hcju/proofwikips huggingface dataset:
Pythagorean Triangle/Examples/3059-8580-9109
== Example of Primitive Pythagorean Triangle ==  The triangle whose sides are of length $3059$, $8580$ and $9109$ is a primitive Pythagorean triangle.  :700px It has generator $\tuple {78, 55}$. == Proof == We have: {{begin-eqn}} {{eqn | l = 78^2 - 55^2       | r = 6084 - 3025       | c =  }} {{eqn | r = 3059       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 2 \times 78 \times 55       | r = 8580       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 78^2 + 55^2       | r = 6084 + 3025       | c =  }} {{eqn | r = 9109       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 3059^2 + 8580^2       | r = 9 \, 357 \, 481 + 73 \, 616 \, 400       | c =  }} {{eqn | r = 82 \, 973 \, 881       | c =  }} {{eqn | r = 9109^2       | c =  }} {{end-eqn}} It follows by Pythagoras's Theorem that $3059$, $8580$ and $9109$  form a Pythagorean triple. We have that: {{begin-eqn}} {{eqn | l = 3059       | r = 7 \times 19 \times 23       | c =  }} {{eqn | l = 8580       | r = 2^2 \times 3 \times 5 \times 11 \times 13       | c =  }} {{end-eqn}} It is seen that $3059$ and $8580$ share no prime factors. That is, $3059$ and $8580$ are coprime. Hence, by definition, $3059$, $8580$ and $9109$ form a primitive Pythagorean triple. The result follows by definition of a primitive Pythagorean triangle.
Pythagorean Triangle/Examples/1380-19,019-19,069
== Example of Primitive Pythagorean Triangle ==  The triangle whose sides are of length $1380$, $19 \, 019$ and $19 \, 069$ is a primitive Pythagorean triangle.  :700px It has generator $\left({138, 5}\right)$. == Proof == We have: {{begin-eqn}} {{eqn | l = 138^2 - 5^2       | r = 19 \, 044 - 25       | c =  }} {{eqn | r = 19 \, 019       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 2 \times 138 \times 5       | r = 1380       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 138^2 + 5^2       | r = 19 \, 044 + 25       | c =  }} {{eqn | r = 19 \, 069       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 1380^2 + 19 \, 019^2       | r = 1 \, 904 \, 400 + 361 \, 722 \, 361       | c =  }} {{eqn | r = 363 \, 626 \, 761       | c =  }} {{eqn | r = 19 \, 069^2       | c =  }} {{end-eqn}} It follows by Pythagoras's Theorem that $1380$, $19 \, 019$ and $19 \, 069$  form a Pythagorean triple. We have that: {{begin-eqn}} {{eqn | l = 1380       | r = 2^2 \times 3 \times 5 \times 23       | c =  }} {{eqn | l = 19 \, 019       | r = 7 \times 11 \times 13 \times 19       | c =  }} {{end-eqn}} It is seen that $1380$ and $19 \, 019$ share no prime factors. That is, $1380$ and $19 \, 019$ are coprime. Hence, by definition, $1380$, $19 \, 019$ and $19 \, 069$ form a primitive Pythagorean triple. The result follows by definition of a primitive Pythagorean triangle.
