hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Let the operation $\cdot$ be closed on $\struct {S, \cdot}$. Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be elements of $\map {\MM_S} {m, n}$. Let $\sqbrk c_{m n} = \sqbrk a_{m n} + \sqbrk b_{m n}$. Then: :$\forall i \in \closedint 1 m, j \in \closedint 1 n: c_{i j} = a_{i j} \circ b_{i j}$ Thus: :$\struct {S, \cdot}$ is closed {{iff}} $c_{i j} \in S$. From the definition of Hadamard product, $\sqbrk c_{m n}$ has the same order as both $\sqbrk a_{m n}$ and $\sqbrk b_{m n}$. Thus it follows that: :$\sqbrk c_{m n} \in \map {\MM_S} {m, n}$ Thus $\struct {\map {\MM_S} {m, n}, +}$, as it is defined, is closed. {{qed|lemma}} === Sufficient Condition === Suppose $\struct {S, \cdot}$ is such that $\cdot$ is not closed on $\struct {S, \cdot}$. Then there exists $a$ and $b$ such that: :$a \cdot b \notin S$ Let $\mathbf A$ and $\mathbf B$ be elements of $\map {\MM_S} {m, n}$ such that: :$a_{i j} = a$, $b_{i j} = b$ where: :$a_{i j}$ is the $\tuple {i, j}$th element of $\mathbf A$ :$b_{i j}$ is the $\tuple {i, j}$th element of $\mathbf B$ Then: :$a_{i j} \cdot b_{i j} \notin S$   That is: :$\mathbf A \circ \mathbf B \notin \map {\MM_S} {m, n}$ because (at least) one element of $\mathbf A \circ \mathbf B$ is not an element of $S$. That is, $\circ$ is not closed on $\map {\MM_S} {m, n}$. {{qed}}
=== Necessary Condition === Let the operation $\cdot$ be associative on $\struct {S, \cdot}$. Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{m n}$ and $\mathbf C = \sqbrk c_{m n}$ be elements of the $m \times n$ matrix space over $S$. Then: {{begin-eqn}} {{eqn | l = \paren {\mathbf A \circ \mathbf B} \circ \mathbf C       | r = \paren {\sqbrk a_{m n} \circ \sqbrk b_{m n} } \circ \sqbrk c_{m n}       | c = Definition of $\mathbf A$, $\mathbf B$ and $\mathbf C$ }} {{eqn | r = \sqbrk {a \cdot b}_{m n} \circ \sqbrk c_{m n}       | c = {{Defof|Hadamard Product}} }} {{eqn | r = \sqbrk {\paren {a \cdot b} \cdot c}_{m n}       | c = {{Defof|Hadamard Product}} }} {{eqn | r = \sqbrk {a \cdot \paren {b \cdot c} }_{m n}       | c = as $\cdot$ is associative }} {{eqn | r = \sqbrk a_{m n} \circ \sqbrk {b \cdot c}_{m n}       | c = {{Defof|Hadamard Product}} }} {{eqn | r = \sqbrk a_{m n} \circ \paren {\sqbrk b_{m n} \circ \sqbrk c_{m n} }       | c = {{Defof|Hadamard Product}} }} {{eqn | r = \mathbf A \circ \paren {\mathbf B \circ \mathbf C}       | c = Definition of $\mathbf A$, $\mathbf B$ and $\mathbf C$ }} {{end-eqn}} That is, $\circ$ is associative on $\map {\MM_S} {m, n}$. {{qed|lemma}} === Sufficient Condition === Suppose $\struct {S, \cdot}$ is such that $\cdot$ is not associative. Then there exists $a$, $b$ and $c$ in $S$ such that: :$a \cdot \paren {b \cdot c} \ne \paren {a \cdot b} \cdot c$ Let $\mathbf A$, $\mathbf B$ and $\mathbf C$ be elements of $\map {\MM_S} {m, n}$ such that: :$a_{i j} = a$, $b_{i j} = b$, $c_{i j} = c$ where: :$a_{i j}$ is the $\tuple {i, j}$th element of $\mathbf A$ :$b_{i j}$ is the $\tuple {i, j}$th element of $\mathbf B$ :$c_{i j}$ is the $\tuple {i, j}$th element of $\mathbf C$ Then: :$a_{i j} \cdot \paren {b_{i j} \cdot c_{i j} } \ne \paren {a_{i j} \cdot b_{i j} } \cdot c_{i j}$   That is: :$\paren {\mathbf A \circ \mathbf B} \circ \mathbf C \ne \mathbf A \circ \paren {\mathbf B \circ \mathbf C}$ because they differ (at least) at indices $\tuple {i, j}$. That is, $\circ$ is not associative on $\map {\MM_S} {m, n}$. {{qed}}
