nyuuzyou/wikis huggingface dataset:
Integers whose Phi times Divisor Count equal Divisor Sum

== Theorem ==
The positive integers whose Euler $\phi$ function multiplied by its divisor count function equals its divisor sum are:
:$1, 3, 14, 42$
== Proof ==
To show that there are no more such integers, we use the formulas:
Which arises from Euler Phi Function is Multiplicative, Divisor Count Function is Multiplicative and Divisor Sum Function is Multiplicative.
We define the mapping $f = \phi \cdot \sigma_0$ as follows:
:$\map f {p^k} = \map \phi {p^k} \map {\sigma_0} {p^k} = p^k \paren {1 - \dfrac 1 p} \paren {k + 1}$
where $p$ is a prime, and $k$ is a non-negative integer.
By Product of Multiplicative Functions is Multiplicative, $f$ and $\dfrac f {\sigma_1}$ are multiplicative functions.
Then $\map \phi n \map {\sigma_0} n = \map {\sigma_1} n$ is equivalent to:
:$\ds \prod_{i \mathop = 1}^r \dfrac {\map f {p_i^{k_i} } } {\map {\sigma_1} {p_i^{k_i} } } = 1$
We consider the cases $k \ge 1$.
Then:
Suppose equality holds.
If $k \ge 2$:
 
which is a contradiction.
This forces $k = 1$.
The equation becomes:
Since $p \ne 1$, we must have $p = 3$.
Indeed:
:$\map f 3 = \map \phi 3 \map {\sigma_0} 3 = \map {\sigma_1} 3$
For the inequality case: $k p^{k + 1} + \paren {k + 1} p^{k - 1} + 1 > 2 \paren {k + 1} p^k$, we check each prime $p$ individually.
For $p = 2$:
For $p \ge 3$, we consider the criterion $k p \ge 2 \paren {k + 1}$.
This criterion implies $k p^{k + 1} \ge 2 \paren {k + 1} p^k$,
which in turn implies our inequality $k p^{k + 1} + \paren {k + 1} + 1 > 2 \paren {k + 1} p^k$.
For $p = 3$ and $k \ge 2$, $3 k = 2 k + k \ge 2 k + 2$.
For $p \ge 5$, $k p \ge 5 k > 2 k + 2 k \ge 2 k + 2$.
Therefore $\map f {p^k} > \map {\sigma_1} {p^k}$, except for:
:$\map f 3 = \map {\sigma_1} 3$
:$\map f 2 < \map {\sigma_1} 2$
:$\map f 4 < \map {\sigma_1} 4$
Now we try to find integers $n$ satisfying $\ds \prod_{i \mathop = 1}^r \dfrac {\map f {p_i^{k_i} } } {\map {\sigma_1} {p_i^{k_i} } } = 1$.
First observe that the sequence $\sequence {\dfrac {\map f {p^k} } {\map {\sigma_1} {p^k} } }_{k \mathop = 1}^\infty$ is strictly increasing:
Now we consider these particular values for $\dfrac f {\sigma_1}$:
:$\dfrac {\map f 2} {\map {\sigma_1} 2 } = \dfrac 2 3$
:$\dfrac {\map f 4} {\map {\sigma_1} 4} = \dfrac 6 7$
:$\dfrac {\map f 3} {\map {\sigma_1} 3} = 1$
:$\dfrac {\map f 5} {\map {\sigma_1} 5} = \dfrac 4 3$
:$\dfrac {\map f {25} } {\map {\sigma_1} {25} } = \dfrac {60} {31} > \dfrac 3 2$
:$\dfrac {\map f 7} {\map {\sigma_1} 7} = \dfrac 3 2$
And for $p \ge 11$:
By the strictly increasing property above, every higher power of these primes has $\dfrac f {\sigma_1} > \dfrac 3 2$.
Since $\dfrac f {\sigma_1}$ is multiplicative, only $n = 3, 14, 42$ can give $\dfrac {\map f n} {\map {\sigma_1} n} = 1$.
Since $\map \phi 1 \map {\sigma_0} 1 = \map {\sigma_1} 1$, the only positive integers whose Euler $\phi$ function multiplied by its divisor count function equals its divisor sum are $1, 3, 14, 42$.
== Sources ==
* : $42$
Category:Euler Phi Function
Category:Divisor Count Function
Category:Divisor Sum Function
Integers whose Ratio between Divisor Sum and Phi is Square