Carmichael Number/Examples/294,409
== Example of Carmichael Number == $294 \, 409$ is a Carmichael number:  :$\forall a \in \Z: a \perp 294 \, 409: a^{294 \, 409} \equiv a \pmod {294 \, 409}$ while $294 \, 409$ is composite.  == Proof == We have that: :$294 \, 409 = 37 \times 73 \times 109$ First note that $294 \, 409$ is square-free. Hence the square of none of its prime factors is a divisor of $294 \, 409$: :$\forall p \divides 294 \, 409: p^2 \nmid 294 \, 409$ We also see that: {{begin-eqn}} {{eqn | l = 294 \, 408       | r = 2^3 \times 3^3 \times 29 \times 47       | c =  }} {{eqn | r = 8178 \times 36       | c =  }} {{eqn | r = 4089 \times 72       | c =  }} {{eqn | r = 2726 \times 108       | c =  }} {{end-eqn}} Thus $294 \, 409$ is a Carmichael number by Korselt's Theorem.  == Sources == * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Carmichael Number/Examples/1729|next = Smallest Integer which is Sum of 2 Fourth Powers in 2 Ways}}: $509,033,161$
Euler Phi Function of 6
== Example of Euler $\phi$ Function of Square-Free Integer ==  :$\map \phi 6 = 2$  where $\phi$ denotes the Euler $\phi$ Function. == Proof == From Euler Phi Function of Square-Free Integer: :$\displaystyle \map \phi n = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$ where $p \divides n$ denotes the primes which divide $n$. We have that: :$6 = 2 \times 3$ and so is square-free. Thus: {{begin-eqn}} {{eqn | l = \map \phi 6       | r = \paren {3 - 1}       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}}
Odd Amicable Pair/Examples/29,912,035,725-34,883,817,075
== Example of Odd Amicable Pair ==  $29 \, 912 \, 035 \, 725$ and $34 \, 883 \, 817 \, 075$ are an odd amicable pair: :$\map \sigma {29 \, 912 \, 035 \, 725} = \map \sigma {34 \, 883 \, 817 \, 075} = 64 \, 795 \, 852 \, 800 = 29 \, 912 \, 035 \, 725 + 34 \, 883 \, 817 \, 075$  == Proof == By definition, $m$ and $n$ form an amicable pair : :$\map \sigma m = \map \sigma n = m + n$ where $\map \sigma n$ denotes the $\sigma$ function of $n$. Thus: {{begin-eqn}} {{eqn | l = \map \sigma {29 \, 912 \, 035 \, 725}       | r = 64 \, 795 \, 852 \, 800       | c = {{SigmaLink|29,912,035,725|29 \, 912 \, 035 \, 725}} }} {{eqn | r = 29 \, 912 \, 035 \, 725 + 34 \, 883 \, 817 \, 075       | c =  }} {{eqn | r = \map \sigma {34 \, 883 \, 817 \, 075}       | c = {{SigmaLink|34,883,817,075|34 \, 883 \, 817 \, 075}} }} {{end-eqn}}  == Sources == * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 64,795,852,800|next = Odd Amicable Pair/Examples/31,695,652,275-33,100,200,525}}: $64,795,852,800$
Odd Amicable Pair/Examples/31,695,652,275-33,100,200,525
Odd Amicable Pair/Examples/31,695,652,275-33,100,200,525 0 64361 324455 2017-11-03T07:11:08Z Prime.mover 59 Created page with "== Example of Odd Amicable Pair ==  $31 \, 695 \, 652 \, 275$ and $33 \, 100 \, 200 \, 525$ are an Defin..." wikitext text/x-wiki == Example of Odd Amicable Pair ==  $31 \, 695 \, 652 \, 275$ and $33 \, 100 \, 200 \, 525$ are an odd amicable pair: :$\sigma \left({31 \, 695 \, 652 \, 275}\right) = \sigma \left({33 \, 100 \, 200 \, 525}\right) = 64 \, 795 \, 852 \, 800 = 31 \, 695 \, 652 \, 275 + 33 \, 100 \, 200 \, 525$  == Proof == By definition, $m$ and $n$ form an amicable pair : :$\sigma \left({m}\right) = \sigma \left({n}\right) = m + n$ where $\sigma \left({n}\right)$ denotes the $\sigma$ function. Thus: {{begin-eqn}} {{eqn | l = \sigma \left({31 \, 695 \, 652 \, 275}\right)       | r = 64 \, 795 \, 852 \, 800       | c = {{SigmaLink|31,695,652,275|31 \, 695 \, 652 \, 275}} }} {{eqn | r = 31 \, 695 \, 652 \, 275 + 33 \, 100 \, 200 \, 525       | c =  }} {{eqn | r = \sigma \left({33 \, 100 \, 200 \, 525}\right)       | c = {{SigmaLink|33,100,200,525|33 \, 100 \, 200 \, 525}} }} {{end-eqn}}  == Sources == * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Odd Amicable Pair/Examples/29,912,035,725-34,883,817,075|next = Odd Amicable Pair/Examples/32,129,958,525-32,665,894,275}}: $64,795,852,800$
Odd Amicable Pair/Examples/32,129,958,525-32,665,894,275
Odd Amicable Pair/Examples/32,129,958,525-32,665,894,275 0 64362 324456 2017-11-03T07:15:34Z Prime.mover 59 Created page with "== Example of Odd Amicable Pair ==  $32 \, 129 \, 958 \, 525$ and $32 \, 665 \, 894 \, 275$ are an Defin..." wikitext text/x-wiki == Example of Odd Amicable Pair ==  $32 \, 129 \, 958 \, 525$ and $32 \, 665 \, 894 \, 275$ are an odd amicable pair: :$\sigma \left({32 \, 129 \, 958 \, 525}\right) = \sigma \left({32 \, 665 \, 894 \, 275}\right) = 64 \, 795 \, 852 \, 800 = 32 \, 129 \, 958 \, 525 + 32 \, 665 \, 894 \, 275$  == Proof == By definition, $m$ and $n$ form an amicable pair : :$\sigma \left({m}\right) = \sigma \left({n}\right) = m + n$ where $\sigma \left({n}\right)$ denotes the $\sigma$ function. Thus: {{begin-eqn}} {{eqn | l = \sigma \left({32 \, 129 \, 958 \, 525}\right)       | r = 64 \, 795 \, 852 \, 800       | c = {{SigmaLink|32,129,958,525|32 \, 129 \, 958 \, 525}} }} {{eqn | r = 32 \, 129 \, 958 \, 525 + 32 \, 665 \, 894 \, 275       | c =  }} {{eqn | r = \sigma \left({32 \, 665 \, 894 \, 275}\right)       | c = {{SigmaLink|32,665,894,275|32 \, 665 \, 894 \, 275}} }} {{end-eqn}}  == Sources == * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Odd Amicable Pair/Examples/31,695,652,275-33,100,200,525|next = Multiply Perfect Number of Order 6}}: $64,795,852,800$
Aurifeuillian Factorization/Examples/2^4n+2 + 1
== Example of Aurifeuillian Factorizations ==  :$2^{4 n + 2} + 1 = \paren {2^{2 n + 1} - 2^{n + 1} + 1} \paren {2^{2 n + 1} + 2^{n + 1} + 1}$  == Proof == From Sum of Squares as Product of Factors with Square Roots: :$x^2 + y^2 = \paren {x + \sqrt {2 x y} + y} \paren {x - \sqrt {2 x y} + y}$ Let $x = 2^{2 n + 1}$ and $y = 1$. Then: {{begin-eqn}} {{eqn | l = 2^{4 n + 2} + 1       | r = \paren {2^{2 n + 1} }^2 + 1^2       | c =  }} {{eqn | r = \paren {2^{2 n + 1} + \sqrt {2 \times \paren {2^{2 n + 1} } \times 1} + 1} \paren {2^{2 n + 1} - \sqrt {2 \times \paren {2^{2 n + 1} } \times 1} + 1}       | c =  }} {{eqn | r = \paren {2^{2 n + 1} + \sqrt {2^{2 n + 2} } + 1} \paren {2^{2 n + 1} - \sqrt {2^{2 n + 2} } + 1}       | c =  }} {{eqn | r = \paren {2^{2 n + 1} - 2^{n + 1} + 1} \paren {2^{2 n + 1} + 2^{n + 1} + 1}       | c =  }} {{end-eqn}}  == Historical Note == {{:Aurifeuillian Factorization/Examples/2^4n+2 + 1/Historical Note}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Prime Decomposition of 2^58+1/Historical Note|next = Aurifeuillian Factorization/Examples/2^4n+2 + 1/Historical Note}}: $2^{58} + 1$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Prime Decomposition of 2^58+1/Historical Note|next = Aurifeuillian Factorization/Examples/2^4n+2 + 1/Historical Note}}: $2^{58} + 1$
