nyuuzyou/wikis huggingface dataset:
Dirichlet's Integral Form of Digamma Function

== Theorem == 
Let $z$ be a complex number with a positive real part.
Then: 
:$\ds \map \psi z = \int_0^\infty \paren {\frac {e^{-t} } t - \frac 1 {t \paren {1 + t}^z } } \rd t$
where $\psi$ is the digamma function.
== Proof == 
We have:
Let:
We also have:
Then:
== Also see ==
* Gauss's Integral Form of Digamma Function
== Sources == 
* : $12.3$: Gauss's expression for the logarithmic derivate of the Gamma function as an infinite integral
Category:Digamma Function
Category:Complex Analysis
Category:Definite Integrals
Dirichlet's Principle for Harmonic Functions

== Theorem ==
Let the function $\map u x$ be the particular solution to Poisson's equation:
:$\Delta u + f = 0$
on a domain $\Omega$ of $\R^n$ with boundary condition:
:$u = g$ on $\partial \Omega$
Then $u$ can be obtained as the minimizer of the Dirichlet's energy:
:$\ds E \sqbrk {\map v x} = \int_\Omega \paren {\frac 1 2 \cmod {\nabla v}^2 - v f} \rd x$
amongst all twice differentiable functions $v$ such that $v = g$ on $\partial \Omega$ .
This result holds provided that there exists at least one function which makes the Dirichlet Integral finite.
== Proof ==
== Also known as ==
== Historical Note ==
Category:Dirichlet's Principle for Harmonic Functions
Category:Dirichlet's Principle
Category:Potential Theory
Dirichlet's Principle for Harmonic Functions/Also known as

== Dirichlet's Principle for Harmonic Functions: Also known as == 
Dirichlet's Principle for Harmonic Functions is also known just as Dirichlet's Principle.
However, there is more than one theorem named such, so its full name is recommended on .
Some sources give this as the Dirichlet Principle.
Category:Dirichlet's Principle for Harmonic Functions
Dirichlet's Principle for Harmonic Functions/Historical Note

== Historical Note on Dirichlet's Principle for Harmonic Functions ==
Dirichlet's Principle for Harmonic Functions was based on work done by during the course of his investigations into electromagnetism.
 used this principle to good effect, and it was he who named it for .
It was finally proved by in $1899$.
== Sources ==
* : Chapter $\text {A}.25$: Gauss ()
* : Chapter $\text {A}.28$: Dirichlet ()
Category:Historical Notes
Dirichlet's Principle for Harmonic Functions/Riemannian Manifold

== Theorem ==
Let $\struct {M, g}$ be a compact connected $n$-dimensional Riemannian manifold with nonempty boundary. 
Let $\map {C^\infty} M$ be the smooth function space.
Let $u \in \map {C^\infty} M$ be a smooth real function.
Let $\rd V_g$ be the Riemannian volume form.
Let $\grad$ be the gradient operator.
Let $\size {\, \cdot \,}$ be the Riemannian inner product norm.
Then $u$ is harmonic $u$ minimizes
:$\ds \int_M \size {\grad u}^2 \rd V_g$
among all smooth real function with the same boundary values.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Problems
Category:Dirichlet's Principle for Harmonic Functions
Category:Riemannian Geometry
Dirichlet's Test for Uniform Convergence

== Theorem ==
Let $D$ be a set.
Let $\struct {V, \norm {\,\cdot\,} }$ be a normed vector space.
Let $a_i, b_i$ be mappings from $D \to M$.
Let the following conditions be satisfied:
:$(1): \quad$ The sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty \map {a_n} x$ be bounded on $D$
:$(2): \quad \sequence {\map {b_n} x}$ be monotonic for each $x \in D$
:$(3): \quad \map {b_n} x \to 0$ converge uniformly on $D$.
Then:
:$\ds \sum_{n \mathop = 1}^\infty \map {a_n} x \map {b_n} x$ converges uniformly on $D$.
== Proof ==
Suppose $\map {b_n} x \ge \map {b_{n + 1} } x$ for each $x \in D$.
All we need to show is that:
:$\ds \sum_{n \mathop = 1}^\infty \size {\map {b_n} x - \map {b_{n + 1} } x}$
converges uniformly on $D$.
To do this we show that the Cauchy criterion holds.
Assign $\epsilon < 0$.
Then by definition of uniform convergence:
:$\exists N \in \N: \forall x \in D: \forall n \ge N: \size {\map {b_n} x} < \dfrac \epsilon 2$
Let $x \in D$ and $n > m \ge N$.
Then:
== Also known as ==
== Sources ==
* : Chapter $\text {A}.28$: Dirichlet ()
* 
* 
* 
Category:Dirichlet's Test for Uniform Convergence
Category:Direct Proofs
Category:Analysis
Category:Convergence
Dirichlet's Theorem on Arithmetic Sequences