== Theorem ==
The sequence of integers whose divisor sum divided by its Euler $\phi$ value is a square begins:
:$1, 14, 30, 105, 248, 264, 418, 714, 1485, 3080, \ldots$
== Examples ==
Category:Integers whose Ratio between Divisor Sum and Phi is Square
Category:Divisor Sum Function
Category:Euler Phi Function
Category:Square Numbers
Integers whose Squares end in 444

== Theorem ==
The sequence of positive integers whose square ends in $444$ begins:
:$38, 462, 538, 962, 1038, 1462, 1538, 1962, 2038, 2462, 2538, 2962, 3038, 3462, \ldots$
== Proof ==
All such $n$ are of the form $500 m + 38$ or $500 m - 38$:
and it is seen that all such numbers end in $444$.
Now we show that all such numbers are so expressed.
In Squares Ending in Repeated Digits, we have shown the only numbers with squares ending in $44$ ends in:
:$12, 38, 62, 88$
hence any number with square ending in $444$ must also end in those numbers.
Suppose $\sqbrk {axy}^2 \equiv 444 \pmod {1000}$, where $a < 10$ and $\sqbrk {xy}$ is in the above list.
For $\sqbrk {xy} = 12$:
This is a contradiction.
Similarly for $\sqbrk {xy} = 88$:
Again, a contradiction.
For $\sqbrk {xy} = 38$:
The solutions to $600 a \equiv 0 \pmod {1000}$ are $a = 0$ or $5$.
Hence:
:$\paren {500 n + 38}^2 \equiv 444 \pmod {1000}$
Similarly, for $\sqbrk {xy} = 62$:
The solutions to $400 a + 400 \equiv 0 \pmod {1000}$ are $a = 4$ or $9$.
Hence:
:$\paren {500 n + 462}^2 \equiv \paren {500 n - 38}^2 \equiv 444 \pmod {1000}$
== Sources ==
* 
* : $462$
* : $1444$
* : $462$
* : $1444$
Category:Square Numbers
Category:Recreational Mathematics
Integers with Metric Induced by P-adic Valuation

== Theorem ==
Let $p \in \N$ be a prime.
Let $d: \Z^2 \to \R_{\ge 0}$ be the mapping defined as:
:$\forall x, y \in \Z: \map d {x, y} = \begin {cases} 0 & : x = y \\ \dfrac 1 r & : x - y = p^{r - 1} k: r \in \N_{>0}, k \in \Z, p \nmid k \end {cases}$
Then $d$ is a metric on $\Z$.
== Proof ==
From Characterization of P-adic Valuation on Integers:
:$d$ is well-defined.
We prove the metric space axioms.
=== ===
This follows immediately from the definition of $d$.
=== ===
Let $x, y, z \in \Z$.
From Characterization of P-adic Valuation on Integers:
:$\exists r, s, t \in \N_{>0}, k, l, m \in \Z, p \nmid k, l , m:$
::$x - y = p^{r - 1} k$
::$y - z = p^{s - 1} l$
::$x - z = p^{t - 1} m$
Let $t' = \min \set{r, s}$.
Then:
Hence:
:$p^{t' - 1} \divides p^{t - 1} m$
Then:
:$t' \le t$
Hence:
=== ===
Let $x, y \in \Z$.
==== Case 1: $x = y$ ====
Let $x = y$.
By definition of $d$:
:$\map d {x, y} = 0 = \map d {y, x}$
==== Case: $x \ne y$ ====
Let $x \neq y$.
Then
Hence:
:$\map d {x, y} = \dfrac 1 r = \map d {y, x}$
=== ===
Let $x, y \in \Z$ such that:
:$x \ne y$
Let $x - y = p^{r - 1} k: r \in \N_{>0}, k \in \Z, p \nmid k$.
We have:
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 23$
Category:Definitions/Examples of Metric Spaces
Integers with Prime Values of Divisor Sum

