hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn	| l = \int_0^\infty \map \exp {-\paren {a x^2 + \frac b {x^2} } } \rd x 	| r = \int_0^\infty \map \exp {-a \paren {x^2 + \frac b {a x^2} } } \rd x }} {{eqn	| r = \int_0^\infty \map \exp {-a \paren {\paren {x - \frac 1 x \sqrt {\frac b a} }^2 + 2 \sqrt {\frac b a} } } \rd x 	| c = Completing the Square }} {{eqn	| r = \map \exp {-2 \sqrt {a b} } \int_0^\infty \map \exp {-a \paren {x - \frac 1 x \sqrt {\frac b a} }^2} \rd x 	| c = Exponential of Sum }} {{eqn	| r = \frac 1 2 \map \exp {-2 \sqrt {a b} } \int_{-\infty}^\infty \map \exp {-a \paren {x - \frac 1 x \sqrt {\frac b a} }^2} \rd x 	| c = Definite Integral of Even Function }} {{eqn	| r = \frac 1 2 \map \exp {-2 \sqrt {a b} } \int_{-\infty}^\infty \map \exp {-a u^2} \rd u 	| c = Glasser's Master Theorem }} {{eqn	| r = \map \exp {-2 \sqrt {a b} } \int_0^\infty \map \exp {-a u^2} \rd u 	| c = Definite Integral of Even Function }} {{eqn	| r = \frac 1 2 \sqrt {\frac \pi a} \map \exp {-2 \sqrt {a b} }  	| c = Definite Integral to Infinity of $\map \exp {-a x^2}$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac x {e^x - 1} \rd x 	| r = \int_0^\infty \frac {x^{2 - 1} } {e^x - 1} \rd x }} {{eqn	| r = \map \zeta 2 \map \Gamma 2 	| c = Integral Representation of Riemann Zeta Function in terms of Gamma Function }} {{eqn	| r = \frac {\pi^2} 6 \times 1! 	| c = Basel Problem, Gamma Function Extends Factorial }} {{eqn	| r = \frac {\pi^2} 6 }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {x^{s - 1} } {e^x + 1} \rd x 	| r = \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 - \paren {-e^{-x} } } \rd x }} {{eqn	| r = \int_0^\infty x^{s - 1} e^{-x} \paren {\sum_{n \mathop = 0}^\infty \paren {-e^{-x} }^n} \rd x 	| c = Sum of Geometric Sequence }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \paren {\paren {-1}^n \int_0^\infty x^{s - 1} e^{-\paren {n + 1} x} \rd x} 	| c = Fubini's Theorem }} {{eqn	| r = \map \Gamma s \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {n + 1}^s} 	| c = Laplace Transform of Complex Power }} {{eqn	| r = \map \Gamma s \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } {n^s} 	| c = shifting the index }} {{eqn	| r = \map \Gamma s \map \eta s 	| c = {{Defof|Dirichlet Eta Function}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac x {e^x + 1} \rd x 	| r = \int_0^\infty \frac {x^{2 - 1} } {e^x + 1} \rd x }} {{eqn	| r = \map \eta 2 \map \Gamma 2 	| c = Integral Representation of Dirichlet Eta Function in terms of Gamma Function }} {{eqn	| r = \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } {n^2} } 1! 	| c = {{Defof|Dirichlet Eta Function}}, Gamma Function Extends Factorial }} {{eqn	| r = \frac {\pi^2} {12} 	| c = Sum of Reciprocals of Squares Alternating in Sign }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^1 \ln x \map \ln {1 - x} \rd x 	| r = -\int_0^1 \ln x \paren {\sum_{n \mathop = 1}^\infty \frac {x^n} n} \rd x 	| c = Power Series Expansion for $\map \ln {1 + x}$: Corollary }} {{eqn	| r = -\sum_{n \mathop = 1}^\infty \frac 1 n \paren {\int_0^1 x^n \ln x \rd x} 	| c = Fubini's Theorem }} {{eqn	| r = \paren {-1}^2 \sum_{n \mathop = 1}^\infty \frac 1 n \paren {\frac {\map \Gamma 2} {\paren {n + 1}^2} } 	| c = Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$ }} {{eqn	| r = 1! \sum_{n \mathop = 1}^\infty \frac 1 {n \paren {n + 1}^2} 	| c = Gamma Function Extends Factorial }} {{eqn	| r = \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {n + 1} } - \sum_{n \mathop = 1}^\infty \paren {\frac 1 {\paren {n + 1}^2} } 	| c = partial fraction expansion }} {{eqn	| r = 1 - \paren {\sum_{n \mathop = 1}^\infty \frac 1 {n^2} - 1} 	| c = Telescoping Series: Example 1 }} {{eqn	| r = 2 - \frac {\pi^2} 6 	| c = Basel Problem }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^1 \ln x \map \ln {1 + x} \rd x 	| r = \int_0^1 \ln x \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n} \rd x 	| c = Power Series Expansion for $\map \ln {1 + x}$ }} {{eqn	| r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {\int_0^1 x^n \ln x \rd x} 	| c = Fubini's Theorem }} {{eqn	| r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} n \paren {\frac {\map \Gamma 2} {\paren {n + 1}^2} } 	| c = Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$ }} {{eqn	| r = 1! \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {n \paren {n + 1}^2} 	| c = Gamma Function Extends Factorial }} {{eqn	| r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} n - \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {n + 1} - \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {\paren {n + 1}^2} 	| c = partial fraction expansion }} {{eqn	| r = -\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n - \sum_{n \mathop = 2}^\infty \frac {\paren {-1}^{n - 1} } n - \sum_{n \mathop = 2}^\infty \frac {\paren {-1}^{n - 1} } {n^2} 	| c = shifting indexes }} {{eqn	| r = -\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n - 1} - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } {n^2} - 1} }} {{eqn	| r = 2 - 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n - \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } {n^2} }} {{eqn	| r = 2 - 2 \ln 2 - \frac {\pi^2} {12} 	| c = Newton-Mercator Series for $\ln 2$, Sum of Reciprocals of Squares Alternating in Sign }} {{end-eqn}} {{qed}}
Let $x \in T$. We are given that: :$\forall S \in \mathbb S: T \subseteq S$  Thus by definition of subset: :$\forall S \in \mathbb S: x \in S$ Hence by definition of intersection: :$x \in \displaystyle \bigcap \mathbb S$ Thus by definition of subset: :$T \subseteq \displaystyle \bigcap \mathbb S$ {{qed}} Category:Set is Subset of Intersection of Supersets rvsddnpa6sh0l40ku0i9rs0cuang369
{{begin-eqn}} {{eqn	| l = \map \erf x 	| r = \frac 2 {\sqrt \pi} \int_0^x e^{-u^2} \rd u 	| c = {{Defof|Error Function}} }} {{eqn	| r = \frac 2 {\sqrt \pi} \int_0^x \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {u^{2 n} } {n!} } \rd u 	| c = {{Defof|Real Exponential Function}} }} {{eqn	| r = \frac 2 {\sqrt \pi} \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {n!} \paren {\int_0^x u^{2 n} \rd u} 	| c = Power Series is Termwise Integrable within Radius of Convergence }} {{eqn	| r = \frac 2 {\sqrt \pi} \sum_{n \mathop= 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {n! \paren {2 n + 1} } 	| c = Primitive of Power }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \map \erf {-x} 	| r = \frac 2 {\sqrt \pi} \int_0^{-x} e^{-u^2} \rd u 	| c = {{Defof|Error Function}} }} {{eqn	| r = -\frac 2 {\sqrt \pi} \int_0^{-\paren {-x} } e^{-\paren {-u}^2} \rd u 	| c = substituting $u \mapsto -u$ }} {{eqn	| r = -\frac 2 {\sqrt \pi} \int_0^x e^{-u^2} \rd u }} {{eqn	| r = -\map \erf x 	| c = {{Defof|Error Function}} }} {{end-eqn}} {{qed}}
