hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn	| l = \int_0^\infty e^{-a x^2} \rd x 	| r = \int_0^\infty e^{-\paren {\sqrt a x}^2} \rd x }} {{eqn	| r = \frac 1 {\sqrt a} \int_0^\infty e^{-t^2} \rd t 	| c = substituting $t = \sqrt a x$ }} {{eqn	| r = \frac 1 2 \sqrt {\frac \pi a} 	| c = Integral to Infinity of $e^{-t^2}$ }} {{end-eqn}} {{qed}}
Let:  :$x = e^{-u}$ We have, by Derivative of Exponential Function: :$\dfrac {\d x} {\d u} = -e^{-u}$  By Exponential Tends to Zero and Infinity:  :as $x \to 0$, $u \to \infty$ By Exponential of Zero:  :as $x \to 1$, $u \to 0$. So: {{begin-eqn}} {{eqn	| l = \int_0^1 \frac {x^m - x^n} {\ln x} \rd x 	| r = -\int_\infty^0 e^{-u} \frac {e^{-m u} - e^{-n u} } {\ln e^{-u} } \rd u 	| c = substituting $x = e^{-u}$ }} {{eqn	| r = -\int_0^\infty \frac {e^{-\paren {m + 1} u} - e^{-\paren {n + 1} u} } u \rd u 	| c = Reversal of Limits of Definite Integral, Exponential of Sum }} {{eqn	| r = \int_0^\infty \frac {e^{-\paren {n + 1} u} - e^{-\paren {m + 1} u} } u \rd u }} {{eqn	| r = \map \ln {\frac {m + 1} {n + 1} } 	| c = Definite Integral to Infinity of $\dfrac {e^{-a x} - e^{-b x} } x$ }} {{end-eqn}} {{qed}}
We have that both $\displaystyle \bigcup_{j \mathop \in J} S_j$ and $\displaystyle \bigcup_{k \mathop \in \relcomp I J} S_k$ are sets. Hence by Union is Commutative we have: :$\bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k = \bigcup_{k \mathop \in \relcomp I J} S_k \cup \bigcup_{j \mathop \in J} S_j$ It remains to be demonstrated that $\displaystyle \bigcup_{i \mathop \in I} S_i = \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k$. So: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \bigcup_{i \mathop \in I} S_i       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \exists i \in I: x       | o = \in       | r = S_i       | c = {{Defof|Union of Family}} }} {{eqn | ll= \leadstoandfrom       | l = \exists j \in J: x       | o = \in       | r = S_j       | c = {{Defof|Relative Complement}} }} {{eqn | lo= \lor       | l = \exists k \in \relcomp I J: x       | o = \in       | r = S_k       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \bigcup_{j \mathop \in J} S_j       | c = {{Defof|Union of Family}} }} {{eqn | lo= \lor       | l = x       | o = \in       | r = \bigcup_{k \mathop \in \relcomp I J} S_k       | c = {{Defof|Union of Family}} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k       | c = {{Defof|Set Union}} }} {{end-eqn}} That is: :$\displaystyle x \in \bigcup_{i \mathop \in I} S_i \iff x \in \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k$ The result follows by definition of set equality. {{qed}}