== Theorem ==
Let $a, q$ be coprime integers.
Let $\PP_{a, q}$ be the set of primes $p$ such that $p \equiv a \pmod q$.
Then $\PP_{a, q}$ has Dirichlet density:
:$\map \phi q^{-1}$
where $\phi$ is Euler's phi function.
In particular, $\PP_{a, q}$ is infinite.
== Proof ==
=== Lemma 1 ===
Define:
:$\eta_{a, q} : n \mapsto \begin {cases} 1 & : n \equiv a \pmod q \\ 0 & : \text {otherwise} \end {cases}$
=== Lemma 2 ===
We have:
By Lemma 1, the first term grows like $\dfrac 1 {\map \phi q} \ln \dfrac 1 {s - 1}$ as $s \to 1$, while all other terms are bounded.
That is:
:$\ds \sum_{p \mathop \in \PP_{a, q} } \frac 1 {p^s} \sim \frac 1 {\map \phi q} \, \map \ln {\dfrac 1 {s - 1} }$
as $s \to 1$. 
== Also known as ==
== Historical Note ==
== Sources ==
* : $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Exercise $9$ (mentioned, but not proved)
* : Chapter $\text {A}.28$: Dirichlet ()
* : Chapter $\text {B}.16$: The Sequence of Primes: Theorem $4$
* 
* 
Category:Dirichlet's Theorem on Arithmetic Sequences
Category:Arithmetic Sequences
Category:Prime Numbers
Category:Analytic Number Theory
Dirichlet's Theorem on Arithmetic Sequences/Also known as

== Dirichlet's Theorem on Arithmetic Sequences: Also known as == 
Dirichlet's Theorem on Arithmetic Sequences is also known just as Dirichlet's Theorem.
However, there is more than one theorem named such, so it is preferable to use the full form.
== Sources ==
* 
* 
Category:Dirichlet's Theorem on Arithmetic Sequences
Dirichlet's Theorem on Arithmetic Sequences/Lemma 1

== Lemma for Dirichlet's Theorem on Arithmetic Sequences ==
Let $a, q$ be coprime integers.
Let $\PP_{a, q}$ be the set of primes $p$ such that $p \equiv a \pmod q$.
Let $\chi$ be a Dirichlet character modulo $q$.
Let:
:$\ds \map f s = \sum_p \map \chi p p^{-s}$
If $\chi$ is non-trivial then $\map f s$ is bounded as $s \to 1$.
If $\chi$ is the trivial character then:
:$\map f s \sim \map \ln {\dfrac 1 {s - 1} }$
as $s \to 1$.
== Proof ==
By Logarithm of Dirichlet L-Functions:
:$(1): \quad \ds \sum_p \map \chi p p^{-s} = \ln \map L {s, \chi} - \sum_p \sum_{n \mathop \ge 2} \frac {\map \chi p^n} {n p^{n s} }$
If $\chi$ is non-trivial, then by L-Function does not Vanish at One, $\ln \map L {s, \chi}$ is bounded as $s \to 1$.
If $\chi$ is trivial, then by Analytic Continuation of Dirichlet L-Function, $\map L {s, \chi}$ has a simple pole at $s = 1$.
Therefore, in this case:
:$\map L {s, \chi} \sim \dfrac \lambda {s - 1}$
where $\lambda$ is the residue of $\map L {s, \chi}$ at $1$, and:
:$\ln \map L {s, \chi} \sim \map \ln {\dfrac \lambda {s - 1} } \sim \map \ln {\dfrac 1 {s - 1} }$
Thus if we can show that the second term of $(1)$ is bounded, the result holds.
On $\map \Re s > 1$:
This last is $\map \zeta 2$ where $\zeta$ is the Riemann zeta function, so is finite.
Category:Dirichlet's Theorem on Arithmetic Sequences
Dirichlet's Theorem on Arithmetic Sequences/Lemma 2