== Theorem ==
The sequence of integer whose divisor sum is prime begins:
== Proof ==
Apart from $2$, all primes are odd.
From Divisor Sum is Odd iff Argument is Square or Twice Square, for $\map {\sigma_1} n$ to be odd it needs to be of the form $m^2$ or $2 m^2$.
Suppose $n$ has two coprime divisors $p$ and $q$, each to power $k_p$ and $k_q$ respectively.
Then $\map {\sigma_1} n$ will have $\map {\sigma_1} {p^{k_p} }$ and $\map {\sigma_1} {q^{k_q} }$ as divisors.
Hence $\map {\sigma_1} n$ will not be prime.
So for $\map {\sigma_1} n$ to be prime, $n$ can have only one prime factor.
This gives possible values for $n$ as:
:powers of $2$, either odd or even
or:
:even powers of a prime number.
These can be investigated in turn, using Divisor Sum of Power of Prime:
:$\map {\sigma_1} {p^k} = \dfrac {p^{k + 1} - 1} {p - 1}$
Note that as $\map {\sigma_1} {2^k} = \dfrac {2^{k + 1} - 1} {2 - 1} = 2^{k + 1} - 1$ it is necessary for powers of $2$ merely to report the appropriate Mersenne prime.
Hence when $k + 1$ is not prime, $\map {\sigma_1} {2^k}$ will not be prime and there is no need to test it.
Thus we test all $n$ such that:
:$n = p^{2 k}$ for prime $p$
:$n = 2^k$ where $k + 1$ is prime
and so:
Hence the sequence as given.
== Sources ==
* : $16$
::but beware a mistake in this sequence.
Category:Divisor Sum Function
Category:Prime Numbers
Integers with Prime Values of Divisor Sum/Mistake

== Source Work ==
:
:The Dictionary
::$16$
== Mistake ==
:$\map \sigma {16} = 31$ is prime. The sequence of $n$ for which $\map \sigma n$ is prime runs: $2, 4, 9, 16, 25, 289 \ldots$
This is incorrect.
The correct sequence is:
:$2, 4, 9, 16, 25, 64, 289 \ldots$
== Sources ==
* : $16$
Category:Mistakes/Curious and Interesting Numbers
Integrability Theorem for Functions Continuous on Open Intervals

== Theorem ==
Let $f$ be a real function defined on a closed interval $\closedint a b$ such that $a < b$.
Let $f$ be continuous on $\openint a b$.
Let the one-sided limits $\ds \lim_{x \mathop \to a^+} \map f x$ and $\ds \lim_{x \mathop \to b^-} \map f x$ exist.
Then $f$ is Darboux integrable on $\closedint a b$.
== Proof ==
We start by showing that $f$ is bounded.
A real function $g$ defined on $\closedint a b$ exists by Extendability Theorem for Function Continuous on Open Interval such that:
:$g$ equals $f$ on $\openint a b$
:$g$ is continuous on $\closedint a b$
Therefore:
:$g$ is bounded on $\closedint a b$ by Continuous Function on Compact Subspace of Euclidean Space is Bounded and Closed Real Interval is Compact in Metric Space
Accordingly, $g$ is bounded on $\openint a b$ as $\openint a b$ is a subset of $\closedint a b$.
Therefore, $f$ is bounded on $\openint a b$ as $f = g$ on $\openint a b$.
Also, $f$ is bounded on the set $\set {a, b}$ by the maximum $\map \max {\size {\map f a}, \size {\map f b} }$.
Therefore:
:$f$ is bounded on $\closedint a b$
We have:
:$f$ is bounded on $\closedint a b$
:$f$ is continuous on $\openint a b$
Therefore:
:$f$ is Darboux integrable on $\closedint a b$ by Bounded Function Continuous on Open Interval is Darboux Integrable.
Category:Integral Calculus
Integrable Function is A.E. Real-Valued