We have that both $\displaystyle \bigcap_{j \mathop \in J} S_j$ and $\displaystyle \bigcap_{k \mathop \in \relcomp I J} S_k$ are sets. Hence by Intersection is Commutative we have: :$\displaystyle \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k = \bigcap_{k \mathop \in \relcomp I J} S_k \cap \bigcap_{j \mathop \in J} S_j$ It remains to be demonstrated that $\displaystyle \bigcap_{i \mathop \in I} S_i = \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k$. So: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \bigcap_{i \mathop \in I} S_i       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \forall i \in I: x       | o = \in       | r = S_i       | c = {{Defof|Intersection of Family}} }} {{eqn | ll= \leadstoandfrom       | l = \forall j \in J: x       | o = \in       | r = S_j       | c = {{Defof|Relative Complement}} }} {{eqn | lo= \land       | l = \forall k \in \relcomp I J: x       | o = \in       | r = S_k       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \bigcap_{j \mathop \in J} S_j       | c = {{Defof|Intersection of Family}} }} {{eqn | lo= \land       | l = x       | o = \in       | r = \bigcap_{k \mathop \in \relcomp I J} S_k       | c = {{Defof|Intersection of Family}} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k       | c = {{Defof|Set Intersection}} }} {{end-eqn}} That is: :$\displaystyle x \in \bigcap_{i \mathop \in I} S_i \iff x \in \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k$ The result follows by definition of set equality. {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {x^{p - 1} \ln x} {1 + x} \rd x 	| r = \int_0^\infty \frac 1 {1 + x} \map {\frac \partial {\partial p} } {x^{p - 1} } \rd x 	| c = Derivative of Power of Constant }} {{eqn	| r = \frac \d {\d p} \int_0^\infty \frac {x^{p - 1} } {1 + x} \rd x 	| c = Definite Integral of Partial Derivative }} {{eqn	| r = \map {\frac \d {\d p} } {\pi \csc p \pi} 	| c = Definite Integral to Infinity of $\dfrac {x^{p - 1} } {1 + x}$ }} {{eqn	| r = -\pi^2 \csc p \pi \cot p \pi 	| c = Chain Rule for Derivatives, Derivative of Cosecant Function }} {{end-eqn}} {{qed}}
By Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$: Lemma, we have:  :$\displaystyle \int_0^\pi \map \ln {\sin x} \rd x = 2 \int_0^{\pi/2} \map \ln {\sin x} \rd x$ By Product of Sines of Fractions of Pi, we have:  :$\displaystyle \prod_{k \mathop = 1}^{n - 1} \map \sin {\frac {k \pi} n} = \frac n {2^{n - 1} }$ Therefore, we have:  {{begin-eqn}} {{eqn	| l = \map \ln {\prod_{k \mathop = 1}^{n - 1} \map \sin {\frac {k \pi} n} } 	| r = \sum_{k \mathop = 1}^{n - 1} \map \ln {\map \sin {\frac {k \pi} n} } 	| c = Sum of Logarithms: Natural Logarithm }} {{eqn	| r = \map \ln {\frac n {2^{n - 1} } } }} {{eqn	| r = \ln n - \paren {n - 1} \ln 2 	| c = Difference of Logarithms }} {{end-eqn}} We have:  {{begin-eqn}} {{eqn	| l = \int_0^\pi \map \ln {\sin x} \rd x 	| r = \lim_{n \mathop \to \infty} \frac \pi n \sum_{k \mathop = 1}^{n - 1} \map \ln {\map \sin {\frac {\pi k} n} }  	| c = {{Defof|Riemann Integral}} }} {{eqn	| r = \lim_{n \mathop \to \infty} \paren {\frac \pi n \ln n - \pi \paren {1 - \frac 1 n} \ln 2} }} {{end-eqn}} We can show the first term to vanish:  {{begin-eqn}} {{eqn	| l = \lim_{n \mathop \to \infty} \frac \pi n \ln n 	| r = \pi \lim_{u \mathop \to \infty} u e^{-u} 	| c = letting $n = e^u$ }} {{eqn	| r = 0 	| c = Limit to Infinity of $x^n e^{-a x}$ }} {{end-eqn}} So:  {{begin-eqn}} {{eqn	| l = \lim_{n \mathop \to \infty} \paren {\frac \pi n \ln n - \paren {1 - \frac 1 n} \ln 2} 	| r = -\pi \ln 2 \lim_{n \mathop \to \infty} \paren {1 - \frac 1 n} 	| c = Combination Theorem for Limits of Functions: Combined Sum Rule }} {{eqn	| r = -\pi \ln 2 }} {{end-eqn}} giving: :$\displaystyle \int_0^{\pi/2} \map \ln {\sin x} \rd x = -\frac \pi 2 \ln 2$ {{qed}}
By Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$: Proof 1 we have:  :$\displaystyle \int_0^{\pi/2} \map \ln {\sin x} \rd x = \int_0^{\pi/2} \map \ln {\cos x} \rd x$ and: :$\displaystyle \int_0^{\pi/2} \map \ln {\sin x} \rd x = -\frac \pi 2 \ln 2$ The result follows. {{qed}}
From: :P-adic Norm is Non-Archimedean Norm :Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary  it follows that: :$\exists N \in \N: \forall n, m \ge N: \norm {x_n}_p = \norm {x_m}_p$ {{qed}}
We can write:  {{begin-eqn}} {{eqn	| l = \int_0^\infty \map \ln {\frac {e^x + 1} {e^x - 1} } \rd x 	| r = \int_0^\infty \map \ln {\frac {e^{x/2} \paren {e^{x/2} + e^{-x/2} } } {e^{x/2} \paren {e^{x/2} - e^{-x/2} } } } \rd x }} {{eqn	| r = \int_0^\infty \map \ln {\coth \frac x 2} \rd x 	| c = {{Defof|Hyperbolic Cotangent}} }} {{end-eqn}} Let:  :$u = \coth \dfrac x 2$ We have, by Derivative of Hyperbolic Cotangent Function: :$\dfrac {\d u} {\d x} = -\dfrac 1 2 \csch^2 \dfrac x 2$ From Difference of Squares of Hyperbolic Cotangent and Cosecant, this can be written:  :$\dfrac {\d u} {\d x} = \dfrac 1 2 \paren {1 - \coth^2 \dfrac x 2} = \dfrac 1 2 \paren {1 - u^2}$ From Limit to Infinity of Hyperbolic Cotangent Function, we have:  :as $x \to \infty$, $u \to 1$. We also have:  :as $x \to 0^+$, $u \to \infty$. With this, we have:  {{begin-eqn}} {{eqn	| l = \int_0^\infty \map \ln {\coth \frac x 2} \rd x 	| r = 2 \int_\infty^1 \frac {\ln u} {1 - u^2} \rd u 	| c = substituting $u = \coth \dfrac x 2$ }} {{eqn	| r = -2 \int_1^\infty \frac {\ln u} {1 - u^2} \rd u 	| c = Reversal of Limits of Definite Integral }} {{eqn	| r = -2 \int_1^0 \paren {-\frac 1 {v^2} } \frac {\map \ln {\frac 1 v} } {1 - \paren {\frac 1 {v^2} }^2} \rd v 	| c = substituting $v = \dfrac 1 v$ }} {{eqn	| r = -2 \int_0^1 \frac {\ln v} {1 - v^2} \rd v 	| c = Reversal of Limits of Definite Integral }} {{eqn	| r = -2 \int_0^1 \ln v \paren {\sum_{n \mathop = 0}^\infty \paren {v^2}^n} \rd v 	| c = Sum of Geometric Sequence }} {{eqn	| r = -2 \sum_{n \mathop = 0}^\infty \int_0^1 v^{2 n} \ln v \rd v 	| c = Fubini's Theorem }} {{eqn	| r = 2 \sum_{n \mathop = 0}^\infty \frac {\map \Gamma 2} {\paren {2 n + 1}^2} 	| c = Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$ }} {{eqn	| r = 2 \times 1! \times \frac {\pi^2} 8 	| c = Gamma Function Extends Factorial, Sum of Reciprocals of Squares of Odd Integers }} {{eqn	| r = \frac {\pi^2} 