== Lemma for Dirichlet's Theorem on Arithmetic Sequences ==
Let $a, q$ be coprime integers.
Let $\PP_{a, q}$ be the set of primes $p$ such that $p \equiv a \pmod q$.
Define:
:$\eta_{a, q} : n \mapsto \begin{cases}
1 & : n \equiv a \pmod q\\
0 & : \text{otherwise}
\end{cases}$
Let $G = \paren {\Z / q \Z}^\times$.
Let $G^*$ be the dual group of Dirichlet characters on $G$.
Then for all $n \in \N$:
:$\ds \map {\eta_{a, q} } n = \sum_{\chi \mathop \in G^*} \frac {\map {\overline \chi} a} {\map \phi q} \, \map \chi n$
== Proof ==
There is only one $x \in G$ such that $\map \eta x \ne 0$, and this equals $\map \eta a = 1$.
So:
:$\ds \sum_{x \mathop \in G} \map {\eta_{a, q} } x \, \map {\overline \chi} x = \map {\overline \chi} a$
Therefore, by Discrete Fourier Transform on Abelian Group we have for all $x \in G$:
:$\ds \map \eta x = \frac 1 {\map \phi q} \sum_{\chi \mathop \in G^*} \map {\overline \chi} a \, \map \chi x$
as required.
Category:Dirichlet's Theorem on Arithmetic Sequences
Dirichlet Beta Function at Odd Positive Integers

== Theorem ==
where:
:$\beta$ denotes the Dirichlet beta function
:$E_n$ denotes the $n$th Euler number
:$n$ is a non-negative integer.
=== Corollary ===
== Proof ==
== Also presented as ==
== Examples ==
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.38$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.38.$
Category:Dirichlet Beta Function at Odd Positive Integers
Category:Dirichlet Beta Function
Category:Euler Numbers
Category:Sums of Sequences
Dirichlet Beta Function at Odd Positive Integers/Also presented as

== Dirichlet Beta Function at Odd Positive Integers: Also presented as == 
This can also be expressed using the alternative form of the Euler numbers in the following form:
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.38$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.38.$
Category:Dirichlet Beta Function at Odd Positive Integers
Dirichlet Beta Function at Odd Positive Integers/Corollary

== Corollary to Dirichlet Beta Function at Odd Positive Integers ==
Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
where $E_n$ is the $n$th Euler number.
== Proof ==
== Also presented as ==
== Sources ==
* : $\S 21$: Series involving Bernoulli and Euler Numbers: $21.11$
* : $\S 23$: Bernoulli and Euler Numbers: Series involving Bernoulli and Euler Numbers: $23.11.$
Category:Dirichlet Beta Function at Odd Positive Integers
Dirichlet Beta Function at Odd Positive Integers/Corollary/Also presented as

== : Also presented as == 
 can also be presented in the form:
where ${E_n}^*$ denotes the alternative form of the Euler numbers.
== Sources ==
* : $\S 21$: Series involving Bernoulli and Euler Numbers: $21.11$
* : $\S 23$: Bernoulli and Euler Numbers: Series involving Bernoulli and Euler Numbers: $23.11.$
Category:Dirichlet Beta Function at Odd Positive Integers
Dirichlet Beta Function at Odd Positive Integers/Examples/Dirichlet Beta Function of 1

== Theorem ==
Let $\beta$ denote the Dirichlet beta function.
Then:
:$\map \beta 1 = \dfrac \pi 4 $
== Proof 1 ==
== Proof 2 ==
Category:Dirichlet Beta Function at Odd Positive Integers
Category:Examples of Dirichlet Beta Function Values
Dirichlet Beta Function at Odd Positive Integers/Examples/Dirichlet Beta Function of 3

== Theorem ==
Let $\beta$ denote the Dirichlet beta function.
Then:
:$\map \beta 3 = \dfrac {\pi^3} {32} $
== Proof ==
Category:Dirichlet Beta Function at Odd Positive Integers
Category:Examples of Dirichlet Beta Function Values
Dirichlet Beta Function at Odd Positive Integers/Examples/Dirichlet Beta Function of 5