By Error Function is Odd, $\erf$ is a odd function.  Therefore, by Odd Function of Zero is Zero: :$\map \erf 0 = 0$ {{qed}}
Let $\sequence {s_N}$ be the sequence of partial sums defined by: :$\forall N \in \Z_{\ge m}: s_N = \displaystyle \sum_{n \mathop = m}^N d_n p^n$ From Corollary of Characterisation of Cauchy Sequence in Non-Archimedean Norm the sequence $\sequence {s_N}$ is a Cauchy sequence if: :$\forall N \in \Z_{\ge m}: s_{N + 1} \equiv s_N \pmod {p^n}$ Now for all $N \in \Z_{\ge m}$: {{begin-eqn}} {{eqn | l = s_{N + 1} - s_N       | r = \sum_{n \mathop = m}^{N + 1} d_n p^ n - \sum_{n \mathop = m}^{N} d_n p^n       | c = {{Defof|Sequence of Partial Sums|Partial Sum}} }} {{eqn | r = d_{N + 1} p^{N + 1} }} {{eqn | ll= \leadsto       | l = s_{N + 1}       | o = \equiv       | r = s_N \pmod {p^N}       | c = {{Defof|Congruence (Number Theory)|subdef = Integer Multiple|Congruence modulo $p^N$}} }} {{end-eqn}} The result follows. {{qed}}
From P-adic Expansion is a Cauchy Sequence in P-adic Norm, the sequence of partial sums of the series: :$\displaystyle \sum_{n \mathop = m}^\infty d_n p^n$ is a Cauchy sequence in the rationals $\Q$ with the $p$-adic norm. From Sequence is Cauchy in P-adic Norm iff Cauchy in P-adic Numbers, the sequence of partial sums of the series: :$\displaystyle \sum_{n \mathop = m}^\infty d_n p^n$ is a Cauchy sequence in the $p$-adic numbers $\Q_p$ with the $p$-adic norm. By definition of the $p$-adic numbers $\Q_p$, the $p$-adic numbers $\Q_p$ form a complete normed division ring. By definition of a complete normed division ring, every Cauchy sequence is convergent in $\struct{\Q_p, \norm{\,\cdot\,}_p}$. {{qed}}
We have, by Derivative of Exponential Function: Corollary 1:  :$\dfrac \d {\d u} \paren {1 - e^{-u} } = e^{-u}$ By Primitive of Reciprocal: :$\displaystyle \int \frac {\d u} u = \ln u + C$ So:  {{begin-eqn}} {{eqn	| l = \int_0^x \frac {1 - e^{-u} } u \rd u 	| r = \intlimits {\paren {1 - e^{-u} } \ln u} 0 x - \int_0^x e^{-u} \ln u \rd u 	| c = Integration by Parts }} {{eqn	| r = \paren {1 - e^{-x} } \ln x - \lim_{x \mathop \to 0^+} \paren {1 - e^{-x} } \ln x - \paren {\int_0^\infty e^{-u} \ln u \rd u - \int_x^\infty e^{-u} \ln u \rd u} 	| c = Sum of Integrals on Adjacent Intervals for Integrable Functions }} {{end-eqn}} We have:  {{begin-eqn}} {{eqn	| l = \lim_{x \mathop \to 0^+} \paren {1 - e^{-x} } \ln x 	| r = \paren {\lim_{x \mathop \to 0^+} \frac {1 - e^{-x} - \paren {1 - e^0} } x} \paren {\lim_{x \mathop \to 0^+} x \ln x} 	| c = Combination Theorem for Limits of Functions: Product Rule }} {{eqn	| r = \frac \d {\d x} \paren {1 - e^{-x} }_{x = 0} \paren {\lim_{x \mathop \to 0^+} x \ln x} 	| c = {{Defof|Derivative}} }} {{eqn	| r = e^0 \times 0 	| c = Derivative of Exponential Function, Limit of $x^m \paren {\ln x}^n$ }} {{eqn	| r = 0 }} {{end-eqn}} Also applying Definite Integral to Infinity of $e^{-x} \ln x$ gives: :$\displaystyle \int_0^x \frac {1 - e^{-u} } u \rd u = \paren {1 - e^{-x} } \ln x + \gamma + \int_x^\infty e^{-u} \ln u \rd u$  We have, by Primitive of $e^{a x}$: :$\displaystyle \int e^{-u} \rd u = -e^{-u} + C$ We have by Derivative of Logarithm Function: :$\dfrac \d {\d u} \paren {\ln u} = \dfrac 1 u$ We therefore have:  {{begin-eqn}} {{eqn	| l = \paren {1 - e^{-x} } \ln x + \gamma + \int_x^\infty e^{-u} \ln u \rd u 	| r = \paren {1 - e^{-x} } \ln x + \gamma + \paren {\intlimits {-e^{-u} \ln u} x \infty + \int_x^\infty \frac {e^{-u} } u \rd u} 	| c = Integration by Parts }} {{eqn	| r = \paren {1 - e^{-x} } \ln x + \gamma - \lim_{u \mathop \to \infty} \paren {e^{-u} \ln u} + e^{-x} \ln x + \map \Ei x 	| c = {{Defof|Exponential Integral Function}} }} {{eqn	| r = \ln x + \gamma + \map \Ei x - \lim_{u \mathop \to \infty} \paren {e^{-u} \ln u} }} {{end-eqn}} It remains to evaluate the limit on the {{RHS}}.  We have, for $u > 1$: :$e^{-u} \ln u > 0$ and by Bounds of Natural Logarithm:  :$e^{-u} \ln u < u e^{-u} - e^{-u}$ We have, from Limit to Infinity of $x^n e^{-a x}$:  :$\displaystyle \lim_{u \mathop \to \infty} \paren {u e^{-u} - e^{-u} } = 0$ So, by the Squeeze Theorem:  :$\displaystyle \lim_{u \mathop \to \infty} \paren {e^{-u} \ln u} = 0$ So:  :$\displaystyle \int_0^x \frac {1 - e^{-u} } u \rd u = \ln x + \gamma + \map \Ei x$ Rearranging gives:  :$\displaystyle \map \Ei x = -\gamma - \ln x + \int_0^x \frac {1 - e^{-u} } u \rd u$ {{qed}}
{{begin-eqn}} {{eqn	| l = \map \erfc 0  	| r = 1 - \map \erf 0 	| c = {{Defof|Complementary Error Function}} }} {{eqn	| r = 1 	| c = Error Function of Zero }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \lim_{x \mathop \to \infty} \map \erfc x 	| r = \lim_{x \mathop \to \infty} \paren {1 - \map \erf x} 	| c = {{Defof|Complementary Error Function}} }} {{eqn	| r = 1 - \lim_{x \mathop \to \infty} \map \erf x 	| c = Combination Theorem for Limits of Functions: Sum Rule }} {{eqn	| r = 1 - 1 	| c = Limit to Infinity of Error Function }} {{eqn	| r = 0 }} {{end-eqn}} {{qed}}
In the extreme pathological edge case: :$S = \set S$ it is seen that: :$S \in S$ and so: :$\forall x \in S: x \in x$ demonstrating that $\RR$ is reflexive in this specific case. However, in this case $\set S$ is a set on which the Axiom of Foundation does not apply. This is seen in Set is Not Element of Itself. Hence this set is not supported by Zermelo-Fraenkel set theory. Consider the set: :$T = \set {\O, \set \O}$ Then we immediately see that: :$\O \notin \O$ and so $\RR$ is seen to be not reflexive. {{qed}}
In the extreme pathological edge case: :$S = \set S$ it is seen that: :$S \in S$ and so: :$\forall x \in S: \tuple {a, b} \in \RR \implies \tuple {b, a} \in \RR$ demonstrating that $\RR$ is symmetric in this specific case. However, in this case $\set S$ is a set on which the Axiom of Foundation does not apply. This is seen in Set is Not Element of Itself. Hence this set is not supported by Zermelo-Fraenkel set theory. Consider the set: :$T = \set {\O, \set \O}$ Then we immediately see that while: :$\O \in \set \O$ we have that: :$\set \O \notin \O$ and so $\RR$ is seen to be not symmetric. {{qed}}