4 }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^{\pi/4} \map \ln {1 + \tan x} \rd x 	| r = \int_0^{\pi/4} \map \ln {1 + \map \tan {\frac \pi 4 - x} } \rd x 	| c = Integral between Limits is Independent of Direction }} {{eqn	| r = \int_0^{\pi/4} \map \ln {1 + \frac {\tan \frac \pi 4 - \tan x} {1 + \tan \frac \pi 4 \tan x} } \rd x 	| c = Tangent of Difference }} {{eqn	| r = \int_0^{\pi/4} \map \ln {1 + \frac {1 - \tan x} {1 + \tan x} } \rd x 	| c = Tangent of 45 Degrees }} {{eqn	| r = \int_0^{\pi/4} \map \ln {\frac 2 {1 + \tan x} } \rd x }} {{eqn	| r = \int_0^{\pi/4} \ln 2 \rd x - \int_0^{\pi/4} \map \ln {1 + \tan x} \rd x 	| c = Difference of Logarithms }} {{eqn	| r = \frac \pi 4 \ln 2 - \int_0^{\pi/4} \map \ln {1 + \tan x} \rd x 	| c = Primitive of Constant }} {{end-eqn}} Therefore:  :$\displaystyle 2 \int_0^{\pi/4} \map \ln {1 + \tan x} \rd x = \frac \pi 4 \ln 2$ giving:  :$\displaystyle \int_0^{\pi/4} \map \ln {1 + \tan x} \rd x = \frac \pi 8 \ln 2$ {{qed}}
We have:  {{begin-eqn}} {{eqn	| l = \map {\frac \d {\d x} } {\map \ln {\sin x} }  	| r = \map {\frac \d {\map \d {\sin x} } } {\map \ln {\sin x} } \map {\frac \d {\d x} } {\sin x} 	| c = Chain Rule for Derivatives }} {{eqn	| r = \frac {\cos x} {\sin x} 	| c = Derivative of Logarithm Function, Derivative of Sine Function }} {{end-eqn}} We also have, by Primitive of Sine Function: :$\ds \int \sin x \rd x = -\cos x + C$ So:  {{begin-eqn}} {{eqn	| l = \int \sin x \map \ln {\sin x} \rd x 	| r = -\cos x \map \ln {\sin x} + \int \frac {\cos^2 x} {\sin x} \rd x 	| c = Integration by Parts }} {{eqn	| r = -\cos x \map \ln {\sin x} + \int \frac {1 - \sin^2 x} {\sin x} \rd x 	| c = Sum of Squares of Sine and Cosine }} {{eqn	| r = -\cos x \map \ln {\sin x} + \int \csc x \rd x - \int \sin x \rd x }} {{eqn	| r = -\cos x \map \ln {\sin x} + \ln \size {\tan \frac x 2} + \cos x + C 	| c = Primitive of Sine Function, Primitive of $\csc x$: Tangent Form }} {{eqn	| r = \cos x \paren {1 - \map \ln {\sin x} } + \ln \size {\tan \frac x 2} + C }} {{end-eqn}} {{qed}} Category:Primitives involving Sine Function Category:Primitives involving Logarithm Function qvetd42p66dfj06kdl355f9xzlg3ah0
From Characterization of Cosine Integral Function, we have:  :$\displaystyle \int_t^\infty \frac {\cos x} x \rd x = -\gamma - \ln t + \int_0^t \frac {1 - \cos x} x \rd x$ for all real $t > 0$. Rearranging:  :$\displaystyle \int_0^t \frac {1 - \cos x} x \rd x - \int_t^\infty \frac {\cos x} x \rd x = \gamma + \ln t$ Setting $t = 1$ gives the result, from Natural Logarithm of 1 is 0. {{qed}}