Tutte's Wheel Theorem/Lemma
  == Lemma == If a graph $G$ is 3-connected with $|V\left({G}\right)| > 4$ then $\exists e, \: e \in E\left({G}\right)$ such that $G \thinspace / \thinspace e$ is also 3-connected. == Proof == Suppose that no such edge $e$ exists.  Then $\forall e, \: e = xy \in E\left({G}\right)$, $\: G \thinspace / \thinspace e$ contains a vertex cut $S, \: |S| ≤ 2$. Since $\kappa\left({G}\right) \geq 3$, the contracted vertex $v_{x,y}$ of $G \thinspace / \thinspace e$ lies in $S$ (i.e. $\exists z, \: z \in G, \: z \notin \{x,y\}$.)   Let $S = \{v_{x,y}, z\}$. Then $T = \{x, y, z\}$ is a vertex cut of $G$.  Thus, every vertex in $T$ has an edge to every component of $G'=G-T$.  Let $C$ be the smallest component of $G'$.  Let $u \in N\left({z}\right) \cap C$ where $N\left({z}\right)$ is the set of all neighbours of the vertex $z$. By assumption, $G \thinspace / \thinspace zu$ is again not 3-connected, so again $\exists w$ such that $\{w, z, u\}$ is a vertex cut of $G$.  It also follows that every vertex in $\{w, z, u\}$ has an edge to every component of $G'' = G - \{w, z, u\}$.  Since $x,y$ are connected, $\exists D$, $D$ is a component of $G''$ and $D \cap \{x,y\} = \emptyset$.  Then, $D \subseteq N\left({z}\right) \cap V\left({C}\right) = \emptyset$.  Hence $D \varsubsetneqq C$ by the choice of $D$, which contradicts the assumption that $C$ was the smallest component.  {{qed|lemma}}
Completion Theorem (Metric Space)/Lemma 1
== Lemma == Let $M = \struct {A, d}$ be a metric space. Let $\CC \sqbrk A$ denote the set of all Cauchy sequences in $A$. Define the equivalence relation $\sim$ on $\CC \sqbrk A$ by: :$\displaystyle \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$ Denote the equivalence class of $\sequence {x_n} \in \CC \sqbrk A$ by $\sqbrk {x_n}$. Denote the set of equivalence classes under $\sim$ by $\tilde A$. Define $\tilde d: \tilde A \to \R_{\ge 0}$ by: :$\displaystyle \map {\tilde d} {\sqbrk {x_n}, \sqbrk {y_n} } = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$ Then:  :$\tilde d$ is well-defined on $\tilde A$.  == Proof == Let $\sequence {x_n}$, $\sequence {\hat x_n}$, $\sequence {y_n}$, $\sequence {\hat y_n} \in \CC \sqbrk A$ be such that: :$\sequence {x_n} \sim \sequence {\hat x_n}$ :$\sequence {y_n} \sim \sequence {\hat y_n}$ We have: {{begin-eqn}} {{eqn | l = \map d {x_n, y_n} - \map d {\hat x_n, \hat y_n}       | o = \le       | r = \map d {x_n, \hat x_n} + \map d {\hat x_n, y_n} - \map d {\hat x_n, \hat y_n}       | c = Triangle Inequality }} {{eqn | o = \le       | r = \map d {x_n, \hat x_n} + \map d {\hat x_n, \hat y_n} + \map d {\hat y_n, y_n} - \map d {\hat x_n, \hat y_n}       | c = Triangle Inequality }} {{eqn | r = \map d {x_n, \hat x_n} + \map d {\hat y_n, y_n}       | c =  }} {{end-eqn}} By an identical argument, we can also show that: :$\map d {\hat x_n, \hat y_n} - \map d {x_n, y_n} \le \map d {x_n, \hat x_n} + \map d {\hat y_n, y_n}$  and therefore: :$\displaystyle 0 \le \size {\map d {x_n, y_n} - \map d {\hat x_n, \hat y_n} } \le \map d {x_n, \hat x_n} + \map d {\hat y_n, y_n}$ Passing to the limit $n \to \infty$ and using the Combination Theorem for Sequences we have shown that: $\displaystyle \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \lim_{n \mathop \to \infty} \map d {\hat x_n, \hat y_n}$ But this precisely means that: :$\map {\tilde d} {\sqbrk {x_n}, \sqbrk {y_n} } = \map {\tilde d} {\sqbrk {\hat x_n}, \sqbrk {\hat y_n} }$