=== Necessary Condition === Let the operation $\cdot$ be commutative on $\struct {S, \cdot}$. Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be elements of the $m \times n$ matrix space over $S$. Then: {{begin-eqn}} {{eqn | l = \mathbf A \circ \mathbf B       | r = \sqbrk a_{m n} \circ \sqbrk b_{m n}       | c = Definition of $\mathbf A$ and $\mathbf B$ }} {{eqn | r = \sqbrk {a \cdot b}_{m n}       | c = {{Defof|Hadamard Product}} }} {{eqn | r = \sqbrk {b \cdot a}_{m n}       | c = as $\cdot$ is commutative }} {{eqn | r = \sqbrk b_{m n} \circ \sqbrk a_{m n}       | c = {{Defof|Hadamard Product}} }} {{eqn | r = \mathbf B \circ \mathbf A       | c = Definition of $\mathbf A$ and $\mathbf B$ }} {{end-eqn}} That is, $\circ$ is commutative on $\map {\MM_S} {m, n}$. {{qed|lemma}} === Sufficient Condition === Suppose $\struct {S, \cdot}$ is such that $\cdot$ is not commutative. Then there exists $a$ and $b$ such that: :$a \cdot b \ne b \cdot a$ Let $\mathbf A$ and $\mathbf B$ be elements of $\map {\MM_S} {m, n}$ such that: :$a_{i j} = a$, $b_{i j} = b$ where: :$a_{i j}$ is the $\tuple {i, j}$th element of $\mathbf A$ :$b_{i j}$ is the $\tuple {i, j}$th element of $\mathbf B$ Then: :$a_{i j} \cdot b_{i j} \ne b_{i j} \cdot a_{i j}$   That is: :$\mathbf A \circ \mathbf B \ne \mathbf B \circ \mathbf A$ because they differ (at least) at indices $\tuple {i, j}$. That is, $\circ$ is not commutative on $\map {\MM_S} {m, n}$. {{qed}}
Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be elements of $\map {\MM_R} {m, n}$. Let $\sqbrk c_{m n} = \sqbrk a_{m n} + \sqbrk b_{m n}$. By definition of matrix entrywise addition: :$\forall i \in \closedint 1 m, j \in \closedint 1 n: a_{i j} + b_{i j} = c_{i j}$ By {{Ring-axiom|A0}}, $R$ is closed under addition. Hence: :$\forall i \in \closedint 1 m, j \in \closedint 1 n: c_{i j} \in R$ From the definition of matrix entrywise addition, $\sqbrk c_{m n}$ has the same order as both $\sqbrk a_{m n}$ and $\sqbrk b_{m n}$. Thus it follows that: :$\sqbrk c_{m n} \in \map {\MM_R} {m, n}$ Thus $\struct {\map {\MM_R} {m, n}, +}$, as it is defined, is closed. {{qed}}
By definition, matrix entrywise addition is the '''Hadamard product''' of $\mathbf A$ and $\mathbf B$ with respect to ring addition. We have from {{Ring-axiom|A0}} that ring addition is closed. The result then follows directly from Closure of Hadamard Product. {{qed}}
=== Necessary Condition === Let $x, y$ be path-connected points in $T$. Let $\gamma: \closedint 0 1 \to T$ be a path from $x$ to $y$. From Image of Path is Path-Connected Set, $\Img \gamma$ is a path-connected set of $T$ containing $x$ and $y$. The result follows. {{qed|lemma}} === Sufficient Condition === Let $B$ be a path-connected set of $T$ containing $x$ and $y$. Then there exists a path $\gamma: \closedint 0 1 \to B$ from $x$ to $y$. Let $i_B: B \to S$ be the inclusion mapping from $B$ into $S$. From Composite of Continuous Mappings is Continuous, $i_B \circ g: \closedint 0 1 \to T$ is continuous. Hence $i_B \circ g$ is a path from $x$ to $y$ in $T$. {{qed}} Category:Path-Connected Spaces 7w3ul7y3km19eb2rbe6ky63nroytkoc
Let $\tilde C$ be any path-connected set such that: :$C \subseteq \tilde C$ Then $x \in \tilde C$. Hence $\tilde C \in \CC_x$. From Set is Subset of Union,  :$\tilde C \subseteq C$. Hence $\tilde C = C$. It follows that $C$ is a maximal path-connected set of $T$ by definition.