By definition, the $p$-adic norm is non-Archimedean on the $p$-adic numbers $\Q_p$. From P-adic Norm is Non-Archimedean Norm, the $p$-adic norm is  non-Archimedean on $\Q$. By Characterisation of Cauchy Sequence in Non-Archimedean Norm: :$\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$ {{iff}} $\displaystyle \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$. Similarly: :$\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q_p, \norm {\,\cdot\,}_p}$ {{iff}} $\displaystyle \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$. By definition, the $p$-adic norm on the $p$-adic numbers is an extension of the $p$-adic norm on the rational numbers. Hence: :$\displaystyle \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$ in $\struct {\Q, \norm {\,\cdot\,}_p}$ {{iff}} $\displaystyle \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$ in $\struct {\Q_p, \norm {\,\cdot\,}_p}$ The result follows. {{qed}} Category:P-adic Number Theory 7eqy8bvirecdp0gpqkl8ggfcnz74c57
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\sin x} {\sqrt x} \rd x 	| r = \int_0^\infty 2 u \frac {\sin u^2} {\sqrt {u^2} } \rd u 	| c = substituting $x = u^2$ }} {{eqn	| r = 2 \int_0^\infty \sin u^2 \rd u }} {{eqn	| r = \sqrt {\frac \pi 2} 	| c = Definite Integral to Infinity of $\sin a x^2$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \map \psi {\frac 1 2} 	| r = -\gamma - \ln 4 - \frac \pi 2 \map \cot {\frac 1 2 \pi} + 2 \sum_{n \mathop = 1}^0 \map \cos {\frac {\pi n} 2} \map \ln {\map \sin {\frac {\pi n} 2} } 	| c = Gauss's Digamma Theorem }} {{eqn	| r = -\gamma - \ln 4 	| c = Cotangent of Half-Integer Multiple of Pi, noting also that the summation is an empty sum }} {{eqn	| r = -\gamma - 2 \ln 2 	| c = Logarithm of Power }} {{end-eqn}} {{qed}} Category:Digamma Function Category:Euler-Mascheroni Constant 0xaveh7rjv89m54ubbhmqle03pzouke
Consider the integral:  :$\displaystyle \int_0^\infty x^t e^{-x^2} \rd x$ for positive real parameter $t$. Using Definite Integral to Infinity of $x^m e^{-a x^2}$, we have:  :$\displaystyle \int_0^\infty x^t e^{-x^2} \rd x = \frac 1 2 \map \Gamma {\frac {1 + t} 2}$ Differentiating the {{LHS}} with respect to $t$ we have:  {{begin-eqn}} {{eqn	| l = \frac \d {\d t} \int_0^\infty t^x e^{-x^2} \rd t 	| r = \int_0^\infty \frac \partial {\partial t} \paren {x^t e^{-x^2} } \rd x 	| c = Definite Integral of Partial Derivative }} {{eqn	| r = \int_0^\infty x^t e^{-x^2} \ln x \rd x 	| c = Derivative of Power of Constant }} {{end-eqn}} Differentiating the {{RHS}} using the Chain Rule for Derivatives gives:  :$\displaystyle \int_0^\infty x^t e^{-x^2} \ln x \rd x = \frac 1 4 \map {\Gamma'} {\frac {1 + t} 2}$ We therefore have:  {{begin-eqn}} {{eqn	| l = \int_0^\infty e^{-x^2} \ln x \rd x 	| r = \int_0^\infty x^0 e^{-x^2} \ln x \rd x }} {{eqn	| r = \frac 1 4 \map {\Gamma'} {\frac 1 2} }} {{eqn	| r = \frac 1 4 \map \Gamma {\frac 1 2} \map \psi {\frac 1 2} 	| c = {{Defof|Digamma Function}} }} {{eqn	| r = -\frac {\sqrt \pi} 4 \paren {\gamma + 2 \ln 2} 	| c = Gamma Function of One Half, Digamma Function of One Half }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\cos x} {\sqrt x} \rd x 	| r = \int_0^\infty 2 u \frac {\cos u^2} {\sqrt {u^2} } \rd u 	| c = substituting $x = u^2$ }} {{eqn	| r = 2 \int_0^\infty \cos u^2 \rd u }} {{eqn	| r = \sqrt {\frac \pi 2} 	| c = Definite Integral to Infinity of $\cos a x^2$ }} {{end-eqn}} {{qed}}