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\mu$-integrable function.
Then $\map f x \in \R$ for almost all $x \in X$.
== Proof ==
From Set of Points for which Measurable Function is Real-Valued is Measurable: Corollary, we have that: 
:$\set {x \in X : \size {\map f x} = +\infty}$ is $\Sigma$-measurable.
We now aim to show that this set is a null set. 
Now note that we have:
:$\set {x \in X : \size {\map f x} = +\infty} \subseteq \set {x \in X : \size {\map f x} \ge n}$
From Measure is Monotone, we therefore have: 
:$\ds \map \mu {\set {x \in X : \size {\map f x} = +\infty} } \le \map \mu {\set {x \in X : \size {\map f x} \ge n} }$
From Markov's Inequality, we have: 
:$\ds \map \mu {\set {x \in X : \size {\map f x} \ge n} } \le \frac 1 n \int \size f \rd \mu$
So:
:$\ds 0 \le \map \mu {\set {x \in X : \size {\map f x} = +\infty} } \le \frac 1 n \int \size f \rd \mu$
for each $n \in \N$.
From Lower and Upper Bounds for Sequences, taking $n \to \infty$, we obtain:
:$\ds \map \mu {\set {x \in X : \size {\map f x} = +\infty} } = 0$ 
That is: 
:$\ds \map \mu {X \setminus \set {x \in X : \map f x \in \R} } = 0$
So: 
:$\map f x \in \R$ for almost all $x \in X$.
== Sources ==
* : $10.13$
Category:Measure Theory
Integrable Function with Zero Integral on Sub-Sigma-Algebra is A.E. Zero

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\GG$ be a sub-$\sigma$-algebra of $\Sigma$.
Let $f : X \to \overline \R$ be a $\GG$-integrable function.
Suppose that, for all $G \in \GG$:
:$\ds \int_G f \rd \mu = 0$
Then $f = 0$ $\mu$-almost everywhere.
== Proof ==
Let: 
:$\ds A = \set {x \in X : \map f x \ge 0}$
so that: 
:$\ds X \setminus A = \set {x \in X : \map f x < 0}$
From Characterization of Measurable Functions:
:$A$ and $X \setminus A$ are $\GG$-measurable.
So:
:$\ds \int_A f \rd \mu = 0$
That is: 
:$\ds \int f \cdot \chi_A \rd \mu = 0$
Note that for $x \in A$, we have: 
:$\map f x \map {\chi_A} x = \map f x \ge 0$ 
and for $x \in X \setminus A$, we have: 
:$\map f x \map {\chi_A} x = 0$
So:
:$f \cdot \chi_A \ge 0$
So that: 
:$f \cdot \chi_A = \size {f \cdot \chi_A}$
This gives: 
:$\ds \int \size {f \cdot \chi_A} = 0$ 
Then, from Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we have: 
:$f \cdot \chi_A = 0$ $\mu$-almost everywhere.
That is, there exists a $\mu$-null set $N_1 \subseteq X$ such that: 
:if $\map f x \map {\chi_A} x \ne 0$ then $x \in N_1$. 
Note that we also have: 
:$\ds \int_{X \setminus A} f \rd \mu = 0$ 
so that: 
:$\ds \int f \cdot \chi_{X \setminus A} \rd \mu = 0$
From Integral of Integrable Function is Homogeneous, we have: 
:$\ds \int \paren {-f \cdot \chi_{X \setminus A} } \rd \mu = 0$
Note that for $x \in A$, we have: 
:$-\map f x \map {\chi_{X \setminus A} } x = 0$
and for $x \in X \setminus A$, we have: 
:$-\map f x \map {\chi_{X \setminus A} } x = -\map f x > 0$
So:
:$-f \cdot \chi_{X \setminus A} \ge 0$
so that: 
:$-f \cdot \chi_{X \setminus A} = \size {f \cdot \chi_{X \setminus A} }$
This gives: 
:$\ds \int \size {f \cdot \chi_{X \setminus A} } \rd \mu = 0$ 
Then, from Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we have: 
:$f \cdot \chi_{X \setminus A} = 0$ $\mu$-almost everywhere.
That is, there exists a $\mu$-null set $N_2 \subseteq X$ such that: 
:if $\map f x \map {\chi_{X \setminus A} } x \ne 0$ then $x \in N_2$.
From Characteristic Function of Set Difference, we have: 
:$\chi_{X \setminus A} = \chi_X - \chi_A$
So:
So if $x \in X$ has $\map f x \ne 0$ then: 
:either $\map f x \map {\chi_{X \setminus A} } x \ne 0$ or $\map f x \map {\chi_A} x \ne 0$.
That is: 
:either $x \in N_1$ or $x \in N_2$.
So if $\map f x \ne 0$, we have that:
:$x \in N_1 \cup N_2$ 
From Null Sets Closed under Countable Union, we have that: 
:$N_1 \cup N_2$ is a $\mu$-null set. 
So:
:$f = 0$ $\mu$-almost everywhere.
== Proof 2 ==
Category:Measure Theory
Integrable Function with Zero Integral on Sub-Sigma-Algebra is A.E. Zero/Proof 2

