hcju/proofwikiqa huggingface dataset:
By Derivative of Error Function, we have: :$\displaystyle \frac \d {\d x} \paren {\map \erf x} = \frac 2 {\sqrt \pi} e^{-x^2}$ So: {{begin-eqn}} {{eqn	| l = \int \map \erf x \rd x 	| r = \int 1 \times \map \erf x \rd x }} {{eqn	| r = x \, \map \erf x - \frac 2 {\sqrt \pi} \int x e^{-x^2} \rd x 	| c = Integration by Parts }} {{eqn	| r = x \, \map \erf x + \frac 1 {\sqrt \pi} \int \paren {-2 x e^{-x^2} } \rd x }} {{eqn	| r = x \, \map \erf x + \frac 1 {\sqrt \pi} \int e^u \rd u 	| c = substituting $u = -x^2$ }} {{eqn	| r = x \, \map \erf x + \frac 1 {\sqrt \pi} e^u + C 	| c = Primitive of Exponential Function }} {{eqn	| r = x \, \map \erf x + \frac 1 {\sqrt \pi} e^{-x^2} + C 	| c = substituting back for $u$ }} {{end-eqn}} {{qed}} Category:Primitives Category:Error Function 26pm880cg2avcbfzg84zbqwr0t8bboq
We have, by the definition of the error function: :$\displaystyle \map \erf x = \frac 2 {\sqrt \pi} \int_0^x e^{-t^2} \rd t$ By Fundamental Theorem of Calculus (First Part): Corollary, we therefore have: :$\displaystyle \frac \d {\d x} \paren {\map \erf x} = \frac 2 {\sqrt \pi} e^{-x^2}$ {{qed}} Category:Error Function Category:Derivatives n5fehqlwytzrt5hcmwcbui8w1pmxs7j
{{begin-eqn}} {{eqn	| l = \lim_{x \mathop \to \infty} \map \erf x 	| r = \lim_{x \mathop \to \infty} \paren {\frac 2 {\sqrt \pi} \int_0^x e^{-t^2} \rd t} 	| c = {{Defof|Error Function}} }} {{eqn	| r = \frac 2 {\sqrt \pi} \int_0^\infty e^{-t^2} \rd t 	| c = Combination Theorem for Limits of Functions: Multiple Rule }} {{eqn	| r = \frac 2 {\sqrt \pi} \times \frac {\sqrt \pi} 2 	| c = Integral to Infinity of Exponential of -t^2 }} {{eqn	| r = 1 }} {{end-eqn}} {{qed}}
=== Sufficient Condition === Let $\RR$ be right-total. Then by definition: :$\forall t \in T: \exists s \in S: \tuple {s, t} \in \RR$ By definition of the inverse of $\RR$, it follows that: :$\forall t \in T: \exists s \in S: \tuple {t, s} \in \RR^{-1}$ So by definition $\RR^{-1}$ is left-total. {{qed|lemma}} === Necessary Condition === Let $\RR^{-1}$ is left-total. Then by definition: :$\forall t \in T: \exists s \in S: \tuple {t, s} \in \RR^{-1}$ and so: :$\forall t \in T: \exists s \in S: \tuple {s, t} \in \RR$ So by definition $\RR$ is right-total. {{qed}} Category:Inverse Relations suxviib9m4sioo5xy1cjieuu7j2xjgy
From Inverse of Inverse Relation, the inverse of $\mathcal R^{-1}$ is $\mathcal R$. From Inverse of Right-Total Relation is Left-Total: :$\mathcal R^{-1}$ is right-total {{iff}} $\mathcal R$ is left-total. Hence the result. {{qed}} Category:Inverse Relations 8kjwaxxqqmf3s7rh3dmy1riukyk0hbw
We have that $f$ is a mapping. Hence $f$ is a fortiori a left-total relation. Then from Inverse of Left-Total Relation is Right-Total, $f^{-1}$ is right-total. {{Qed}} Category:Inverse Relations Category:Inverse Mappings 8z3bk7s5ooiityhy08py1bfdkd5nu1q
Let $\RR$ be a one-to-one relation. Let $\RR^{-1}$ denote its inverse By definition, $\RR$ is a relation which is both many-to-one and one-to-many. From Inverse of Many-to-One Relation is One-to-Many: :$\RR^{-1}$ is both one-to-many and many-to-one. Hence the result by definition of one-to-one relation. {{qed}} Category:Inverse Relations 38i2x22s1p8077snqedjfr59djnaphc
We are given that $f$ is an injective mapping. Hence by definition $f$ is a one-to-one relation. The result follows from from Inverse of One-to-One Relation is One-to-One. {{Qed}} Category:Inverse Relations Category:Injections Category:Inverse Mappings jujitmkas2e8cfyh5sv1yrpnynt98g1