Let:  :$\displaystyle \map I \alpha = \int_0^\infty \frac {\map {\sin^3} {\alpha x} } {x^3} \rd x$ for positive real parameter $\alpha$. We have:  {{begin-eqn}} {{eqn	| l = \map I 0 	| r = \int_0^\infty \frac {\map {\sin^3} {0 x} } {x^3} \rd x }} {{eqn	| r = \int_0^\infty \frac 0 {x^3} \rd x 	| c = Sine of Zero is Zero }} {{eqn	| r = 0 }} {{end-eqn}} We aim to evaluate explicitly:  :$\displaystyle \int_0^\infty \frac {\sin^3 x} {x^3} \rd x = \map I 1$ Differentiating with respect to $\alpha$ we have:  {{begin-eqn}} {{eqn	| l = \map {I'} \alpha 	| r = \frac \d {\d \alpha} \int_0^\infty \frac {\map {\sin^3} {\alpha x} } {x^3} \rd x }} {{eqn	| r = \int_0^\infty \frac \partial {\partial \alpha} \paren {\frac {\map {\sin^3} {\alpha x} } {x^3} } \rd x 	| c = Definite Integral of Partial Derivative }} {{eqn	| r = \frac 1 4 \int_0^\infty \frac \partial {\partial \alpha} \paren {\frac {3 \map \sin {\alpha x} - \map \sin {3 \alpha x} } {x^3} } \rd x 	| c = Cube of Sine }} {{eqn	| r = \frac 3 4 \int_0^\infty \frac {\map \cos {\alpha x} - \map \cos {3 \alpha x} } {x^2} \rd x 	| c = Derivative of Cosine Function, Chain Rule for Derivatives }} {{eqn	| r = \frac 3 4 \times \frac \pi 2 \paren {3 \alpha - \alpha} 	| c = Definite Integral to Infinity of $\dfrac {\cos p x - \cos q x} {x^2}$ }} {{eqn	| r = \frac {3 \alpha \pi} 4 }} {{end-eqn}} We therefore have:  {{begin-eqn}} {{eqn	| l = \int_0^1 \map {I'} \alpha \rd \alpha 	| r = \intlimits {\frac {3 \alpha^2 \pi} 8} 0 1 	| c = Primitive of Power }} {{eqn	| r = \frac {3 \pi} 8 }} {{end-eqn}} By Fundamental Theorem of Calculus: Second Part, we also have:  :$\displaystyle \int_0^1 \map {I'} \alpha \rd \alpha = \map I 1 - \map I 0 = \map I 1$ giving:  :$\displaystyle \int_0^\infty \frac {\sin^3 x} {x^3} \rd x = \frac {3 \pi} 8$  {{qed}}
Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings such that: :$\Dom {f_2} = \Cdm {f_1}$ === $(1)$ implies $(2)$ === Let $f_2 \circ f_1$ be a composite mapping by definition 1. Then by definition: :$\forall x \in S_1: \map {\paren {f_2 \circ f_1} } x := \map {f_2} {\map {f_1} x}$ We have that $f_1$ is a mapping, and so: :$\forall x \in S_1: \exists \map {f_1} x \in S_2$ We have that $f_2$ is a mapping, and so: :$\forall \map {f_1} x \in S_2: \exists \map {f_2} {\map {f_1} x} \in S_3$ Then: {{begin-eqn}} {{eqn | l = z       | r = \map {f_2} {\map {f_1} x}       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {\map {f_1} x, z}       | o = \in       | r = f_2       | c = {{Defof|Mapping}} as a Relation }} {{end-eqn}} and so: :$f_2 \circ f_1 = \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$ Thus $f_2 \circ f_1$ is a composite mapping by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $f_2 \circ f_1$ be a composite mapping by definition 2. Then by definition: :$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$ We have that $f_1$ is a mapping, and so: :$\forall x \in S_1: \exists \map {f_1} x \in S_2$ and so by definition of a {{Defof|Mapping}} as a Relation: {{begin-eqn}} {{eqn | lo= \forall x \in S_1:       | l = \tuple {\map {f_1} x, z}       | o = \in       | r = f_2       | c =  }} {{eqn | ll= \leadsto       | l = z       | r = \map {f_2} {\map {f_1} x}       | c =  }} {{end-eqn}} Thus $f_2 \circ f_1$ is a composite mapping by