Let:  :$x = \sin \theta$ By Derivative of Sine Function, we have:  :$\dfrac {\d x} {\d \theta} = \cos \theta$ We have, by Arcsine of Zero is Zero:  :as $x \to 0$, $\theta \to \arcsin 0 = 0$. By Arcsine of One is Half Pi, we have:  :as $x \to 1$, $\theta \to \arcsin 1 = \dfrac \pi 2$. We have: {{begin-eqn}} {{eqn	| l = \int_0^1 \frac {\arcsin x} x \rd x 	| r = \int_0^{\pi/2} \frac {\cos \theta \map \arcsin {\sin \theta} } {\sin \theta} \rd \theta 	| c = substituting $x = \sin \theta$  }} {{eqn	| r = \int_0^{\pi/2} \theta \cot \theta \rd \theta 	| c = {{Defof|Arcsine}}, {{Defof|Real Cotangent Function}} }} {{end-eqn}} By Primitive of Cotangent Function: :$\displaystyle \int \cot \theta \rd \theta = \map \ln {\sin \theta} + C$ So: {{begin-eqn}} {{eqn	| l = \int_0^{\pi/2} \theta \cot \theta \rd \theta 	| r = \intlimits {\theta \map \ln {\sin \theta} } 0 {\frac \pi 2} - \int_0^{\pi/2} \map \ln {\sin \theta} \rd \theta 	| c = Integration by Parts }} {{eqn	| r = \frac \pi 2 \map \ln {\sin \frac \pi 2} - \lim_{\theta \to 0^+} \paren {\theta \map \ln {\sin \theta} } + \frac \pi 2 \ln 2 	| c = Definite Integral from 0 to $\dfrac \pi 2$ of $\map \ln {\sin x}$ }} {{eqn	| r = -\lim_{\theta \to 0^+} \paren {\theta \map \ln {\sin \theta} } + \frac \pi 2 \ln 2 	| c = Sine of Right Angle, Natural Logarithm of 1 is 0 }}  {{end-eqn}} We have:  {{begin-eqn}} {{eqn	| l = \lim_{\theta \to 0^+} \paren {\theta \map \ln {\sin \theta} } 	| r = \lim_{\theta \to 0^+} \paren {\theta \map \ln {\frac {\sin \theta} \theta \theta} } }} {{eqn	| r = \lim_{\theta \to 0^+} \paren {\theta \map \ln {\frac {\sin \theta} \theta} } + \lim_{\theta \to 0^+} \theta \ln \theta 	| c = Sum of Logarithms, Combination Theorem for Limits of Functions: Sum Rule }} {{eqn	| r = \paren {\lim_{\theta \to 0^+} \theta} \paren {\map \ln {\lim_{\theta \to 0^+} \frac {\sin \theta} \theta} } + \lim_{\theta \to 0^+} \theta \ln \theta 	| c = Combination Theorem for Limits of Functions: Product Rule }} {{eqn	| r = 0 \ln 1 + 0 	| c = Limit of $\dfrac {\sin x} x$, Limit of $x^n \paren {\ln x}^m$ }} {{eqn	| r = 0 }} {{end-eqn}} giving:  :$\displaystyle \int_0^{\pi/2} \theta \cot \theta \rd \theta = \frac \pi 2 \ln 2$ hence the result. {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^1 \frac {\arctan x} x \rd x 	| r = \int_0^1 \frac 1 x \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {2 n + 1} } \rd x 	| c = Power Series Expansion for Real Arctangent Function }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {2 n + 1} \int_0^1 x^{2 n} \rd x 	| c = Power Series is Termwise Integrable within Radius of Convergence }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^2} 	| c = Primitive of Power }} {{eqn	| r = G 	| c = {{Defof|Catalan's Constant}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k       | r = \map \complement {\map \complement {\bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k} }       | c = Complement of Complement }} {{eqn | r = \map \complement {\map \complement {\bigcap_{j \mathop \in J} S_j} \cup \map \complement {\bigcap_{k \mathop \in \relcomp I J} S_k} }       | c = De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection }} {{eqn | n = 1       | r = \map \complement {\bigcup_{j \mathop \in J} \map \complement {S_j} \cup \bigcup_{k \mathop \in \relcomp I J} \map \complement {S_k} }       | c = De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Intersection }} {{eqn | r = \map \complement {\bigcup_{k \mathop \in \relcomp I J} \map \complement {S_k} \cup \bigcup_{j \mathop \in J} \map \complement {S_j} }       | c = General Commutativity of Set Union }} {{eqn | r = \map \complement {\map \complement {\bigcap_{k \mathop \in \relcomp I J} S_k}\cup \map \complement {\bigcap_{j \mathop \in J} S_j} }       | c = De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Intersection }} {{eqn | r = \map \complement {\map \complement {\bigcap_{k \mathop \in \relcomp I J} \cap S_k \bigcap_{j \mathop \in J} S_j} }       | c = De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection }} {{eqn | r = \bigcap_{k \mathop \in \relcomp I J} S_k \cap \bigcap_{j \mathop \in J} S_j       | c = Complement of Complement }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \bigcap_{i \mathop \in I} S_i       | r = \map \complement {\map \complement {\bigcap_{i \mathop \in I} S_i} }       | c = Complement of Complement }} {{eqn | r = \map \complement {\bigcup_{i \mathop \in I} \map \complement {S_i} }       | c = De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Intersection }} {{eqn | r = \map \complement {\bigcup_{j \mathop \in J} \map \complement {S_j} \cup \bigcup_{k \mathop \in \relcomp I J} \map \complement {S_k} }       | c = General Commutativity of Set Union }} {{eqn | r = \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k       | c = from $(1)$ above }} {{end-eqn}} {{qed}}
From Definite Integral from $0$ to $1$ of $\dfrac {\arctan x} x$, we have:  :$\displaystyle \int_0^1 \frac {\arctan x} x \rd x = G$ Let:  :$x = \tan \theta$  By Derivative of Tangent Function, we have:  :$\displaystyle \frac {\d x} {\d \theta} = \sec^2 \theta$ We have, by Arctangent of Zero is Zero:  :as $x \to 0$, $\theta \to 0$. We also have, by Arctangent of One: :as $x \to 1$, $\theta \to \dfrac \pi 4$ We therefore have:  {{begin-eqn}} {{eqn	| l = \int_0^1 \frac {\arctan x} x \rd x 	| r = \int_0^{\pi/4} \frac {\sec^2 \theta \map \arctan {\tan \theta} } {\tan \theta} \rd \theta 	| c = substituting $x = \tan \theta$ }} {{eqn	| r = \int_0^{\pi/4} \frac \theta {\cos^2 \theta} \times \frac {\cos \theta} {\sin \theta} \rd \theta 	| c = {{Defof|Real Secant Function}}, {{Defof|Real Tangent Function}} }} {{eqn	| r = \int_0^{\pi/4} \frac \theta {\sin \theta \cos \theta} \rd \theta }} {{eqn	| r = \int_0^{\pi/4} \frac \theta {\frac 1 2 \sin 2 \theta} \rd \theta 	| c = Double Angle Formula for Sine }} {{eqn	| r = \frac 1 2 \int_0^{\pi/2} \frac \phi {\sin \phi} \rd \phi 	| c = substituting $\phi = 2 \theta$ }} {{end-eqn}} giving:  :$\displaystyle \int_0^{\pi/2} \frac \phi {\sin \phi} \rd \phi = 2 G$ {{qed}}