== Theorem ==
== Proof ==
In view of Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we shall show:
:$\ds \int \size f \rd \mu = 0$
Since $f$ is $\GG$-measurable:
:$G_+ := \set { x \in X : f(x) > 0 } \in \GG$
and:
:$G_- := \set { x \in X : f(x) \le 0 } \in \GG$
On the one hand:
On the other hand, from $f \cdot \chi _{ G_- } \le 0$ follows:
and:
Thus:
which implies:
Therefore:
:$\ds \int \size f \rd \mu = \int f^+ \rd \mu + \int f^- \rd \mu = 0$
Category:Measure Theory
Integral Closure is Integrally Closed

== Theorem ==
Let $A \subseteq B$ be an extension of commutative rings with unity.
Let $C$ be the integral closure of $A$ in $B$.
Then $C$ is integrally closed.
== Proof ==
Suppose $x \in B$ is integral over $C$.
Certainly $C$ is integral over $A$, so by Transitivity of Integrality, $C \sqbrk x$ is integral over $A$.
In particular, $x$ is integral over $A$, so $x \in C$.
Category:Algebraic Number Theory
Category:Commutative Algebra
Integral Closure is Subring

== Theorem ==
Let $A$ be an extension of a commutative ring with unity $\struct {R, +, \circ}$.
Let $C$ be the integral closure of $R$ in $A$.
Then $C$ is a subring of $A$.
== Proof ==
For each $r \in R$, we have $r$ is a zero of $a - r = 0$.
Hence $R \subseteq C$, and in particular, $C$ is non-empty.
By Subring Test it is sufficient to prove that if $x, y \in C$ then $x + \paren {-y}, x \circ y \in C$.
Let $x, y \in C$.
By Equivalent Definitions of Integral Dependence: $(1) \implies (2)$, $R \sqbrk x$ and $R \sqbrk y$ are finitely generated over $R$.
Because $R \subseteq R \sqbrk x$, $R \sqbrk x \sqbrk y$ is also finitely generated over $R \sqbrk x$.
Therefore, we have finitely generated extensions:
:$R \hookrightarrow R \sqbrk x \hookrightarrow R \sqbrk x \sqbrk y = R \sqbrk {x, y}$
Hence, by Transitivity of Finite Generation, $R \sqbrk {x, y}$ is finitely generated over $R$.
So by Equivalent Definition of Integral Dependence: $(2) \implies (1)$, every element of $R \sqbrk {x, y}$ is integral over $R$.
In particular, $x + \paren {-y}$ and $x \circ y$ are integral over $R$.
Category:Commutative Algebra
Category:Algebraic Number Theory
Integral Domain iff Zero Ideal is Prime

== Theorem ==
Let $A$ be a commutative ring with unity.
:$(1): \quad A$ is an integral domain
:$(2): \quad$ the zero ideal $\ideal 0 \subseteq A$ is prime
== Proof ==
By Prime Ideal iff Quotient Ring is Integral Domain, $\ideal 0$ is prime the quotient ring $A / \ideal 0$ is an integral domain.
By Quotient Ring by Null Ideal, $A \cong A / \ideal 0$.
Category:Integral Domains
Category:Prime Ideals of Rings
Integral Domain is Reduced Ring

== Theorem ==
Let $\left({D, +, \circ}\right)$ be an integral domain.
Then $D$ is reduced.
== Proof ==
Let $x \in D$ be a nilpotent element.
Then by Nilpotent Element is Zero Divisor, $x$ is a zero divisor in $D$.
By the definition of an integral domain, this means that $x = 0$.
Therefore the only nilpotent element of $D$ is $0$.
That is, $D$ is reduced.
Category:Ring Theory
Integral Domain of Prime Order is Field