By definition: :$\tilde C \in \CC_x$ From Set is Subset of Union: :$\tilde C \subseteq C$ By maximality of $\tilde C$: :$\tilde C = C$
Let $A \in \powerset S$. Then by the definition of power set, $A \subseteq S$. By definition of relative complement: :$\relcomp S A = \set {x \in S: x \notin A}$ Hence $\relcomp S A$ is a subset of $S$. That is: :$\relcomp S A \in \powerset S$ and closure is proved. {{qed}}
From Alternating Group is Normal Subgroup of Symmetric Group, $A_n$ is seen to be normal in $S_n$. It remains to be shown that $A_n$ is the only such normal subgroup of $S_n$. {{ProofWanted}}
By definition, the path components of $T$ are a partition of $S$. Hence $S$ is the union of the open path components of $T$. Since a path component is a maximal path-connected set by definition, then $S$ is a union of open path-connected sets of $T$
Let $S = \bigcup \{ U \subseteq S : U \in \tau \text { and } U \text { is path-connected} \}$. Let $C$ be a path component of $T$. ==== Lemma ==== :For any path-connected set $U$ then: {{:Path Components are Open iff Union of Open Path-Connected Sets/Lemma 1}} Then: {{begin-eqn}} {{eqn | l = C       | r = C \cap S       | c = Intersection with Subset is Subset }} {{eqn | r = C \cap \bigcup \{ U \subseteq S : U \in \tau \text { and } U \text { is path-connected} \} }} {{eqn | r = \bigcup \{ C \cap U : U \in \tau \text { and } U \text { is path-connected} \}       | c = Intersection Distributes over Union }} {{eqn | r = \bigcup \{ C \cap U : U \in \tau, U \cap C \neq \O \text { and } U \text { is path-connected} \}       | c = Union with Empty Set }} {{eqn | r = \bigcup \{ U \subseteq C : U \in \tau, U \neq \O \text { and } U \text { is path-connected} \}       | c = Lemma }} {{end-eqn}}  Hence $C$ is the union of open sets. By definition of a topology then $C$ is an open set. The result follows.
=== Necessary Condition === Let $U \cap C \ne \O$. From Union of Path-Connected Sets with Common Point is Path-Connected, $U \cup C$ is a path-connected set of $T$. From Set is Subset of Union, $C \subseteq U \cup C$. By definition of a path component, $C$ is a maximal path-connected set. Hence $C = U \cup C$. From Union with Superset is Superset, $U \subseteq C$. Then: {{begin-eqn}} {{eqn | r = U \cap C       | l = U       | c = Intersection with Subset is Subset }} {{eqn | o = \ne       | r = \O       | c = Assumption }} {{end-eqn}} The result follows. {{qed|lemma}} === Sufficient Condition === Let $U \ne \O$ and $U \subseteq C$. Then: {{begin-eqn}} {{eqn | l = U \cap C       | r = U       | c = Intersection with Subset is Subset }} {{eqn | o = \ne       | r = \O       | c = Assumption }} {{end-eqn}} The result follows. {{qed}} Category:Path-Connected Spaces Category:Path Components obgdceufhjsnjjziia26r0j0q7btiew
Let the components of $T$ be open. By definition, the components of $T$ are a partition of $S$. Hence $S$ is the union of the open components of $T$. Since a component is a maximal connected set by definition, then $S$ is a union of open connected sets of $T$.