We are given that $f$ is a surjective mapping. By Inverse of Mapping is Right-Total Relation, $f^{-1}$ is a right-total relation. By definition of surjection, $f$ is itself a right-total relation. From Inverse of Right-Total Relation is Left-Total, $f^{-1}$ is a left-total relation. Hence the result. {{Qed}} Category:Inverse Relations Category:Surjections Category:Inverse Mappings bkmk09kupdupdsdbbzy0pibrd208gjx
Let $f^{-1}: T \to S$ be a mapping. Let $\map f {x_a} = y$ and $\map f {x_b} = y$. Then: {{begin-eqn}} {{eqn | l = \tuple {x_a, y}       | o = \in       | r = f       | c = {{Defof|Mapping}} }} {{eqn | lo= \land       | l = \tuple {x_b, y}       | o = \in       | r = f       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {y, x_a}       | o = \in       | r = f^{-1}       | c = {{Defof|Inverse of Mapping}} }} {{eqn | lo= \land       | l = \tuple {y, x_b}       | o = \in       | r = f^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = x_a       | r = x_b       | c = {{Defof|Mapping|index = 4}}: $f^{-1}$ is many-to-one }} {{end-eqn}} Thus, by definition, $f$ is an injection. {{qed}}
Let $f^{-1}: T \to S$ be a mapping. {{begin-eqn}} {{eqn | l = \map f x       | r = \map f y       | c =  }} {{eqn | ll= \leadsto       | l = \map {f^{-1} } {\map f x}       | r = \map {f^{-1} } {\map f y}       | c = as $f^{-1}$ is a mapping }} {{eqn | ll= \leadsto       | l = x       | r = y       | c = {{Defof|Inverse Mapping|index = 1}}: as $\map {f^{-1} } {\map f x}$ and so on }} {{end-eqn}} Thus $f$ is by definition an injection. {{qed}}
Let $f^{-1}: T \to S$ be a mapping. We have: {{begin-eqn}} {{eqn | l = t       | o = \in       | r = T       | c =  }} {{eqn | ll= \leadsto       | l = \map {f^{-1} } t       | o = \in       | r = S       | c = as $f^{-1}$ is a mapping }} {{eqn | ll= \leadsto       | l = \map f {\map {f^{-1} } t}       | r = t       | c = {{Defof|Inverse Mapping|index = 1}} }} {{eqn | ll= \leadsto       | lo= \exists s \in S:       | l = \map f s       | r = t       | c = setting $s = \map {f^{-1} } t$ }} {{end-eqn}} Thus $f$ is by definition a surjection. {{qed}}
By definition, $d$ is invertible {{iff}} $d$ is a bijection. It is sufficient to demonstrate that $d$ is not an injection. Hence a fortiori $d$ is shown to not be a bijection. Consider a differentiable function $f \in B$. Then consider the function $g \in B$ defined as: :$\forall x \in \Bbb I: \map g x = \map f x + C$ where $C \in \R$ is a constant. Then we have that: {{begin-eqn}} {{eqn | l = \map {D_x} g       | r = \map {D_x} f       | c = Derivative of Function plus Constant }} {{eqn | ll= \leadsto       | l = \map d f       | r = \map d g       | c = Definition of $d$ }} {{end-eqn}} demonstrating that $d$ is not an injection. The result follows. {{qed}}
=== Definition 1 iff Definition 2 === {{:Equivalence of Definitions of Convergent Sequence in Metric Space/Definition 1 iff Definition 2}}{{qed|lemma}} === Definition 1 iff Definition 3 === {{:Equivalence of Definitions of Convergent Sequence in Metric Space/Definition 1 iff Definition 3}}{{qed|lemma}} === Definition 2 implies Definition 4 === {{:Equivalence of Definitions of Convergent Sequence in Metric Space/Definition 2 implies Definition 4}}{{qed|lemma}} === Definition 4 implies Definition 2 === {{:Equivalence of Definitions of Convergent Sequence in Metric Space/Definition 4 implies Definition 2}}{{qed}} Category:Metric Spaces Category:Convergence Category:Sequences Category:Convergent Sequences (Metric Space) Category:Equivalence of Definitions of Convergent Sequence in Metric Space 864bpf0i2do43t0vip30y80zhx1hbd6
Let a fixed $\epsilon \in \R{>0}$ be selected. Then: :$\exists N \in \R_{>0}: \forall n \in \N: n > N \implies x_n \in \map {B_\epsilon} l$ Hence the only $x_k$ that cannot be in the open $\epsilon$-ball $\map {B_\epsilon} l$ of $l$ are those for which $n \le N$. There are finitely many of these.