== Theorem ==
Let $\beta$ denote the Dirichlet beta function.
Then:
:$\map \beta 5 = \dfrac {5 \pi^5} {1536} $
== Proof ==
Category:Dirichlet Beta Function at Odd Positive Integers
Category:Examples of Dirichlet Beta Function Values
Dirichlet Beta Function in terms of Hurwitz Zeta Function

== Theorem ==
:$\map \beta s = \dfrac 1 {4^s} \paren {\map \zeta {s, \dfrac 1 4} - \map \zeta {s, \dfrac 3 4} }$
where:
:$\map \beta s$ is the Dirichlet beta function
:$\map \zeta {s, x}$ is the Hurwitz zeta function
:$s$ is a complex number with $\map \Re s > 1$
== Proof ==
Our reliance upon the Hurwitz zeta function requires that $s \in \C$ and $\map \Re s > 1$.
Category:Dirichlet Beta Function
Category:Hurwitz Zeta Function
Dirichlet Conditions/Examples/Reciprocal of 4 minus x squared

== Example of Dirichlet Conditions ==
The function:
:$\map f x = \dfrac 1 {4 - x^2}$
does not satisfy the Dirichlet conditions on the real interval $\openint 0 {2 \pi}$.
== Proof ==
Recall the Dirichlet conditions:
:
At the point $x = 2$, $\dfrac 1 {4 - x^2}$ is not defined.
Let $\epsilon \in \R_{>0}$.
Then:
:$\map f {2 - \dfrac \epsilon 2} > \map f {2 - \epsilon}$
and so $x = 2$ is not a finite discontinuity.
Hence $f$ does not satisfy Dirichlet condition $(\text D 3)$.
== Sources ==
* : Chapter One: $\S 2$. Fourier Series
Category:Dirichlet Conditions
Dirichlet Conditions/Examples/Sine of Reciprocal of x - 1

== Example of Dirichlet Conditions ==
The function:
:$\map f x = \map \sin {\dfrac 1 {x - 1} }$
does not satisfy the Dirichlet conditions on the real interval $\openint 0 {2 \pi}$.
== Proof ==
Recall the Dirichlet conditions:
:
Around the point $x = 1$, $\map f x$ has an infinite number of local maxima and local minima.
Hence it does not satisfy Dirichlet condition $(\text D 2)$.
== Sources ==
* : Chapter One: $\S 2$. Fourier Series
Category:Dirichlet Conditions
Dirichlet Convolution Preserves Multiplicativity

== Theorem ==
Let $f, g: \N \to \C$ be multiplicative arithmetic functions.
Then their Dirichlet convolution $f * g$ is again multiplicative.
=== General Result ===
== Proof ==
Let $m, n$ be coprime integers.
By definition of multiplicative functions, we have:
:$(1): \quad \map f {m n} = \map f m \map f n$
:$(2): \quad \map g {m n} = \map g m \map g n$
== Also see ==
* Properties of Dirichlet Convolution
Category:Dirichlet Convolution
Category:Multiplicative Functions
Dirichlet Convolution Preserves Multiplicativity/General Result

== Theorem ==
Let $f, g$ be multiplicative functions.
Let $S \subset \N$ be a set of natural numbers with the property:
:$m n \in S, \map \gcd {m, n} = 1 \implies m, n \in S$
Define:
:$\map {\paren {f*_S g} } n = \ds \sum_{\substack {d \mathop \divides n \\ d \mathop \in S} } \map f d \map g {n / d}$
Then $f*_S g$ is multiplicative.
== Proof ==
Category:Dirichlet Convolution
Category:Multiplicative Functions
Dirichlet Convolution is Associative

== Theorem ==
Let $f, g, h$ be arithmetic functions.
Let $*$ denote Dirichlet convolution.
Then:
: $\paren {f * g} * h = f * \paren {g * h}$
== Proof ==
We have:
and
and associativity follows.
== Also see ==
* Properties of Dirichlet Convolution
Category:Dirichlet Convolution
Dirichlet Convolution is Commutative