Heine-Borel Theorem/Euclidean Space/Necessary Condition
Heine-Borel Theorem/Euclidean Space/Necessary Condition 0 67083 342472 2018-02-12T08:09:32Z Prime.mover 59 Created page with "== Theorem ==  For any natural number $n \ge 1$, a closed and Definition:Bounded Metric Spac..." wikitext text/x-wiki == Theorem ==  For any natural number $n \ge 1$, a closed and bounded subspace of the Euclidean space $\R^n$ is compact.  == Proof 1 == {{:Heine-Borel Theorem/Euclidean Space/Necessary Condition/Proof 1}} == Proof 2 == {{:Heine-Borel Theorem/Euclidean Space/Necessary Condition/Proof 2}} {{Namedfor|Heinrich Eduard Heine|name2 = Émile Borel}}
Separated Subsets of Linearly Ordered Space under Order Topology/Lemma
== Lemma == Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space. Let $A$ and $B$ be separated sets of $T$. Let $A^*$ and $B^*$ be defined as: :$A^* := \displaystyle \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$ :$B^* := \displaystyle \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$ where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$. Then: :$(1): \quad A \subseteq A^*$ :$(2): \quad B \subseteq B^*$ :$(3): \quad A^* \cap B^* = \O$ == Proof == Let $a \in A$. Then: {{begin-eqn}} {{eqn | l = \closedint a a       | r = \set a       | c =  }} {{eqn | ll= \leadsto       | l = \closedint a a \cap B^-       | r = \O       | c = {{Defof|Separated Sets}} }} {{eqn | ll= \leadsto       | l = \closedint a a       | o = \subseteq       | r = A^*       | c = Definition of $A^*$ }} {{eqn | ll= \leadsto       | l = a       | o = \in       | r = A^*       | c = Definition of $\closedint a a$ }} {{eqn | ll= \leadsto       | l = A       | o = \subseteq       | r = A^*       | c = {{Defof|Subset}} }} {{end-eqn}} Similarly, $B \subseteq B^-$. {{qed|lemma}}  $A^* \cap B^* \ne \O$. Then: :$\exists p: p \in A^* \cap B^*$ Hence: :$\exists a, b \in A, c, d \in B: p \in \closedint a b \cap \closedint c d$ But because $A$ and $B$ are separated sets: :$c, d \notin A$ and: :$a, b \notin B$ and so: :$\closedint a b \cap \closedint c d = \O$ Thus $p \notin \closedint a b \cap \closedint c d$ It follows by Proof by Contradiction that $A^* \cap B^* = \O$.  == Sources == * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Definition:Convex Component|next = Separated Subsets of Linearly Ordered Space under Order Topology}}: Part $\text {II}$: Counterexamples: $39$. Order Topology: $3$
Power Series Expansion for Logarithm of 1 + x/Corollary
== Corollary to Power Series Expansion for $\ln \paren {1 + x}$ ==  {{begin-eqn}} {{eqn | l = \ln \paren {1 - x}       | r = -\sum_{n \mathop = 1}^\infty \frac {x^n} n       | c =  }} {{eqn | r = -x - \frac {x^2} 2 - \frac {x^3} 3 - \frac {x^4} 4 - \cdots       | c =  }} {{end-eqn}} valid for $-1 < x < 1$.  == Proof == By Power Series Expansion for $\ln \paren {1 + x}$: :$\displaystyle \ln \paren {1 + x} = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n$ Then: {{begin-eqn}} {{eqn | l = \ln \paren {1 - x}       | r = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\paren {-x}^n} n       | c = substituting $x \to -x$ }} {{eqn | r = -\sum_{n \mathop = 1}^\infty \paren {-1}^{2 n} \frac {x^n} n  }} {{eqn | r = -\sum_{n \mathop = 1}^\infty \frac {x^n} n  }} {{end-eqn}}