Let $\map {B_\epsilon} l$ be any open $\epsilon$-ball of $l$. Let $A = \set {n: x_n \notin \map {B_\epsilon} l}$. By assumption $A$ is finite. From Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements, any finite subset of $\N$ has a maximum. Let $N$ be the maximum of $A$. Then for every $n > N$, $x_n$ must be in the open $\epsilon$-ball $\map {B_\epsilon} l$.
{{begin-eqn}} {{eqn	| l = \int_0^{2 \pi} \frac {\d x} {a + b \cos x} 	| r = \int_0^\pi \frac {\d x} {a + b \cos x} + \int_\pi^{2 \pi} \frac {\d x} {a + b \cos x} 	| c = Sum of Integrals on Adjacent Intervals for Integrable Functions }} {{eqn	| r = \intlimits {\frac 2 {\sqrt {a^2 - b^2} } \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} } 0 \pi + \intlimits {\frac 2 {\sqrt {a^2 - b^2} } \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} } \pi {2 \pi} 	| c = Primitive of $\dfrac 1 {p + q \cos a x}$ }} {{eqn	| r = \frac 2 {\sqrt {a^2 - b^2} } \paren {\lim_{x \mathop \to \pi^+} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} - \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan 0} } }} {{eqn	| o =  	| ro= + 	| r = \frac 2 {\sqrt {a^2 - b^2} } \paren {\map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \pi} - \lim_{x \mathop \to \pi^-} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} } }} {{eqn	| r = \frac 2 {\sqrt {a^2 - b^2} } \paren {\lim_{x \mathop \to \pi^+} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} - \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan 0} } 	| c = Tangent Function is Periodic on Reals }} {{eqn	| o =  	| ro= + 	| r = \frac 2 {\sqrt {a^2 - b^2} } \paren { \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan 0} - \lim_{x \mathop \to \pi^-} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} } 	| c =  }} {{eqn	| r = \frac 2 {\sqrt {a^2 - b^2} } \paren {\lim_{x \mathop \to \pi^+} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} - \lim_{x \mathop \to \pi^-} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} } 	| c = simplfying }} {{eqn	| r = \frac 2 {\sqrt {a^2 - b^2} } \paren {\lim_{u \mathop \to \infty} \map \arctan {\sqrt {\frac {a - b} {a + b} } u} - \lim_{u \mathop \to -\infty} \map \arctan {\sqrt {\frac {a - b} {a + b} } u} } 	| c = letting $u = \tan \dfrac x 2$: Tangent Function $\to \pm \infty$ }} {{eqn	| r = \frac 2 {\sqrt {a^2 - b^2} } \paren {\frac \pi 2 - \paren {-\frac \pi 2} } 	| c = Limit to Positive and Negative Infinity of Arctangent Function }} {{eqn	| r = \frac {2 \pi} {\sqrt {a^2 - b^2} }  }} {{end-eqn}} {{qed}}
Let $f \in A$ be an arbitrary continuous real function $f: \Bbb I \to \R$. From Continuous Real Function is Darboux Integrable, $\map {\paren {\map h f} } x$ exists and is continuous on $\Bbb I$. Let $x = a$. Then we have: {{begin-eqn}} {{eqn | l = \map {\paren {\map h f} } x       | r = \int_a^x \map f t \rd t       | c = Definition of $h$ }} {{eqn | ll= \leadsto       | l = \map {\paren {\map h f} } a       | r = \int_a^a \map f t \rd t       | c =  }} {{eqn | r = 0       | c = Definite Integral on Zero Interval }} {{end-eqn}} Thus $h$ is a continuous real function on $\Bbb I$ such