 2: n \in \N}: \map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
where $\N$ denotes the natural numbers.

Proof:
From the definition of the Beta function:
{{begin-eqn}}
{{eqn | l = \map \Beta {z_1, z_2}
      | r = \frac {\map \Gamma {z_1} \map \Gamma {z_2} } {\map \Gamma {z_1 + z_2} } 
      | c = 
}}
{{eqn | r = \int_0^1 u^{z_1 - 1} \paren {1 - u}^{z_2 - 1} \rd u
      | c = Equivalence of Definitions of Beta Function
}}
{{end-eqn}}
Letting $z_1 = z_2 = z$ gives:
{{begin-eqn}}
{{eqn | l = \frac {\map \Gamma z \map \Gamma z} {\map \Gamma {2 z} }
      | r = \int_0^1 u^{z - 1} \paren {1 - u}^{z - 1} \rd u
      | c = 
}}
{{eqn | r = \frac 1 2 \int_{-1}^1 \paren {\frac {1 + x} 2 }^{z - 1} \paren {\frac {1 - x} 2}^{z - 1} \rd x
      | c = Substitute $u = \dfrac {1 + x} 2$
}}
{{eqn | r = \frac 1 {2^{2 z - 1} } \int_{-1}^1 \paren {1 - x^2}^{z - 1} \rd x
      | c = 
}}
{{eqn | ll= \leadsto
      | n = 1
      | l = 2^{2 z - 1} \map \Gamma z \map \Gamma z
      | r = 2 \map \Gamma {2 z} \int_0^1 \paren {1 - x^2}^{z - 1} \rd x
      | c = by Definite Integral of Even Function
}}
{{end-eqn}}
Now substituting $u = x^2$ into the Beta function:
:$\ds \map \Beta {z_1, z_2} = \int_0^1 x^{2 z_1 - 2} \paren {1 - x^2}^{z_2 - 1} 2 x \rd x$
Letting $z_1 = \dfrac 1 2$ and $z_2 = z$ gives:
:$(2): \quad \ds \map \Beta {\frac 1 2, z} = 2 \int_0^1 \paren {1 - x^2}^{z - 1} \rd x$
Combining results $(1)$ and $(2)$:
{{begin-eqn}}
{{eqn | l = 2^{2 z - 1} \map \Gamma z \map \Gamma z 
      | r = \map \Gamma {2 z} \map \Beta {\frac 1 2, z}
      | c =
}}
{{eqn | r = \map \Gamma {2 z} \frac {\map \Gamma {\frac 1 2} \map \Gamma z} {\map \Gamma {\frac 1 2 + z} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \Gamma z \map \Gamma {z + \dfrac 1 2} 
      | r = 2^{1 - 2 z} \map \Gamma {\frac 1 2} \map \Gamma {2 z}
}}
{{end-eqn}}
From Gamma Function of One Half:
:$\map \Gamma {\dfrac 1 2} = \sqrt \pi$
It follows that:
:$\map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
{{qed}}
Theorem: Legendre's Duplication Formula

:$\forall z \notin \set {-\dfrac n 2: n \in \N}: \map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
where $\N$ denotes the natural numbers.

Proof:
From Gauss Multiplication Formula:
:$\ds \prod_{k \mathop = 0}^{n - 1} \map \Gamma {z + \frac k n} = \paren {2 \pi}^{\paren {n - 1} / 2} n^{1/2 - n z} \map \Gamma {n z}$
Substituting $n = 2$ yields:
{{begin-eqn}}
{{eqn | l = \map \Gamma z \map \Gamma {z + \frac 1 2}
      | r = \paren {2 \pi}^{1 / 2} 2^{1/2 - 2 z} \map \Gamma {2 z}
      | c = 
}}
{{eqn | r = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Approximation to Reciprocal times Derivative of Gamma Function

For all $z \in \C$ such that $\cmod {\map \arg z} < \pi - \epsilon, \cmod z > 1$:
:$\dfrac {\map {\Gamma'} z} {\map \Gamma z} = \ln z + \map {\OO_\epsilon} {z^{-1} }$
where:
:$\map \OO {z^{-1} }$ denotes big-$\OO$ notation
:the implied constant depends on $\epsilon$.