Let $S = \bigcup \set {U \subseteq S : U \in \tau \text { and } U \text { is connected} }$. Let $C$ be a component of $T$. ==== Lemma ==== :For any connected set $U$ then: {{:Components are Open iff Union of Open Connected Sets/Lemma 1}} Then: {{begin-eqn}} {{eqn | l = C       | r = C \cap S       | c = Intersection with Subset is Subset }} {{eqn | r = C \cap \bigcup \set { U \subseteq S : U \in \tau \text { and } U \text { is connected} } }} {{eqn | r = \bigcup \set {C \cap U : U \in \tau \text { and } U \text { is connected} }       | c = Intersection Distributes over Union }} {{eqn | r = \bigcup \set {C \cap U : U \in \tau, U \cap C \ne \empty \text { and } U \text { is connected} }       | c = Union with Empty Set }} {{eqn | r = \bigcup \set {U \subseteq C : U \in \tau \text { and } U \text { is connected} }       | c = Lemma }} {{end-eqn}}  Hence $C$ is the union of open sets. By definition of a topology then $C$ is an open set. The result follows.
=== Necessary Condition === Let $U \cap C \neq \O$. From Union of Connected Sets with Common Point is Connected, $U \cup C$ is a connected set of $T$. From Set is Subset of Union, $C \subseteq U \cup C$. By definition of a component, $C$ is a maximal connected set. Hence $C = U \cup C$. From Union with Superset is Superset, $U \subseteq C$. Then: {{begin-eqn}} {{eqn | r = U \cap C       | l = U       | c = Intersection with Subset is Subset }} {{eqn | o = \neq       | r = \O       | c = Assumption }} {{end-eqn}} The result follows. {{qed|lemma}} === Sufficient Condition === Let $U \neq \O$ and $U \subseteq C$. Then: {{begin-eqn}} {{eqn | l = U \cap C       | r = U       | c = Intersection with Subset is Subset }} {{eqn | o = \neq       | r = \O       | c = Assumption }} {{end-eqn}} The result follows. {{qed}} Category:Connected Spaces Category:Components bvnxmo4vajb3f0paz5zofkpa95c25id
We have:  {{begin-eqn}} {{eqn	| l = \map {\frac \d {\d x} } {\sin^2 p x} 	| r = \map {\frac \d {\d x} } {\sin p x} \map {\frac \d {\map \d {\sin p x} } } {\sin^2 p x} 	| c = Chain Rule for Derivatives }} {{eqn	| r = 2 p \cos p x \sin p x 	| c = Derivative of Sine of a x, Derivative of Power }} {{eqn	| r = p \sin 2 p x 	| c = Double Angle Formula for Sine }} {{end-eqn}} We also have, by Primitive of Power:  :$\displaystyle \int \frac {\d x} {x^2} = -\frac 1 x + C$ So:  {{begin-eqn}} {{eqn	| l = \int_0^\infty \paren {\frac {\sin p x} x}^2 \rd x 	| r = \intlimits {-\frac {\sin^2 p x} x} 0 \infty - \int_0^\infty \paren {-\frac {p \sin 2 p x} x} \rd x 	| c = Integration by Parts }} {{eqn	| r = - \lim_{x \mathop \to \infty} \paren {\frac {\sin^2 p x} x} + \lim_{x \mathop \to 0} \paren {\frac {\sin^2 p x} x} + p \int_0^\infty \frac {\sin 2 p x} x \rd x }} {{end-eqn}} We have, from Real Sine Function is Bounded, that:  :$0 \le \sin^2 p x \le 1$ for all real $x$.  Therefore:  :$0 \le \dfrac {\sin^2 p x} x \le \dfrac 1 x$ for all strictly positive real $x$. We have:  :$\displaystyle \lim_{x \mathop \to \infty} \frac 1 x = 0$ Therefore, by the Squeeze Theorem, we have:  :$\displaystyle \lim_{x \mathop \to \infty} \paren {\frac {\sin^2 p x} x} = 0$ As to the other limit, note that:  {{begin-eqn}} {{eqn	| l = \lim_{x \mathop \to 0} \frac {\sin p x} x 	| r = \lim_{p x \mathop \to 0} \frac {\sin px} {\frac {p x} p} }} {{eqn	| r = p \lim_{u \mathop \to 0} \frac {\sin u} u 	| c = letting $u = p x$ }} {{eqn	| r = p 	| c = Limit of Sine of X over X }} {{end-eqn}} So: {{begin-eqn}} {{eqn	| l = \lim_{x \mathop \to 0} \frac {\sin^2 p x} x 	| r = \paren {\lim_{x \mathop \to 0} \sin p x} \paren {\lim_{x \mathop \to 0} \frac {\sin p