Fourier's Theorem/Lemma 1
== Lemma for Fourier's Theorem ==  Let $\psi$ be a real function defined on a closed interval $\closedint a b$. Let $\psi$ be piecewise continuous with one-sided limits on $\closedint a b$. Then: :$\displaystyle \lim_{N \mathop \to \infty} \int_a^b \map \psi u \sin N u \rd u = 0$  == Proof == We are given that $\psi$ is piecewise continuous with one-sided limits on $\closedint a b$. Therefore, there exists a finite subdivision $\set {x_0, x_1, \ldots, x_m}$ of $\closedint a b$, where $x_0 = a$ and $x_m = b$, such that for all $i \in \set {1, 2, \ldots, m}$: :$\psi$ is continuous on $\openint {x_{i - 1} } {x_i}$  :$\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \map \psi x$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} \map \psi x$ exist. From the corollary to Sum of Integrals on Adjacent Intervals for Integrable Functions: :$\displaystyle \int_a^b \map \psi u \sin N u \rd u = \sum_{r \mathop = 0}^{m - 1} \int_{x_r}^{x_{r + 1} } \map \psi u \sin N u \rd u$ Then: {{begin-eqn}} {{eqn | n = 1       | l = \int_{x_r}^{x_{r + 1} } \map \psi u \sin N u \rd u       | r = \intlimits {-\map \psi u \frac {\cos N u} N} {x_r} {x_{r + 1} }       | c = Integration by Parts }} {{eqn | o =        | ro= +       | r = \frac 1 N \int_{x_r}^{x_{r + 1} } \map {\psi'} u \cos N u \rd u       | c =  }} {{end-eqn}} The last integral is bounded. Thus $(1)$ is less than $\dfrac {M_r} N$ for $M_r \in \R$. Let $M = \max \set {\size {M_0}, \size {M_1}, \dotsc, \size {M_{m - 1} } }$. Then: :$\displaystyle \size {\int_a^b \map \psi u \sin N u \rd u} < \dfrac {M m} N$ As $M$ and $m$ are finite: :$\displaystyle \lim_{N \mathop \to \infty} \dfrac {M m} N = 0$ Hence the result.  == Sources == * {{BookReference|Fourier Series|1961|I.N. Sneddon|prev = Definition:Real Left-Hand Derivative|next = Fourier's Theorem/Lemma 1/Mistake 1}}: Chapter Two: $\S 2$. Some Important Limits: Lemma $(1)$
Fourier's Theorem/Lemma 2
== Lemma for Fourier's Theorem ==  Let $\psi$ be a real function defined on a half-open interval $\hointl 0 a$. Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\hointl 0 a$. Let $\map \psi u$ have a right-hand derivative at $u = 0$. Then: :$\displaystyle \lim_{N \mathop \to \infty} \int_0^a \map \psi u \frac {\sin N u} u \rd u = \frac \pi 2 \map \psi {0^+}$ where $\map \psi {0^+}$ denotes the limit of $\psi$ at $0$ from the right.  == Proof == We have: :$\map \psi u = \map \psi {0^+} + \paren {\map \psi u - \map \psi {0^+} }$ from which: :$(1): \quad \displaystyle \int_0^a \map \psi u \frac {\sin N u} u \rd u = \map \psi {0^+} \int_0^a \frac {\sin N u} u \rd u + \int_0^a \map \phi u \sin N u \rd u$ where: :$\map \phi u = \dfrac {\map \psi u - \map \psi {0^+} } u$ Let $\xi = N u$. Then: {{begin-eqn}} {{eqn | l = \int_0^a \frac {\sin N u} u \rd u       | r = \int_0^{N a} \frac {\sin \xi} \xi \rd \xi       | c =  }} {{eqn | o = \to       | r = \int_0^\infty \frac {\sin \xi} \xi \rd \xi       | c =  }} {{eqn | r = \frac \pi 