== Theorem ==
Let $f, g$ be arithmetic functions.
Let $*$ denote Dirichlet convolution.
Then:
:$f * g = g * f$
== Proof ==
From the definition of the Dirichlet convolution:
:$\ds \map {\paren {f * g} } n = \sum_{a b \mathop = n} \map f a \map g b$
By definition, arithmetic functions are mappings from the natural numbers $\N$ to the complex numbers $\C$.
Thus $\map f a, \map g b \in \C$ and commutativity follows from Complex Multiplication is Commutative.
== Also see ==
* Properties of Dirichlet Convolution
Category:Dirichlet Convolution
Dirichlet Eigenvalue is Strictly Positive

== Theorem ==
Let $\struct {M, g}$ be a compact connected Riemannian manifold with non-empty boundary $\partial M$.
Let $\lambda$ be a Dirichlet eigenvalue of $M$.
Then $\lambda$ is strictly positive.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Problems
Category:Riemannian Geometry
Dirichlet Eta Function at Non-Positive Integers

== Theorem == 
 
Let $n \ge 0$ be a integer. 
Then: 
:$\map \eta {-n} = \dfrac {\paren {-1}^{n + 1} \paren {2^{n + 1 } - 1} B_{n + 1} } {n + 1}$ 
where:
:$B_n$ is the $n$th Bernoulli number
:$\eta$ is the Dirichlet eta function
== Proof == 
== Also see ==
* Riemann Zeta Function at Non-Positive Integers
Category:Dirichlet Eta Function
Dirichlet Function has no Period

== Theorem ==
The Dirichlet functions are periodic by Dirichlet Function is Periodic.
However, they do not admit a period.
That is, there does not exist a smallest value $L \in \R_{> 0}$ such that:
:$\forall x \in \R: \map D x = \map D {x + L}$
== Proof ==
Let $D: \R \to \R$ be a Dirichlet function.
In proving that the Dirichlet Function is Periodic, it was shown that every non-zero rational number is a periodic element of $D$.
Therefore, the period of $D$ must be the smallest element of $\Q_{> 0}$.
But from Rational Numbers are not Well-Ordered under Conventional Ordering there cannot exist such an element.
Hence the result.
Category:Periodic Functions
Category:Dirichlet Functions
Dirichlet Function is Discontinuous

== Theorem ==
Let $D: \R \to \R$ denote the Dirichlet function:
:$\forall x \in \R: \map D x = \begin {cases} c & : x \in \Q \\ d & : x \notin \Q \end {cases}$
where $\Q$ denotes the set of rational numbers.
Then $D$ is discontinuous at every $x \in \R$.
== Proof ==
Let $\epsilon = \dfrac {\size {c - d} } 2$.
Let $x \in \Q$.
Let $\delta \in \R_{>0}$ be arbitrary.
Let $y \in \Q$ such that $\size {x - y} < \delta$.
, let $y > x$.
From Between two Rational Numbers exists Irrational Number:
:$\exists z \in \R \setminus \Q: x < z < y$
and so:
:$\size {\map D x - \map D z} = \size {c - d} > \epsilon$
Similarly if $y < x$:
:$\exists z \in \R \setminus \Q: y < z < x: \size {\map D x - \map D z} > \epsilon$
and by definition of continuity $D$ is discontinuous at $x$.
Let $x \in \R \setminus \Q$.
Let $\delta \in \R_{>0}$ be arbitrary.
Let $y \in \R \setminus \Q$ such that $\size {x - y} < \delta$.
, let $y > x$.
From Between two Real Numbers exists Rational Number:
:$\exists z \in \Q: x < z < y$
and so:
:$\size {\map D x - \map D z} = \size {c - d} > \epsilon$
Similarly if $y < x$:
:$\exists z \in \Q: y < z < x: \size {\map D x - \map D z} > \epsilon$
and by definition of continuity $D$ is discontinuous at $x$.
$D$ has been shown to be discontinuous at all $x \in \R$ whether $x$ is rational or irrational.
Hence the result.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 20$
Category:Dirichlet Functions
Dirichlet Function is Periodic