x} x} 	| c = Combination Theorem for Limits of Functions: Product Rule }} {{eqn	| r = p \map \sin {0 p} }} {{eqn	| r = 0 	| c = Sine of Zero is Zero }} {{end-eqn}} So:  :$\displaystyle \int_0^\infty \paren {\frac {\sin p x} x}^2 \rd x = p \int_0^\infty \frac {\sin 2 p x} x \rd x$ By Integral to Infinity of Sine p x over x, we have:  :$\displaystyle \int_0^\infty \frac {\sin 2 p x} x \rd x = \begin{cases} \dfrac \pi 2 & : p > 0 \\ 0 & : p = 0 \\ -\dfrac \pi 2 & : p < 0 \end{cases}$ So: :$\displaystyle p \int_0^\infty \frac {\sin 2 p x} x \rd x = \begin{cases} \dfrac \pi 2 p & : p > 0 \\ 0 & : p = 0 \\ -\dfrac \pi 2 p & : p < 0 \end{cases}$ Hence: :$\displaystyle p \int_0^\infty \frac {\sin 2 p x} x \rd x = \frac {\pi \size p} 2$ Hence the result. {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {1 - \cos p x} {x^2} \rd x 	| r = \int_0^\infty \frac {2 \sin^2 \paren {\frac {p x} 2} } {x^2} \rd x 	| c = Square of Sine }} {{eqn	| r = 2 \times \frac {\pi \size p} {2 \times 2} 	| c = Integral to Infinity of $\paren {\dfrac {\sin p x} x}^2$ }} {{eqn	| r = \frac {\pi \size p} 2 }} {{end-eqn}} {{qed}}
From Sum of 4 Unit Fractions that equals 1: {{:Sum of 4 Unit Fractions that equals 1}} This includes repeated unit fractions. The full list is: {{begin-eqn}} {{eqn | n = 1       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}       | o =  }} {{eqn | n = 2       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}       | o =  }} {{eqn | n = 3       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}       | o =  }} {{eqn | n = 4       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15}       | o =  }} {{eqn | n = 5       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}       | o =  }} {{eqn | n = 6       | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}       | o =  }} {{eqn | n = 7       | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}       | o =  }} {{eqn | n = 8       | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8       | o =  }} {{eqn | n = 9       | r = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}       | o =  }} {{eqn | n = 10       | r = \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6       | o =  }} {{eqn | n = 11       | r = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}       | o =  }} {{eqn | n = 12       | r = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}       | o =  }} {{eqn | n = 13       | r = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6       | o =  }} {{eqn | n = 14       | r = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4       | o =  }} {{end-eqn}} The result follows from inspection: solutions $(1)$, $(2)$, $(3)$, $(4)$, $(6)$ and $(7)$ are the only solutions of the above such that the denominators of the summands are distinct. {{qed}}
=== Definition 1 implies Definition 2 === {{:Equivalence of Definitions of Weakly Locally Connected at Point/Definition 1 implies Definition 2}}{{qed|lemma}} === Definition 2 implies Definition 1 === {{:Equivalence of Definitions of Weakly Locally Connected at Point/Definition 2 implies Definition 1}}{{qed}} Category:Locally Connected Spaces Category:Equivalence of Definitions of Weakly Locally Connected at Point 2kc8t20qoxazlh9fmry2k5s0p0pkj04
Let $U$ be an open neighborhood of $x$. By assumption there exists a connected neighborhood $C$ of $x$ such that $C \subseteq U$. By definition of a neighborhood there exists an open neighborhood $V$ of $x$ such that $V \subseteq C$. From Subset Relation is Transitive, $V \subseteq U$. By definition of a subset: :$\forall y, z \in V: y, z \in C$ The result follows.