Note that the integrand is of the form:  :$\displaystyle \int_0^\infty \frac {\map f {p x} - \map f {q x} } x \rd x$ where: :$\map f x = \arctan x$ We have, by Derivative of Arctangent Function: :$\map {f'} x = \dfrac 1 {1 + x^2}$ which is continuous on $\R$.  By Limit to Infinity of Arctangent Function:  :$\displaystyle \lim_{x \mathop \to \infty} \map f x =  \lim_{x \mathop \to \infty} \arctan x = \frac \pi 2$ As $f$ is continuously differentiable and $\displaystyle \lim_{x \mathop \to \infty} \map f x$ exists and is finite, we may apply Frullani's Integral, giving: {{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\arctan p x - \arctan q x} x \rd x 	| r = \paren {\lim_{x \mathop \to \infty} \arctan x - \arctan 0} \ln \frac p q }} {{eqn	| r = \frac \pi 2 \ln \frac p q 	| c = Arctangent of Zero is Zero }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^{\pi/2} \sin x \map \ln {\sin x} \rd x 	| r = \intlimits {\cos x \paren {1 - \map \ln {\sin x} } + \map \ln {\tan \frac x 2} } 0 {\pi/2} 	| c = Primitive of $\sin x \map \ln {\sin x}$ }} {{eqn	| r = \cos \frac \pi 2 \paren {1 - \map \ln {\sin \frac \pi 2} } + \map \ln {\tan \frac \pi 4} - \cos 0 + \lim_{x \mathop \to 0^+} \paren {\map \ln {\sin x} - \map \ln {\tan \frac x 2} } }} {{eqn	| r = -1 + \lim_{x \mathop \to 0^+} \paren {\map \ln {\sin x} - \map \ln {\tan \frac x 2} } 	| c = Cosine of Right Angle, Tangent of 45 Degrees, Natural Logarithm of 1 is 0, Cosine of Zero is One }} {{end-eqn}} It remains to compute:  :$\displaystyle \lim_{x \mathop \to 0^+} \paren {\map \ln {\sin x} - \map \ln {\tan \frac x 2} }$ We have: {{begin-eqn}} {{eqn	| l = \lim_{x \mathop \to 0^+} \paren {\map \ln {\sin x} - \map \ln {\tan \frac x 2} } 	| r = \lim_{x \mathop \to 0^+} \paren {\map \ln {2 \sin \frac x 2 \cos \frac x 2} - \map \ln {\sin \frac x 2} + \map \ln {\cos \frac x 2} } 	| c = Double Angle Formula for Sine, {{Defof|Real Tangent Function}}, Difference of Logarithms }} {{eqn	| r = \lim_{x \mathop \to 0^+} \paren {\ln 2 + \map \ln {\sin \frac x 2} + \map \ln {\cos \frac x 2} - \map \ln {\sin \frac x 2} + \map \ln {\cos \frac x 2} } 	| c = Sum of Logarithms }} {{eqn	| r = \ln 2 + 2 \lim_{x \mathop \to 0^+} \map \ln {\cos \frac x 2} 	| c = Combination Theorem for Limits of Functions: Sum Rule }} {{eqn	| r = \ln 2 + 2 \map \ln {\lim_{x \mathop \to 0^+} \cos \frac x 2} }} {{eqn	| r = \ln 2 + 2 \ln 1 	| c = Cosine of Zero is One }} {{eqn	| r = \ln 2 	| c = Natural Logarithm of 1 is 0 }} {{end-eqn}} giving:  :$\displaystyle \int_0^{\pi/2} \sin x \map \ln {\sin x} \rd x = \ln 2 - 1$ {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\pi x \, \map \ln {\sin x} \rd x 	| r = \int_0^\pi \paren {\pi - x} \, \map \ln {\map \sin {\pi - x} } \rd x 	| c = Integral between Limits is Independent of Direction }} {{eqn	| r = \pi \int_0^\pi \map \ln {\sin x} - \int_0^\pi x \, \map \ln {\sin x} \rd x 	| c = Sine of Supplementary Angle, Linear Combination of Definite Integrals }} {{end-eqn}} So:  {{begin-eqn}} {{eqn	| l = 2 \int_0^\pi x \, \map \ln {\sin x} \rd x 	| r = \pi \int_0^\pi \map \ln {\sin x} \rd x }} {{eqn	| r = 2 \pi \int_0^{\pi/2} \map \ln {\sin x} \rd