== Theorem ==
Let $\struct {\Z_p, +_p, \times_p}$‎ be the ring of integers modulo $p$.
The following statements are equivalent:
:$(1): \quad p$ is a prime.
:$(2): \quad \struct {\Z_p, +_p, \times_p}$ is an integral domain.
:$(3): \quad \struct {\Z_p, +_p, \times_p}$ is a field.
== Proof ==
By Prime Number iff Generates Principal Maximal Ideal and Maximal Ideal iff Quotient Ring is Field, $(1)$ implies $(3)$, and from Field is Integral Domain, $(3)$ implies $(2)$.
By definition of Integral Domain, $\Z_p$ is an integral domain $\struct {\Z_p^*, \times_p}$ is a semigroup.
Let $\ideal p$ be the principal ideal of $\struct {\Z, +, \times}$ generated by $p$.
From Quotient Epimorphism from Integers by Principal Ideal:
:$\struct {\Z_p, +_p, \times_p}$ is isomorphic to $\struct {\Z, +, \times} / \ideal p$.
So, we can let $\map {q_p} m: \Z \to \Z_p$ be the quotient mapping from $\struct {\Z, +, \times}$ to $\struct {\Z_p, +_p, \times_p}$.
Let $0_p$ denote the zero of $\Z_p$.
Suppose $p = m n$ where $1 < m < p, 1 < n < p$.
Then in the ring $\Z_p$ we have $\map {q_p} m \ne 0_p, \map {q_p} n \ne 0_p$.
But as $q_p$ is an epimorphism and therefore obeys the morphism property, $\map {q_p} m \times_p \map {q_p} n = \map {q_p} {m n} = \map {q_p} p = 0_p$.
But by definition, $0_p \notin \Z_p^*$.
Thus if $p = m n$, then $\struct {\Z_p^*, \times_p}$ is not a semigroup.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.6$
Category:Integral Domains
Category:Galois Fields
Integral Domain with Characteristic Zero

== Theorem ==
In an integral domain with characteristic zero, every non-zero element has infinite order under ring addition.
== Proof ==
Let $\struct {D, +, \circ}$ be an integral domain, whose zero is $0_D$ and whose unity is $1_D$, such that $\Char D = 0$.
Let $x \in D, x \ne 0_D$.
Then:
Then:
That is, $x$ has infinite order in $\struct {D, +}$.
== Sources ==
* : $\S 61.2$ Characteristic of an integral domain or field
Category:Integral Domains
Integral Expression of Harmonic Number

== Theorem ==
Let $\sequence {H_n}_{n \mathop \in \N}$ be the harmonic numbers.
Then:
:$\ds H_n = 1 + \int_1 ^n \dfrac {\floor u} {u^2} \rd u$
where $\floor u$ denotes the floor of $u$.
== Proof ==
Observe that:
:$\ds \forall x \in \R_{\ge 1} : \floor x = \sum_{1 \mathop \le k \mathop \le x} 1$
Let $f: \R_{\ge 1} \to \R$ be defined as:
:$\ds \map f x := \dfrac 1 x$
Then:
Category:Harmonic Numbers
Category:Analytic Number Theory
Integral Form of Gamma Function equivalent to Euler Form/Lemma

== Lemma for Integral Form of Gamma Function equivalent to Euler Form ==
Let $0 \le t \le m$.
Then:
:$0 \le e^{-t} - \paren {1 - \dfrac t m}^m \le t^2 \dfrac {e^{-t} } m$
== Proof ==
From Exponential of x not less than 1+x:
:$1 + x \le e^x$
Let $x = \pm \dfrac t m$.
Then:
and:
Then:
Hence:
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $20$
Category:Integral Form of Gamma Function equivalent to Euler Form
Integral Form of Gamma Function equivalent to Euler Form/Proof 1

== Theorem ==
== Proof ==
It is taken for granted that the Gamma function increases monotonically on $\R_{\ge 1}$.
We begin with an inequality that can easily be verified using cross multiplication.
Let $x$ be a real number between $0$ and $1$.
Let $n$ is a positive integer.
Then:
:$\ds \frac {\log \map \Gamma {n - 1} - \log \map \Gamma n} {\paren {n - 1} - n} \le \frac {\log \map \Gamma {x + n} - \log \map \Gamma n} {\paren {x + n} - n} \le \frac {\log \map \Gamma {n + 1} - \log \map \Gamma n} {\paren {n + 1} - n}$
Since n is a positive integer, we can make use of the identity:
:$\map \Gamma n = \paren {n - 1}!$
Simplifying, we get:
:$\map \log {n - 1} \le \dfrac {\log \map \Gamma {x + n} - \map \log {\paren {n - 1}!} } x \le \map \log n$
We now make use of the identity:
:$\ds \map \Gamma {x + n} = \prod_{k \mathop = 1}^n \paren {x + n - k} \map \Gamma x$
along with the fact that the Gamma Function is Log-Convex, to simplify the inequality:
:$\ds \paren {n - 1}^x \paren {n - 1}! \prod_{k \mathop = 1}^n \paren {x + n - k}^{-1} \le \map \Gamma x \le n^x \paren {n - 1}!\prod_{k \mathop = 1}^n \paren {x + n - k}^{-1}$
Taking the limit as $n$ goes to infinity and using the Squeeze Theorem:
:$\ds \map \Gamma x = \lim_{n \mathop \to \infty} n^x n! \prod_{k \mathop = 0}^n \paren {x + n - k}^{-1}$
which is another representation of Euler's form.
This proves equivalence for $x$ between $0$ and $1$.
The result follows from the Gamma Difference Equation.
Category:Integral Form of Gamma Function equivalent to Euler Form
Integral Form of Gamma Function equivalent to Euler Form/Proof 2

