hcju/proofwikiqa huggingface dataset:
From the definition of the Beta function: :$\Beta \left({x, y}\right) := \dfrac {\Gamma \left({x}\right) \Gamma \left({y}\right)} {\Gamma \left({x + y}\right)}$ Setting $x = y = \dfrac 1 2$: {{begin-eqn}} {{eqn | l = \Beta \left({\dfrac 1 2, \dfrac 1 2}\right)       | r = \frac {\Gamma \left({\dfrac 1 2}\right) \Gamma \left({\dfrac 1 2}\right)} {\Gamma \left({\dfrac 1 2 + \dfrac 1 2}\right)}       | c =  }} {{eqn | r = \left({\Gamma \left({\dfrac 1 2}\right)}\right)^2       | c =  }} {{end-eqn}} Then from Beta Function of Half with Half: :$\Beta \left({\dfrac 1 2, \dfrac 1 2}\right) = \pi$ Hence the result. {{qed}}
From Euler's Reflection Formula: :$\forall z \notin \Z: \Gamma \left({z}\right) \Gamma \left({1 - z}\right) = \dfrac \pi {\sin \left({\pi z}\right)}$ Setting $z = \dfrac 1 2$: {{begin-eqn}} {{eqn | l = \Gamma \left({\frac 1 2}\right) \Gamma \left({\frac 1 2}\right)       | r = \frac \pi {\sin \left({\frac \pi 2}\right)}       | c =  }} {{eqn | r = \frac \pi 1       | c = Sine of Right Angle }} {{eqn | ll= \implies       | l = \Gamma \left({\frac 1 2}\right)       | r = \pm \sqrt \pi       | c =  }} {{end-eqn}} By definition of the $\Gamma$ function: :$\forall z \in \R_{\ge 0}: \Gamma \left({z}\right) > 0$ and so the negative square root can be discarded. Hence: :$\Gamma \left({\dfrac 1 2}\right) = \sqrt \pi$ as required. {{qed}}
{{begin-eqn}} {{eqn | l = \Gamma \left({\dfrac 1 2}\right)       | r = \int_0^{\to \infty} t^{-\frac 1 2} e^{-t} \ \mathrm d t       | c = {{Defof|Integral Form of Gamma Function|Gamma Function}} }} {{eqn | r = \int_0^{\to \infty} u^{-1} e^{-u^2} 2 u \ \mathrm d u       | c = Integration by Substitution, $\phi \left({u}\right) = u^2$ }} {{eqn | r = 2 \int_0^{\to \infty} e^{-u^2} \ \mathrm d u       | c = Linear Combination of Integrals }} {{eqn | r = \int_{\to -\infty}^{\to \infty} e^{-u^2} \ \mathrm d u       | c = Definite Integral of Even Function }} {{eqn | r = \sqrt {\pi}       | c = Gaussian Integral }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \Gamma \left({\frac 1 2}\right)        | r = \frac {0!} {2^0 0!} \sqrt \pi       | c = Gamma Function of Positive Half-Integer }} {{eqn | r = \sqrt \pi       | c = Factorial of Zero }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \Gamma 1 \, \map \Gamma {\frac 1 2}        | r = 2^0 \sqrt \pi \ \map \Gamma 1        | c = Legendre's Duplication Formula  }}  {{eqn | ll= \leadsto        | l = 0! \, \map \Gamma {\frac 1 2}        | r = 0! \, \sqrt \pi       | c = {{Defof|Gamma Function}} }} {{eqn | ll= \leadsto        | l = \map \Gamma {\frac 1 2}        | r = \sqrt \pi       | c = Factorial of Zero }} {{end-eqn}} {{qed}}