2       | c = Integral to Infinity of $\dfrac {\sin \xi} \xi$ }} {{end-eqn}} We have that $\map \psi u$ is piecewise continuous with one-sided limits on $\hointl 0 a$. Hence it follows that $\map \phi u = \dfrac {\map \psi u - \map \psi {0^+} } u$ is also piecewise continuous with one-sided limits on $\hointl 0 a$. We also have that $\map \psi u$ has a right-hand derivative at $u = 0$. It follows that $\map \phi u$ is piecewise continuous with one-sided limits on $\hointl 0 a$. Thus from Lemma 1 for Fourier's Theorem: :$\displaystyle \lim_{N \mathop \to \infty} \int_0^a \map \phi u \sin N u \rd u = 0$ and letting $N \to \infty$ in $(1)$ above: :$\displaystyle \lim_{N \mathop \to \infty} \int_0^a \map \psi u \frac {\sin N u} u \rd u = \frac \pi 2 \map \psi {0^+}$   == Sources == * {{BookReference|Fourier Series|1961|I.N. Sneddon|prev = Fourier's Theorem/Lemma 1/Mistake 2|next = Fourier's Theorem/Lemma 2/Mistake}}: Chapter Two: $\S 2$. Some Important Limits: Lemma $(2)$
Fourier's Theorem/Lemma 3
== Lemma for Fourier's Theorem ==  Let $\psi$ be a real function defined on an open interval $\openint a b$. Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\openint a b$. Let $\map \psi u$ have both right-hand derivative and left-hand derivative at a point $u = x$ where $x \in \openint a b$. Then: :$\displaystyle \lim_{N \mathop \to \infty} \int_a^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \frac \pi 2 \paren {\map \psi {x^+} + \map \psi {x^-} }$ where: :$\map \psi {x^+}$ denotes the limit of $\psi$ at $x$ from the right :$\map \psi {x^-}$ denotes the limit of $\psi$ at $x$ from the left.  == Proof == From Sum of Integrals on Adjacent Intervals for Integrable Functions, we have: :$\displaystyle \int_a^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \int_a^x \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u + \int_x^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u$ Let $u = x - \xi$. Then by Integration by Substitution: :$\displaystyle \int_a^x \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \int_0^{x - a} \map \phi \xi \frac {\sin N \xi} \xi \rd \xi$ where: :$\map \phi \xi = \map \psi {u - \xi}$ {{explain|It needs to be established how the limits change from $a \to x$ to $0 \to x - a$ as this is not obvious.}} By Fourier's Theorem: Lemma 2: {{begin-eqn}} {{eqn | l = \lim_{N \mathop \to \infty} \int_0^{x - a} \map \phi \xi \frac {\sin N \xi} \xi \rd \xi       | r = \frac \pi 2 \map \phi {0^+}       | c =  }} {{eqn | r = \frac \pi 2 \map \psi {x^-}       | c =  }} {{end-eqn}} Similarly, substituting $u = x + \eta$: :$\displaystyle \int_x^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \int_0^{b - x} \map \chi \eta \frac {\sin N \eta} \eta \rd \eta$ where: :$\map \chi \xi = \map \psi {x + \eta}$ By Fourier's Theorem: Lemma 2: {{begin-eqn}} {{eqn | l = \lim_{N \mathop \to \infty} \int_0^{b - x} \map \chi \eta \frac {\sin N \eta} \eta \rd \eta       | r = \frac \pi 2 \map \phi {0^+}       | c =  }} {{eqn | r = \frac \pi 2 \map \psi {x^+}       | c =  }} {{end-eqn}} The result follows by adding the two limits.   == Sources == * {{BookReference|Fourier Series|1961|I.N. Sneddon|prev = Fourier's Theorem/Lemma 2/Mistake|next = Fourier's Theorem}}: Chapter Two: $\S 2$. Some Important Limits: Lemma $(3)$