Proof:
From Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function:
:$\ln \map \Gamma z = \paren {z - \dfrac 1 2} \ln z - z + \dfrac {\ln 2 \pi} 2 + \map \OO {z^{-1} }$
Taking the derivative {{WRT|Differentiation}} $z$:
:$(1): \quad \dfrac {\map {\Gamma'} z} {\map \Gamma z} = \ln z - \dfrac 1 {2 z} + \dfrac \d {\d z} \map \OO {z^{-1} }$
Since there exists $\map c \epsilon > 0$ such that:
:$\forall \size z > 1: -\dfrac c {\size {z^{-1} } } < \size {\map \OO {z^{-1} } } < \dfrac c {\size {z^{-1} } }$
{{explain|The above statement needs to be explained and justified.}}
it follows directly that the third term in $(1)$ is $\map \OO {z^{-1} }$.
{{questionable|Why? $\dfrac \d {\d z} \map \OO {z^{-1} }$ is not estimated at all}}
{{qed}}
Category:Gamma Function
Category:Reciprocals
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Theorem: Complex Conjugate of Gamma Function

:$\forall z \in \C \setminus \set {0, -1, -2, \ldots}: \map \Gamma {\overline z} = \overline {\map \Gamma z}$

Proof:
This is immediate from, say, the Euler form of $\Gamma$ and the fact that complex conjugation preserves products and sums.
Category:Gamma Function
Category:Complex Conjugates
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Theorem: Linear Combination of Integers is Ideal

Let $a, b$ be any integers.
Let $\Bbb S = \set {a x + b y: x, y \in \Z}$.
Then the algebraic structure:
:$\struct {\Bbb S, +, \times}$
is an ideal of $\Z$.

Proof:
From the definition of integer combination (or just because it's obvious):
:$\Bbb S \subseteq \Z$
and clearly:
:$\Bbb S \ne \O$
as $a 0 + b 0 = 0 \in \Bbb S$.
Let $w_1, w_2 \in \Bbb S$:
:$w_1 = a x_1 + b y_1, w_2 = a x_2 + b y_2$
Then $w_1 + \paren {-w_2} = a \paren {x_1 - x_2} + b \paren {y_1 - y_2} \in \Bbb S$ as $x_1 - x_2 \in \Z$ and $y_1 - y_2 \in \Z$.
Let $r \in \Z$.
Then:
:$w_1 \times r = a x_1 r + b y_1 r = a \paren {x_1 r} + b \paren {y_1 r} \in \Bbb S$
:$r \times w_1 = r a x_1 + r b y_1 = \paren {r x_1} a + \paren {r y_1} b \in \Bbb S$
The result follows from Test for Ideal.
{{qed}}
Theorem: Intersection of Integer Ideals is Lowest Common Multiple

Let $\ideal m$ and $\ideal n$ be ideals of the integers $\Z$.
Let $\ideal k$ be the intersection of $\ideal m$ and $\ideal n$.
Then $k = \lcm \set {m, n}$.

Proof:
By Intersection of Ring Ideals is Ideal we have that $\ideal k = \ideal m \cap \ideal n$ is an ideal of $\Z$.
By Ring of Integers is Principal Ideal Domain we have that $\ideal m$, $\ideal n$ and $\ideal k$ are all necessarily principal ideals.
By Subrings of Integers are Sets of Integer Multiples we have that:
:$\ideal m = m \Z, \ideal n = n \Z$
Thus:
:$\ideal k = \set {x \in \Z: n \divides x \land m \divides x}$
The result follows by LCM iff Divides All Common Multiples.
{{qed}}
Theorem: LCM iff Divides All Common Multiples

Let $a, b \in \Z$ such that $a b \ne 0$.
Let $m \in \Z$ such that $m > 0$.
Then $m = \lcm \set {a, b}$ {{iff}}:
: $(1): \quad a \divides m \land b \divides m$
: $(2): \quad a \divides n \land b \divides n \implies m \divides n$
That is, in the set of positive integers, $m$ is the LCM of $a$ and $b$ {{iff}} $m$ is a common multiple of $a$ and $b$, and $m$ also divides any other common multiple of $a$ and $b$.