that $\map h a = 0$. Further, by definition, $\int \map f t \rd t$ is the primitive of $f$, and exists by dint of the above. It follows from the Fundamental Theorem of Calculus that $f = \map {\dfrac \d {\d x} } h$ and so $\map h x$ is functions differentiable on $\Bbb I$. Thus it is shown that $C \subseteq B \subseteq A$ and the result follows. {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^{\pi/2} \frac {\d x} {1 + \tan^m x}  	| r = \int_0^{\pi/2} \frac {\cos^m x} {\cos^m x + \sin^m x} \rd x 	| c = multiplying by $\dfrac {\cos^m x} {\cos^m x}$ }} {{eqn	| r = \int_0^{\pi/2} \frac {\map {\cos^m} {\frac \pi 2 - x} } {\map {\cos^m} {\frac \pi 2 - x} + \map {\sin^m} {\frac \pi 2 - x} } \rd x 	| c = Integral between Limits is Independent of Direction }} {{eqn	| r = \int_0^{\pi/2} \frac {\sin^m x} {\sin^m x + \cos^m x} \rd x 	| c = Cosine of Complement equals Sine, Sine of Complement equals Cosine }} {{end-eqn}} So: {{begin-eqn}} {{eqn	| l = 2 \int_0^{\pi/2} \frac {\d x} {1 + \tan^m x} 	| r = \int_0^{\pi/2} \frac {\sin^m x + \cos^m x} {\sin^m x + \cos^m x} \rd x }} {{eqn	| r = \int_0^{\pi/2} \rd x }} {{eqn	| r = \frac \pi 2 	| c = Primitive of Constant }} {{end-eqn}} giving:  :$\displaystyle \int_0^{\pi/2} \frac {\d x} {1 + \tan^m x} = \frac \pi 4$ {{qed}}
From Primitive of Power, we have:  :$\displaystyle \int_0^1 x^m \rd x = \frac 1 {m + 1}$ We have:  {{begin-eqn}} {{eqn	| l = \frac {\d^n} {\d m^n} \int_0^1 x^m \rd x 	| r = \int_0^1 \frac {\partial^n} {\partial m^n} x^m \rd x 	| c = Definite Integral of Partial Derivative }} {{eqn	| r = \int_0^1 x^m \paren {\ln x}^n \rd x 	| c = Derivative of Power of Constant }} {{end-eqn}} So:  {{begin-eqn}} {{eqn	| l = \int_0^1 x^m \paren {\ln x}^n \rd x 	| r = \frac {\d^n} {\d m^n} \paren {\frac 1 {m + 1} }  }} {{eqn	| r = \frac {\d^{n + 1} } {\d m^{n + 1} } \paren {\map \ln {m + 1} }  	| c = Derivative of Natural Logarithm }} {{eqn	| r = \frac {\paren {-1}^n \map \Gamma {n + 1} } { \paren {m + 1}^{n + 1} } 	| c = $n$th Derivative of Natural Logarithm }} {{end-eqn}} {{qed}}
We can write:  {{begin-eqn}} {{eqn	| l = x^x 	| r = \map \exp {x \ln x} 	| c = {{Defof|Power to Real Number}} }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac {x^n \paren {\ln x}^n} {n!} 	| c = {{Defof|Exponential Function}} }} {{end-eqn}} So:  {{begin-eqn}} {{eqn	| l = \int_0^1 x^x \rd x 	| r = \int_0^1 \paren {\sum_{n \mathop = 0}^\infty \frac {x^n \paren {\ln x}^n} {n!} }\rd x }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {\int_0^1 x^n \paren {\ln x}^n \rd x} }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {\frac {\paren {-1}^n \map \Gamma {n + 1} } {\paren {n + 1}^{n + 1} } } 	| c = Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$ }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {n + 1}^{n + 1} } 	| c = Gamma Function Extends Factorial }} {{eqn	| r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } {n^n} 	| c = shifting the index and using $\paren {-1}^{n + 1} = \paren {-1}^2 \paren {-1}^{n - 1} = \paren {-1}^{n - 1}$ }} {{eqn	| r = -\sum_{n \mathop = 1}^\infty \paren {-\frac 1 n}^n }} {{eqn	| r = -\sum_{n \mathop = 1}^\infty \paren {-n}^{-n} }} {{end-eqn}} Numerical computation of partial sums gives the decimal approximation. {{qed}} Category:Definite Integrals qt7qek77whynt75f2n8nqxdz25m39a0