== Theorem ==
Let $D: \R \to \R$ be a Dirichlet function:
:$\forall x \in \R: \map D x = \begin{cases} c & : x \in \Q \\ d & : x \notin \Q \end{cases}$
Then $D$ is periodic.
Namely, every non-zero rational number is a periodic element of $D$.
== Proof ==
Let $x \in \R$.
Let $L \in \Q$.
If $x \in \Q$, then:
If $x \notin \Q$, then:
Combining the above two shows that:
:$\forall x \in \R: \map D x = \map D {x + L}$
Hence the result.
Category:Periodic Functions
Category:Dirichlet Functions
Dirichlet Integral

== Theorem ==
:$\ds \int_0^\infty \frac {\sin x} x \rd x = \frac \pi 2$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
Category:Integral Calculus
Category:Dirichlet Integral
Category:Definite Integrals involving Sine Function
Dirichlet Integral/Proof 1

== Theorem ==
== Proof ==
By Fubini's Theorem:
:$\ds \int_0^\infty \paren {\int_0^\infty e^{- x y} \sin x \rd y} \rd x = \int_0^\infty \paren {\int_0^\infty e^{- x y} \sin x \rd x} \rd y$
Then:
and:
Hence:
Category:Dirichlet Integral
Dirichlet Integral/Proof 3

== Theorem == 
== Proof == 
Let:
:$\map f x = \begin {cases} \dfrac {e^{i x} - 1} x & x \ne 0 \\ i & x = 0 \end {cases}$
We have, by Euler's Formula, for $x \in \R$:
:$\map \Im {\map f x} = \begin {cases} \dfrac {\sin x} x & x \ne 0 \\ 1 & x = 0 \end {cases}$
So: 
:$\ds \map \Im {\int_0^\infty \dfrac {e^{i x} - 1} x \rd x} = \int_0^\infty \dfrac {\sin x} x \rd x$
Let $C_R$ be the arc of the circle of radius $R$ centred at the origin connecting $R$ and $-R$ anticlockwise.
Let $\Gamma_R = C_R \cup \closedint {-R} R$.
Then, by Contour Integral of Concatenation of Contours:
:$\ds \oint_{\Gamma_R} \frac {e^{i x} - 1} x \rd x = \int_{C_R} \frac {e^{i x} - 1} x \rd x + \int_{-R}^R \frac {e^{i x} - 1} x \rd x$
From Linear Combination of Contour Integrals, we write: 
:$\ds \oint_{\Gamma_R} \frac {e^{i x} - 1} x \rd x = \int_{C_R} \frac {e^{i x} } x \rd x - \int_{C_R} \frac {\rd x} x + \int_{-R}^R \frac {e^{i x} - 1} x \rd x$
Note that $f$ is holomorphic inside our contour. 
It then follows from the Cauchy-Goursat Theorem, that:
:$\ds \oint_{\Gamma_R} \frac {e^{i x} - 1} x \rd x = 0$
We also have: 
Therefore: 
:$\ds \lim_{R \mathop \to \infty} \int_{C_R} \frac {\rd x} x = \lim_{R \mathop \to \infty} \int_{-R}^R \frac {e^{i x} - 1} x \rd x = \int_{-\infty}^\infty \frac {e^{i x} - 1} x \rd x$
Evaluating the integral on the : 
So:
:$\ds \int_{-\infty}^\infty \frac {e^{i x} - 1} x \rd x = \pi i$
Taking the imaginary part:
:$\ds \int_{-\infty}^\infty \frac {\sin x} x \rd x = \pi$
From Definite Integral of Even Function:
:$\ds \int_{-\infty}^\infty \frac {\sin x} x \rd x = 2 \int_0^\infty \frac {\sin x} x \rd x$
Hence: 
:$\ds \int_0^\infty \frac {\sin x} x \rd x = \frac \pi 2$
Category:Dirichlet Integral
Dirichlet Integral/Proof 4

== Theorem == 
== Proof == 
From Integral to Infinity of Function over Argument:
:$\ds \int_0^\infty {\dfrac {\map f x} x} = \int_0^{\to \infty} \map F u \rd u$
for a real function $f$ and its Laplace transform $\laptrans f = F$, provided they exist.
Let $\map f x := \sin x$.
Then from Laplace Transform of Sine:
:$\laptrans {\map f x} = \dfrac 1 {s^2 + 1}$
Hence:
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Division by $t$: $22 \ \text{(b)}$
Category:Dirichlet Integral
Dirichlet Integral/Proof 5