Let $\mathcal B = \set {B \subseteq S: B \text{ is a connected neighborhood of } x}$ It will be shown that $\mathcal B$ is a neighborhood basis consisting of connected sets. Let $N$ be any neighborhood of $x$. By definition of a neighborhood there exists an open neighborhood $U$ of $x$ such that $U \subseteq N$ By assumption there exists an open neighborhood $V$ of $x$ such that every two points of $V$ lie in some connected subset of $U$. For each $y \in V, y \ne x$, let $C_y$ be a connected subset of $U$ such that $x, y \in C_y \subseteq U$. Let $C = \set x \cup \bigcup \set {C_y:y \in V, y \ne x}$ From Singleton is Connected in Topological Space, $\set x$ is connected. From Union of Connected Sets with Common Point is Connected, $C$ is a connected subset. From Union is Smallest Superset, $C \subseteq U \subseteq N$. Now $x \in C$ and for each $y \in V, y \ne x$ then $y \in C_y \subseteq C$. Hence $V \subseteq C$. By definition of a neighborhood, $C$ is a connected neighborhood of $x$. That is, $C \in \mathcal B$. Since $N$ was an arbitrary neighborhood of $x$, it follows that $\mathcal B$ is a neighborhood basis of $x$ consisting of connected sets.
Consider the set of sets $\mathbb A = \set {A_1, A_2, \dotsc, A_n}$ Consider the relational structure $S = \struct {\mathbb A, \subseteq}$. We have from Subset Relation is Ordering that $S$ is an ordered structure. The result follows from Ordering Cycle implies Equality. {{qed}}
We have, by Laplace Transform of Power:  :$\displaystyle \frac {\paren {-1}^n \map \Gamma s} {\paren {2 n + 1}^s} = \paren {-1}^n \int_0^\infty x^{s - 1} e^{-\paren {2 n + 1} x} \rd x$ for $\map \Re s > 0$. Summing, we have:   {{begin-eqn}} {{eqn	| l = \map \Gamma s \sum_{n \mathop = 0}^N \frac {\paren {-1}^n} {\paren {2 n + 1}^s} 	| r = \sum_{n \mathop = 0}^N \paren {-1}^n \int_0^\infty x^{s - 1} e^{-\paren {2 n + 1} x} \rd x }} {{eqn	| r = \int_0^\infty x^{s - 1} \sum_{n \mathop = 0}^N \paren {-1}^n e^{-\paren {2 n + 1} x} \rd x 	| c = Linear Combination of Definite Integrals }} {{eqn	| r = \int_0^\infty x^{s - 1} e^{-x} \paren {\sum_{n \mathop = 0}^N \paren {-e^{-2 x} }^n} \rd x }} {{end-eqn}} We have:  {{begin-eqn}} {{eqn	| l = \lim_{N \mathop \to \infty} \paren {\map \Gamma s \sum_{n \mathop = 0}^N \frac {\paren {-1}^n} {\paren {2 n + 1}^s} } 	| r = \map \Gamma s \lim_{N \mathop \to \infty} \paren {\sum_{n \mathop = 0}^N \frac {\paren {-1}^n} {\paren {2 n + 1}^s} } 	| c = Combination Theorem for Limits of Functions: Multiple Rule }} {{eqn	| r = \map \Gamma s \map \beta s 	| c = {{Defof|Dirichlet Beta Function}} }} {{end-eqn}} Therefore:  {{begin-eqn}} {{eqn	| l = \map \Gamma s \map \beta s 	| r = \lim_{N \mathop \to \infty} \paren {\int_0^\infty x^{s - 1} e^{-x} \paren {\sum_{n \mathop = 0}^N \paren {-e^{-2 x} }^n} \rd x} }} {{eqn	| r = \int_0^\infty x^{s - 1} e^{-x} \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-e^{-2 x} }^n} \rd x 	| c = Lebesgue's Dominated Convergence Theorem }} {{eqn	| r = \int_0^\infty x^{s - 1} e^{-x} \paren {\sum_{n \mathop = 0}^\infty \paren {-e^{-2 x} }^n} \rd x }} {{eqn	| r = \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 - \paren {-e^{- 2x} } } \rd x 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 + e^{-2 x} } \rd x }} {{end-eqn}} giving:  :$\displaystyle \map \beta s = \frac 1 {\map \Gamma s} \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 + e^{-2 x} } \rd x$ {{qed}} Category:Dirichlet Beta Function Category:Gamma Function kpr57owmp33juk5buanht2wus09exkg