x 	| c = Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$: Lemma }} {{eqn	| r = -\pi^2 \ln 2 	| c = Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$ }} {{end-eqn}} giving:  :$\displaystyle \int_0^\pi x \, \map \ln {\sin x} \rd x = -\frac {\pi^2} 2 \ln 2$ {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \map \exp {-\paren {a x^2 + bx + c} } \rd x 	| r = \int_0^\infty \map \exp {-a \paren {x + \frac b {2 a} }^2 + \frac {b^2} {4 a} - c} \rd x 	| c = Completing the Square }} {{eqn	| r = \map \exp {\frac {b^2 - 4 a c} {4 a} } \int_0^\infty \map \exp {-a \paren {x + \frac b {2 a} }^2} \rd x 	| c = Exponential of Sum }} {{eqn	| r = \map \exp {\frac {b^2 - 4 a c} {4 a} } \int_0^\infty \map \exp {-\paren {\sqrt a x + \frac b {2 \sqrt a} }^2} \rd x }} {{eqn	| r = \frac 1 {\sqrt a} \map \exp {\frac {b^2 - 4 a c} {4 a} } \int_{\frac b {2 \sqrt a} }^\infty \map \exp {-u^2} \rd u 	| c = substituting $u = \sqrt a x + \dfrac b {2 \sqrt a}$ }} {{eqn	| r = \frac 1 2 \sqrt {\frac \pi a} \map \exp {\frac {b^2 - 4 a c} {4 a} } \paren {\frac 2 {\sqrt \pi} \int_{\frac b {2 \sqrt a} }^\infty \map \exp {-u^2} \rd u} }} {{eqn	| r = \frac 1 2 \sqrt {\frac \pi a} \map \exp {\frac {b^2 - 4 a c} {4 a} } \map \erfc {\frac b {2 \sqrt a} } 	| c = {{Defof|Complementary Error Function}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = S       | o = \subseteq       | r = T_1 }} {{eqn | lo= \land       | l = S       | o = \subseteq       | r = T_2 }} {{eqn | ll= \leadsto       | l = S \cap S       | o = \subseteq       | r = S \cap T_2       | c = Set Intersection Preserves Subsets }} {{eqn | ll= \leadsto       | l = S       | o = \subseteq       | r = T_1 \cap T_2       | c = Intersection is Idempotent }} {{end-eqn}} {{Qed}}
{{begin-eqn}} {{eqn	| l = \int_{-\infty}^\infty \map \exp {-\paren {a x^2 + b x + c} } \rd x 	| r = \int_{-\infty}^\infty \map \exp {-a \paren {x + \frac b {2 a} }^2 + \frac {b^2} {4 a} - c} \rd x 	| c = Completing the Square }} {{eqn	| r = \map \exp {\frac {b^2 - 4 a c} {4 a} } \int_{-\infty}^\infty \map \exp {-a \paren {x + \frac b {2 a} }^2} \rd x 	| c = Exponential of Sum }} {{eqn	| r = \map \exp {\frac {b^2 - 4 a c} {4 a} } \int_{-\infty}^\infty \map \exp {-\paren {\sqrt a x + \frac b {2 \sqrt a} }^2} \rd x }} {{eqn	| r = \frac 1 {\sqrt a} \map \exp {\frac {b^2 - 4 a c} {4 a} } \int_{-\infty}^\infty \map \exp {-u^2} \rd u 	| c = substituting $u = \sqrt a + \dfrac b {2 \sqrt a}$ }} {{eqn	| r = \sqrt {\frac \pi a} \map \exp {\frac {b^2 - 4 a c} {4 a} } 	| c = Gaussian Integral }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty x^m e^{-a x^2} \rd x 	| r = \int_0^\infty \frac 1 {2 \sqrt t} \paren {\sqrt t}^m e^{-a t} \rd t 	| c = substituting $t = x^2$ }} {{eqn	| r = \frac 1 2 \int_0^\infty t^{\paren {\paren {m + 1}/2} - 1} e^{-a t} \rd t }} {{eqn	| r = \frac {\map \Gamma {\paren {m + 1}/2} } {2 a^{\paren {m + 1}/2} } 	| c = Laplace Transform of Real Power }} {{end-eqn}} {{qed}}