definition 1. {{qed|lemma}} === $(2)$ implies  $(3)$ === Let $f_2 \circ f_1$ be a composite mapping by definition 2. Then by definition: :$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$ Because $f_1$ is a mapping, it follows that: :$\forall x \in S_1: \exists y \in S_2: \map {f_1} x = y$ Similarly: :$\forall y \in S_2: \exists z \in S_3: \map {f_2} y = z$ Hence: :$\tuple {\map {f_1} x, z} \in f_2 \implies \exists y \in S_2: \map {f_1} x = y \land \map {f_2} y = z$ and so: :$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \map {f_1} x = y \land \map {f_2} y = z}$ Thus $f_2 \circ f_1$ is a composite mapping by definition 3. {{qed|lemma}} === $(3)$ implies  $(2)$ === Let $f_2 \circ f_1$ be a composite mapping by definition 3. Then by definition: :$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \map {f_1} x = y \land \map {f_2} y = z}$ We have that: :$\forall x \in S_1: \exists y \in S_2: \map {f_1} x = y$ and that: :$\forall y \in S_2: \exists z \in S_3: \map {f_2} y = z$ Hence: :$\forall x \in S_1: \exists z \in S_3: \map {f_1} x = y \land \map {f_2} y = z$ Thus: :$\forall x \in S_1: \exists z \in S_3: \tuple {\map {f_1} x, y} \in f_2$ and so: :$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$ Thus $f_2 \circ f_1$ is a composite mapping by definition 2. {{qed}}
{{begin-eqn}} {{eqn	| l = \map \Ei x 	| r = -\gamma - \ln x + \int_0^x \frac {1 - e^{-u} } u \rd u 	| c = Characterization of Exponential Integral Function }} {{eqn	| r = -\gamma - \ln x + \int_0^x \frac 1 u \paren {1 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {u^n} {n!} } \rd u 	| c = {{Defof|Real Exponential Function}} }} {{eqn	| r = -\gamma - \ln x + \int_0^x \frac 1 u \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^{n + 1} \frac {u^n} {n!} } \rd u }} {{eqn	| r = -\gamma - \ln x + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } {n!} \paren {\int_0^x u^{n - 1} \rd u} 	| c = Power Series is Termwise Integrable within Radius of Convergence }} {{eqn	| r = -\gamma - \ln x + \sum_{n \mathop = 1}^\infty \paren {-1}^{n + 1} \frac {x^n} {n \times n!} 	| c = Primitive of Power }} {{end-eqn}} {{qed}}
Let $a \in \Q_p$. From P-adic Number times Integer Power of p is P-adic Integer, there exists $n \in \N$ such that $p^n a \in \Z_p$. Since $n \in \N$ and $p^n a \in \Z_p$, let $\map f X \in \Z_p \sqbrk X$ be the polynomial: :$\paren {p^n a} X^n$ Then: :$\map f {1 / p} = \paren {p^n a} \paren {1 / p}^n = a$. Hence: :$a \in \Z_p \sqbrk {1 / p}$. Since $a \in \Q_p$ was arbitrary it follows that $\Q_p \subseteq \Z_p \sqbrk {1 / p }$. By definition of a generated ring extension, $\Z_p \sqbrk {1 / p } \subseteq \Q_p$. The result follows. {{qed}}
{{begin-eqn}} {{eqn	| l = \lim_{x \mathop \to \infty} \map \Ei x 	| r = \lim_{x \mathop \to \infty} \int_x^\infty \frac {e^{-u} } u \rd u 	| c = {{Defof|Exponential Integral Function}} }} {{eqn	| r = \lim_{x \mathop \to \infty} \int_1^\infty \frac {e^{-x t} } {x t} x \rd t 	| c = substituting $u = x t$ }} {{eqn	| r = \int_1^\infty \lim_{x \mathop \to \infty} \paren {\frac {e^{-x t} } t} \rd t 	| c = Lebesgue's Dominated Convergence Theorem }} {{eqn	| r = \int_1^\infty \frac 0 t \rd t 	| c = Exponential Tends to Zero and Infinity }} {{eqn	| r = 0 }} {{end-eqn}} {{qed}}