{{begin-eqn}} {{eqn 	| l = \map \beta {2 n + 1} 	| r = \dfrac 1 {4^{2 n + 1} } \paren {\map \zeta {2 n + 1, \frac 1 4} - \map \zeta {2 n + 1, \frac 3 4} } 	| c = Dirichlet Beta Function in terms of Hurwitz Zeta Function }} {{eqn 	| r = \dfrac 1 {4^{2 n + 1} } \paren { \dfrac {\map {\psi_{2 n} } {\dfrac 1 4} - \map {\psi_{2 n} } {\dfrac 3 4} } { \paren {-1}^{2 n + 1} \map \Gamma {2 n + 1} } } 	| c = Polygamma Function in terms of Hurwitz Zeta Function }} {{eqn 	| r = -\dfrac 1 {4^{2 n + 1} \paren {2 n}!} \paren {\map {\psi_{2 n} } {\dfrac 1 4} - \map {\psi_{2 n} } {\dfrac 3 4} } 	| c = }} {{eqn 	| r = - \dfrac 1 {4^{2 n + 1} \paren {2 n}!} \paren {-\pi \valueat {\frac {\d^{2 n} } {\d z^{2 n} } \cot \pi z} {z \mathop = \frac 1 4} } 	| c = Polygamma Reflection Formula for $z = \dfrac 1 4$ }} {{eqn 	| r = \dfrac \pi {4^{2 n + 1} \paren {2 n}!} \valueat {\dfrac {\d^{2 n} } {\d z^{2 n} } \cot \pi z} {z \mathop = \frac 1 4} 	| c =  }} {{eqn 	| r = \dfrac \pi {4^{2 n + 1} \paren {2 n}!} \paren {-1}^n \paren {2 \pi}^{2 n} E_{2 n} 	| c = $\cot \pi z$ $2n$th derivative at $\dfrac 1 4$ }} {{eqn 	| r = \paren {-1}^n \dfrac { E_{2 n} \pi^{2 n + 1} } {4^{n + 1} \paren {2 n}!} 	| c = }} {{end-eqn}} {{qed}} Category:Dirichlet Beta Function Category:Euler Numbers Category:Polygamma Function 7hd4v2s6xwd670l99213bopu791r0eu
First it is to be shown that $j_t$ is left-total. This follows from the fact that $j_t$ is defined for all $s$: :$\map {j_t} s = \tuple {s, t}$ {{qed|lemma}} Next it is to be shown that $j_t$ is many-to-one, that is: :$\forall s_1, s_2 \in S: \map {j_t} {s_1} \ne \map {j_t} {s_2} \implies s_1 \ne s_2$ We have that: {{begin-eqn}} {{eqn | l = \map {j_t} {s_1}       | o = \ne       | r = \map {j_t} {s_2}       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {s_1, t}       | o = \ne       | r = \tuple {s_2, t}       | c = Definition of $j_t$ }} {{eqn | ll= \leadsto       | l = s_1       | o = \ne       | r = s_2       | c = {{Defof|Ordered Pair}} }} {{end-eqn}} Hence the result. {{qed}}
It has been shown in Correspondence between Set and Ordinate of Cartesian Product is Mapping that $j_t$ is a mapping. Now it is to be shown that $j_t$ is injective, that is: :$\forall s_1, s_2 \in S: \map {j_t} {s_1} = \map {j_t} {s_2} \implies s_1 = s_2$ We have that: {{begin-eqn}} {{eqn | l = \map {j_t} {s_1}       | o = =       | r = \map {j_t} {s_2}       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {s_1, t}       | o = =       | r = \tuple {s_2, t}       | c = Definition of $j_t$ }} {{eqn | ll= \leadsto       | l = s_1       | o = =       | r = s_2       | c = {{Defof|Ordered Pair}} }} {{end-eqn}} Hence the result. {{qed}}
By Derivative of Sine Integral Function, we  have:  :$\displaystyle \frac \d {\d x} \paren {\map \Si x} = \frac {\sin x} x$ So: {{begin-eqn}} {{eqn	| l = \int \map \Si x \rd x 	| r = \int 1 \times \map \Si x \rd x }} {{eqn	| r = x \map \Si x - \int x \frac {\sin x} x \rd x 	| c = Integration by Parts }} {{eqn	| r = x \map \Si x - \int \sin x \rd x }} {{eqn	| r = x \map \Si x + \cos x + C 	| c = Primitive of Sine Function }} {{end-eqn}} {{qed}} Category:Primitives Category:Sine Integral Function r25o0orymphqj2y701esrh7ugpyc8d5