== Theorem == 
== Proof == 
Let $M \in \R_{>0}$.
Define a real function $I_M : \R \to \R$ by:
:$\ds \map {I_M} \alpha := \int_0^M \dfrac {\sin x} x e^{-\alpha x} \rd x$
Then, for $\alpha > 0$:
On the other hand:
Thus:
Thus:
Therefore:
As:
:$\ds \map {I_M} 0 = \int_0^M \dfrac {\sin x} x \rd x$
we have shown:
:$\ds \forall M \in \R_{>0} : \size {\int_0^M \dfrac {\sin x} x \rd x - \dfrac \pi 2} \le \dfrac 2 M$
In particular:
Category:Dirichlet Integral
Dirichlet L-Function from Trivial Character

== Theorem ==
Let $\chi_0$ be the trivial Dirichlet character modulo $q$.
Let $\map L {s, \chi_0}$ be the Dirichlet $L$-function associated to $\chi_0$.
Let $\map \zeta s$ be the Riemann zeta function.
Then:
:$\ds \map L {s, \chi_0} = \map \zeta s \cdot \prod_{p \mathop \divides q} \paren {1 - p^{-s} }$
where $\divides$ denotes divisibility.
== Proof ==
By definition:
:$\map {\chi_0} a = \begin{cases} 1 & : \gcd \set {a, q} = 1 \\ 0 & : \text{otherwise} \end{cases}$
Therefore:
Hence the result.
Category:Analytic Number Theory
Dirichlet Series Absolute Convergence Lemma

== Theorem ==
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a Dirichlet series.
Suppose that $f$ converges absolutely at $s_0 = \sigma_0 + i t_0 \in \C$.
Then $f$ converges absolutely at all points $s = \sigma + i t \in \C$ with $\sigma \ge \sigma_0$.
== Proof ==
Suppose that $f$ converges absolutely at $\sigma_0 + i t_0$.
If $\sigma \ge \sigma_0$, then:
Therefore absolute convergence of $\map f {s_0}$ directly implies absolute convergence of $\map f s$ for all $s = \sigma + i t$ with $\sigma > \sigma_0$.
== Sources ==
* : $\S 11.2$
Category:Dirichlet Series
Dirichlet Series Convergence Lemma

== Theorem ==
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a Dirichlet series.
Let $\map f s$ converge at $s_0 = \sigma_0 + i t_0$.
Then $\map f s$ converge for all $s = \sigma + i t$ where $\sigma > \sigma_0$.
=== General Dirichlet Series ===
== Proof ==
We begin with a lemma:
=== Lemma ===
Suppose that $\map f s$ converges at $s_0 = \sigma_0 + it_0$.
Then by Convergent Sequence in Metric Space is Bounded, $\size {\ds \sum_{k \mathop = 1}^n \frac {a_k} {k^{s_0} } }$ is bounded.
Thus the results of the lemma hold.
Choose any $s = \sigma + it$ with $\sigma > \sigma_0$.
By Cauchy's Convergence Criterion, it suffices to show that for all $\epsilon>0$, there exists an $N>0$ such that for all $m>n>N$
:$\ds \size {\sum_{k \mathop = 1}^m \frac {a_k} {k^s} - \sum_{k \mathop = 1}^n \frac {a_k} {k^s} } = \size {\sum_{k \mathop = n + 1}^m \frac {a_k} {k^s} } < \epsilon$
The lemma shows that for a given $s$ there exists a constant $C$ independent of $N$ such that:
:$\ds \size {\sum_{k \mathop = n + 1}^n a_n n^{-s} } \le \map C {n + 1}^{\sigma_0 - \sigma}$
Since $\sigma_0 - \sigma <0$, the tends to zero as $n \to \infty$.
Thus we may choose $N$ large enough so that for $n > N$.
:$\ds \paren {n + 1}^{\sigma_0 - \sigma} < \dfrac \epsilon C$
which gives us:
:$\ds \size {\sum_{k \mathop = n + 1}^n a_n n^{-s} } < \epsilon$
== Sources ==
* : $\S 11.6$: Lemma $2$, Theorem $11.8$
Category:Dirichlet Series