== Theorem ==
== Proof ==
First we present a lemma:
Recall the definition of the partial Gamma function:
:$\ds \map {\Gamma_m} x := \frac {m^x m!} {x \paren {x + 1} \paren {x + 2} \dotsm \paren {x + m} }$
We have that:
We have that for large $t$:
:$t^{x - 1} < t^{t / 2}$
and so as $m \to \infty$:
:$\ds \int_m^\infty e^{-t} t^{x - 1} \rd t \to 0$
Then:
Now we have that as $m \to \infty$:
:$\ds \frac 1 m \int_0^\infty t^{x + 1} e^{-t} \rd t \to 0$
so:
:$\ds \int_0^\infty e^{-t} t^{x - 1} \rd t - \map {\Gamma_m} x = 0$
leading to:
:$\ds \int_0^\infty e^{-t} t^{x - 1} \rd t = \lim_{m \mathop \to \infty} \dfrac {m^x m!} {x \paren {x + 1} \paren {x + 2} \dotsm \paren {x + m} }$
as was to be demonstrated.
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $20$
Category:Integral Form of Gamma Function equivalent to Euler Form
Integral Form of Polygamma Function

== Theorem == 
Let $z$ be a complex number with a positive real part.
Then: 
:$\ds \map {\psi_n} z = \paren {-1}^{n + 1} \int_0^\infty \frac {t^n e^{-z t} } {1 - e^{-t} } \rd t$
where $\map {\psi_n} z$ denotes the $n$th polygamma function.
=== Corollary ===
== Proof == 
From Gauss's Integral Form of Digamma Function, we have:
:$\ds \map \psi z = \int_0^\infty \paren {\frac {e^{-t} } t - \frac {e^{-z t} } {1 - e^{-t} } } \rd t$
Therefore:
== Also see ==
* Gauss's Integral Form of Digamma Function
Category:Integral Form of Polygamma Function
Category:Polygamma Function
Category:Complex Analysis
Category:Definite Integrals
Integral Form of Polygamma Function/Corollary

== Corollary to Integral Form of Polygamma Function ==
Let $z$ be a complex number with a positive real part.
Then: 
:$\ds \map {\psi_n} z= -\int_0^1 \frac {u^{z - 1} \paren {\ln u}^n } {1 - u} \rd u$
where $\map {\psi_n} z$ denotes the $n$th polygamma function.
== Proof ==
Category:Integral Form of Polygamma Function
Category:Primitives involving Logarithm Function
Integral Form of Ramanujan Phi Function

== Theorem == 
:$\ds \map \phi x = \int_{\to 0}^{\to 1} \paren {\dfrac {u^{x - 2} \paren {1 - u}^2} {1 - u^x} } \rd u + 1$
where:
:$\map \phi x$ denotes the Ramanujan phi function
:$x \in \C$
:$\map \Re x > 1$
== Proof == 
== Sources == 
* : Chapter $8$. Analogues of the Gamma Function
Category:Ramanujan Phi Function
Category:Definite Integrals
Integral Ideal iff Set of Integer Multiples

== Theorem ==
Let $J$ be a non-empty subset of the set of integers $\Z$.
Then:
:$J$ is an integral ideal
:
:$\exists m \in \Z: J = m \Z$.
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
Category:Integral Ideals
Category:Sets of Integer Multiples