We can write:  {{begin-eqn}} {{eqn	| l = x^x 	| r = \map \exp {-x \ln x} 	| c = {{Defof|Power to Real Number}} }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac {\paren {-x}^n \paren {\ln x}^n} {n!} 	| c = {{Defof|Exponential Function}} }} {{end-eqn}} So:  {{begin-eqn}} {{eqn	| l = \int_0^1 x^x \rd x 	| r = \int_0^1 \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^n \paren {\ln x}^n} {n!} }\rd x }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {n!} \paren {\int_0^1 x^n \paren {\ln x}^n \rd x} }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {n!} \paren {\frac {\paren {-1}^n \map \Gamma {n + 1} } {\paren {n + 1}^{n + 1} } } 	| c = Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$ }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac 1 {\paren {n + 1}^{n + 1} } 	| c = Gamma Function Extends Factorial }} {{eqn	| r = \sum_{n \mathop = 1}^\infty n^{-n} 	| c = shifting the index }} {{end-eqn}} Numerical computation of partial sums gives the decimal approximation. {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty e^{-a x} \sin b x \rd x 	| r = \intlimits {\frac {e^{-a x} \paren {-a \sin b x - b \cos b x} } {a^2 + b^2} } 0 \infty 	| c = Primitive of $e^{a x} \sin b x$ }} {{eqn	| r = -\lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} \paren {a \sin b x + b \cos b x} } {a^2 + b^2} } + \frac {e^0 \paren {a \sin 0 + b \cos 0} } {a^2 + b^2}  }} {{eqn	| r = -\lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} \paren {a \sin b x + b \cos b x} } {a^2 + b^2} } +\frac b {a^2 + b^2} 	| c = Exponential of Zero, Sine of Zero is Zero, Cosine of Zero is One }} {{end-eqn}} Note that we have, by Linear Combination of Sine and Cosine:  :$\displaystyle 0 \le \size {\frac {e^{-a x} \paren {a \sin b x + b \cos b x} } {a^2 + b^2} } \le \frac {e^{-a x} \sqrt {a^2 + b^2} } {a^2 + b^2} = \frac {e^{-a x} } {\sqrt {a^2 + b^2} }$ By Exponential Tends to Zero and Infinity:  :$\displaystyle \lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} } {\sqrt {a^2 + b^2} } } = 0$ So by the Squeeze Theorem:  :$\displaystyle \lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} \paren {a \sin b x + b \cos b x} } {a^2 + b^2} } = 0$ So:  :$\displaystyle \int_0^\infty e^{-a x} \sin b x \rd x = \frac b {a^2 + b^2}$ {{qed}}
{{begin-eqn}} {{eqn	| l = \sum_{k \mathop = 0}^n \map \exp {i k x}  	| r = \frac {\map \exp {i x \paren {n + 1} } - 1} {\map \exp {i x} - 1} 	| c = Sum of Geometric Sequence }} {{eqn	| r = \frac {\map \exp {\frac {i x \paren {n + 1} } 2} \paren {\map \exp {\frac {i x \paren {n + 1} } 2} - \map \exp {\frac {-i x \paren {n + 1} } 2} } } {\map \exp {\frac {i x} 2} \paren {\map \exp {\frac {i x} 2} - \map \exp {-\frac {i x} 2} } } 	| c = factoring $\dfrac {\map \exp {\frac {i x \paren {n + 1} } 2} } {\map \exp {\frac {i x} 2} }$ }} {{eqn	| r = \map \exp {\frac {i x n} 2} \frac {2 i \map \sin {\frac {\paren {n + 1} x} 2} } {2 i \sin \frac x 2} 	| c = Exponential of Sum: Real Numbers: Corollary, Sine Exponential Formulation }} {{eqn	| r = \paren {i \sin \frac {n x} 2 + \cos \frac {n x} 2} \frac {\map \sin {\frac {\paren {n + 1} x} 2} } {\sin \frac x 2} 	| c = Euler's Formula }} {{end-eqn}} {{qed}} Category:Exponential Function 1q29cjv15x4ykjsurfjiw9nusndunzf