Proof:
=== Necessary Condition ===
Suppose $m = \lcm \set {a, b}$.
Then from LCM Divides Common Multiple:
: $a \divides n \land b \divides n \implies m \divides n$
{{qed|lemma}}
Theorem: LCM Divides Common Multiple

Let $a, b \in \Z$ such that $a b \ne 0$.
Let $n$ be any common multiple of $a$ and $b$.
That is, let $n \in \Z: a \divides n, b \divides n$.
Then:
:$\lcm \set {a, b} \divides n$
where $\lcm \set {a, b}$ is the lowest common multiple of $a$ and $b$.
{{:Euclid:Proposition/VII/35}}

Proof:
Let $m = \lcm \set {a, b}$.
Then $a \divides m$ and $b \divides m$ by definition.
Suppose $n$ is some other common multiple of $a$ and $b$ such that $m \nmid n$ ($m$ does not divide $n$).
Then from the Division Theorem:
:$n = k m + r$
for some integer $k$ and with $0 < r < m$.
Then since $r = n - k m$, using $a \divides n$ and $a \divides m$:
: $a \divides r$
Similarly:
: $b \divides r$
Then $r$ is a common multiple of $a$ and $b$.
But we have that $r < m$.
This contradicts the fact that $m$ is the lowest common multiple of $a$ and $b$.
So, by contradiction, it follows that $m \divides n$.
{{qed}}
{{Euclid Note|35|VII|}}
Theorem: Sum of Integer Ideals is Greatest Common Divisor

Let $\ideal m$ and $\ideal n$ be ideals of the integers $\Z$.
Let $\ideal d = \ideal m + \ideal n$.
Then $d = \gcd \set {m, n}$.

Proof:
By Sum of Ideals is Ideal we have that $\ideal d = \ideal m + \ideal n$ is an ideal of $\Z$.
By Ring of Integers is Principal Ideal Domain we have that $\ideal m$, $\ideal n$ and $\ideal d$ are all necessarily principal ideals.
By Subrings of Integers are Sets of Integer Multiples we have that:
:$\ideal m = m \Z, \ideal n = n \Z$
Thus:
:$\ideal d = \ideal m + \ideal n = \set {x \in \Z: \exists a, b \in \Z: x = a m + b n}$
That is, $\ideal d$ is the set of all integer combinations of $m$ and $n$.
The result follows by Bézout's Identity.
{{qed}}
Theorem: Quotient Ring by Null Ideal

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\struct {\set {0_R}, +, \circ}$ be the null ideal of $\struct {R, +, \circ}$.
Let $\struct {R / \set {0_R}, +, \circ}$ be the quotient ring of $R$ defined by $\set {0_R}$.
Then $\struct {R / \set {0_R}, +, \circ}$ is isomorphic to $\struct {R, +, \circ}$.

Proof:
Consider the additive group $\struct {R, +}$.
From Trivial Quotient Group is Quotient Group:
:$\struct {R, +} / \set {0_R} \cong \struct {R, +}$
{{finish}}
Theorem: Ring Homomorphism Preserves Negatives

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Then:
:$\forall x \in R_1: \map \phi {-x} = -\paren {\map \phi x}$

Proof:
We have that Ring Homomorphism of Addition is Group Homomorphism.
The result follows from Group Homomorphism Preserves Inverses.
{{qed}}
Theorem: Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function

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 ===
{{:Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function/Lemma 1}}{{qed|lemma}}
Theorem: Mapping on Quadratic Integers over 2 to Conjugate is Automorphism

Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Then the mapping $\phi: \Z \sqbrk {\sqrt 2} \to \Z \sqbrk {\sqrt 2}$ defined as:
:$\forall x = a + b \sqrt 2 \in \Z \sqbrk {\sqrt 2}: \map \phi {a + b \sqrt 2} = a - b \sqrt 2$
is a ring automorphism.

Proof:
We have Quadratic Integers over 2 form Subdomain of Reals.
First we note that:
:$\forall x \in \Z \sqbrk {\sqrt 2}: \map \phi x \in \Z \sqbrk {\sqrt 2}$