First we note that as Real Addition is Closed and Real Multiplication is Closed: :$\Dom {\R^*} \subseteq \R^*$ Recall from Möbius Transformation is Bijection that the Möbius transformation on the extended complex plane is a bijection {{iff}} $a c - b d \ne 0$. From Restriction of Injection is Injection, if $a c - b d \ne 0$ {{iff}} $f$ is an injection. As the inverse $f^{-1}$ of $f$ is also the restriction of a Möbius transformation, it follows that $f^{-1}$ is also an injection. Hence the result from Injection is Bijection iff Inverse is Injection. {{qed}}
By definition of extension: :$\forall x \in A: \map f x = \map g x$ and: :$\forall x \in B: \map g x = \map h x$ and so: :$\forall x \in A: \map g x = \map h x$ from which it follows that: :$\forall x \in A: \map f x = \map h x$ {{qed}}
Recall the definition of the cartesian space of $S$ indexed by $I$: {{:Definition:Indexed Cartesian Space}} {{ProofWanted|This will probably end up being part of the body of material which will eventually be put into place which finally rationalises and competently explains the concept of cartesian product, resolving the two definitions of Definition:Cartesian Product of Family (equivalence proof still needed). Can't get my head round this today, or ever, it seems.}}
{{begin-eqn}} {{eqn	| l = \int_0^1 \frac {\ln x} {1 + x} \rd x 	| r = \int_0^1 \frac {\ln x} {1 - \paren {-x} } \rd x }} {{eqn	| r = \int_0^1 \ln x \paren {\sum_{n \mathop = 0}^\infty \paren {-x}^n} \rd x 	| c = Sum of Geometric Sequence }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 x^n \ln x \rd x 	| c = Fubini's Theorem }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^{n + 1} \map \Gamma 2} {\paren {n + 1}^2} 	| c = Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$ }} {{eqn	| r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n 1!} {n^2} 	| c = Gamma Function Extends Factorial, shifting the index }} {{eqn	| r = -\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } {n^2} 	| c = writing $-\paren {-1}^{n + 1} = \paren {-1}^{n + 2} = \paren {-1}^2 \paren {-1}^n = \paren {-1}^n$ }} {{eqn	| r = -\frac {\pi^2} {12} 	| c = Sum of Reciprocals of Squares Alternating in Sign }} {{end-eqn}} {{qed}}
By definition of an open ball: :$\forall n \in \N: \map d {x_n, l} < \epsilon \iff x_n \in \map {B_\epsilon} l$ The result follows.
By definition of a convergent real sequence: :$\displaystyle \lim_{n \mathop \to \infty} \map d {x_n, l} = 0$ {{iff}} :$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: n > N \implies \size {\map d {x_n, l} - 0} < \epsilon$  From Distance in Pseudometric is Non-Negative: :$\forall x, y \in A: \map d {x, y} \ge 0$ Hence: :$\forall n \in \N: \map d {x_n, l} = \size {\map d {x_n, l}} = \size {\map d {x_n, l} - 0}$ The result follows.
{{begin-eqn}} {{eqn	| l = \int_0^1 \frac {\ln x} {1 - x} \rd x 	| r = \int_0^1 \ln x \paren {\sum_{n \mathop = 0}^\infty x^n} \rd x 	| c = Sum of Geometric Sequence }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \paren {\int_0^1 x^n \ln x \rd x} 	| c = Fubini's Theorem }} {{eqn	| r = -\sum_{n \mathop = 0}^\infty \frac {\map \Gamma 2} {\paren {n + 1}^2} 	| c = Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$ }} {{eqn	| r = -\sum_{n \mathop = 1}^\infty \frac 1 {n^2} 	| c = shifting the index, Gamma Function Extends Factorial }} {{eqn	| r = -\frac {\pi^2} 6 	| c = Basel Problem }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^1 \frac {\map \ln {1 + x} } x \rd x 	| r = \int_0^1 \frac 1 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 \paren {\frac {\paren {-1}^{n - 1} } n \int_0^1 x^{n - 1} \rd x} 	| c = Power Series is Termwise Integrable within Radius of Convergence }} {{eqn	| r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } {n^2}  	| c = Primitive of Power }} {{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 \frac {\map \ln {1 - x} } x \rd x 	| r = -\int_0^1 \frac 1 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 \paren {\frac 1 n  \int_0^1 x^{n - 1} \rd x} 	| c = Power Series is Termwise Integrable within Radius of Convergence }} {{eqn	| r = -\sum_{n \mathop = 1}^\infty \frac 1 {n^2} 	| c = Primitive of Power }} {{eqn	| r = -\frac {\pi^2} 6 	| c = Basel Problem }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty e^{-a x} \cos b x \rd x 	| r = \intlimits {\frac {e^{-a x} \paren {-a \cos b x + b \sin b x} } {a^2 + b^2} } 0 \infty 	| c = Primitive of $e^{a x} \cos b x$ }} {{eqn	| r = \lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} \paren {-a \cos b x + b \sin b x} } {a^2 + b^2} } + \frac {e^0 \paren {a \cos 0 - b \sin 0} } {a^2 + b^2} }} {{eqn	| r = \lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} \paren {-a \cos b x + b \sin b x} } {a^2 + b^2} } + \frac a {a^2 + b^2} 	| c = Exponential of Zero, Cosine of Zero is One, Sine of Zero is Zero }} {{end-eqn}} Note that we have, by Linear Combination of Sine and Cosine:  :$\displaystyle 0 \le \size {\frac {e^{-a x} \paren {-a \cos b x + b \sin x} } {a^2 + b^2} } \le \frac {e^{-a x} \sqrt {a^2 + b^2} } {a^2 + b^2} = \frac {e^{-a x} } {\sqrt {a^2 + b^2} }$ By Exponential Tends to Zero and Infinity:  :$\displaystyle \lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} } {\sqrt {a^2 + b^2} } } = 0$ So by the Squeeze Theorem:  :$\displaystyle \lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} \paren {-a \cos b x + b \sin x} } {a^2 + b^2} } = 0$ So:  :$\displaystyle \int_0^\infty e^{-a x} \cos b x \rd x = \frac a {a^2 + b^2}$ {{qed}}
Let $S: \Bbb S \to \Bbb S$ denote the identity mapping on $\Bbb S$: :$\forall i \in \Bbb S: S_i = i$ where we use $S_i$ to mean the image of $i$ under $S$: :$S_i := \map S i$ Then we can consider $S$ as an indexing function from $\Bbb S$ to $\Bbb S$. Hence in this case $\Bbb S$ is at the same time both: :an indexing set and: :the set indexed by itself. It follows that each of the sets $i \in \Bbb S$ is both: :an index and: :a term $S_i$ of the family of elements of $\Bbb S$ indexed by $\Bbb S$. Thus we would write $\Bbb S$ as: :$\family {S_i}_{i \mathop \in \Bbb S}$ {{qed}}
Note that the integrand is of the form:  :$\displaystyle \int_0^\infty \frac {\map f {a x} - \map f {b x} } x \rd x$ where: :$\map f x = e^{-x}$ We have, by Derivative of Exponential Function: :$\map {f'} x = -e^{-x}$ which is continuous on $\R$. We also have, by Exponential Tends to Zero and Infinity:  :$\displaystyle \lim_{x \mathop \to \infty} \map f x = \lim_{x \mathop \to \infty} e^{-x} = 0$ 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 {e^{-a x} - e^{-b x} } x \rd x 	| r = \paren {\lim_{x \mathop \to \infty} e^{-x} - e^0} \ln \frac a b }} {{eqn	| r = -\ln \frac a b 	| c = Exponential of Zero }} {{eqn	| r = \ln \frac b a 	| c = Logarithm of Reciprocal }} {{end-eqn}} {{qed}}