Let $t = \dfrac {x^2} 2$. Then: {{begin-eqn}} {{eqn | l = \int_0^{\mathop \to +\infty} e^{- x^2 / 2} \rd x       | r = \int_0^{\mathop \to +\infty} \left({2 t}\right) e^{-t} \rd t       | c = Integration by Substitution }} {{eqn | r = \frac 1 {\sqrt 2} \Gamma \left({\frac 1 2}\right)       | c = {{Defof|Integral Form of Gamma Function|Gamma Function}} }} {{eqn | r = \frac 1 {\sqrt 2} \sqrt \pi       | c = Gamma Function of One Half }} {{eqn | r = \frac {\sqrt {2 \pi} } 2       | c = multiplying top and bottom by $\sqrt 2$ }} {{end-eqn}} We have that $e^{- x^2 / 2}$ is an even function. From Definite Integral of Even Function: Corollary: :$\displaystyle \int_{\mathop \to -\infty}^{\mathop \to +\infty} e^{- x^2 / 2} \rd x = 2 \int_0^{\mathop \to +\infty} e^{- x^2 / 2} \rd x$ Hence the result. {{qed}}
Fix $c \in \R$. It is sufficient to show there exists $N \in \N$, such that: :$(1): \quad n \ge N \implies \map \ln {\ln n} - C > c$ Proceed as follows: {{begin-eqn}} {{eqn | l = \map \ln {\ln n} - C       | o = >       | r = c }} {{eqn | ll= \leadstoandfrom       | l = \ln n       | o = >       | r = \map \exp {c + C}       | c = {{Defof|Exponential Function/Real|subdef = Sum of Series|Exponential}} }} {{eqn | ll= \leadstoandfrom       | l = n       | o = >       | r = \map \exp {\map \exp {c + C} }       | c = {{Defof|Exponential Function/Real|subdef = Sum of Series|Exponential}} }} {{end-eqn}} Let $N \in \N$ such that $N > \map \exp {\map \exp {c + C} }$. By Logarithm is Strictly Increasing it follows that $N$ satisfies condition $(1)$. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \left({\mathcal O\mathcal E}\right)\left({-x}\right)       | r = \mathcal O\left({-x}\right)\mathcal E\left({-x}\right)       | c = Definition of Pointwise Multiplication of Real-Valued Functions }} {{eqn | r = -\mathcal O\left({x}\right)\mathcal E\left({x}\right)       | c = as $\mathcal O$ is odd and $\mathcal E$ is even }} {{eqn | r = -\left({\mathcal O\mathcal E}\right)\left({x}\right) }} {{end-eqn}} The result follows from the definition of an odd function. {{qed}}
First we will prove that: :$\displaystyle \sum_{r \mathop = 0}^n \frac 1 {F_{2^r}} = 3 - \frac {F_{2^n - 1}} {F_{2^n}}$ for $n \ge 1$. We see that: :$\dfrac 1 {F_1} + \dfrac 1 {F_2} = 2 = 3 - \dfrac {F_1} {F_2}$  so the proposition holds for $n = 1$. Suppose the proposition is true for $n = k$. Then: {{begin-eqn}} {{eqn |l = \sum_{r \mathop = 0}^{k + 1}  \frac 1 {F_{2^r} }        |r = 3 - \frac {F_{2^k - 1} } {F_{2^k} } + \frac 1 {F_{2^{k + 1} } }       |c =  }} {{eqn |r = 3 - \frac {F_{2^k - 1} F_{2^{k + 1} } - F_{2^k} } {F_{2^k} F_{2^{k + 1} } }       |c =  }} {{eqn |r = 3 - \frac {F_{2^k - 1} F_{2^k} \left({F_{2^k + 1} + F_{2^k - 1} }\right) - F_{2^k} } {F_{2^k} F_{2^{k + 1} } }       |c = by repeated Fibonacci recurrence formula on $F_{2^{k + 1} }$  }} {{eqn |r = 3 - \frac {F_{2^k - 1} F_{2^k + 1} + F_{2^k - 1}^2 - 1} {F_{2^{k + 1} } }       |c =  }} {{eqn |r = 3 - \frac {F_{2^k}^2 + F_{2^k - 1}^2} {F_{2^{k + 1} } }       |c = by Cassini's Identity }} {{eqn |r = 3 - \frac {F_{2^{k + 1} - 1} } {F_{2^{k + 1} } }       |c = by repeated Fibonacci recurrence formula on $F_{2^{k + 1} - 1}$  }} {{end-eqn}} Thus by Principle of Mathematical Induction, the proof is complete.  {{qed|lemma}} Now taking the limit, we have: {{begin-eqn}} {{eqn |l = \sum_{r \mathop = 0}^\infty \frac 1 {F_{2^r} }        |r = \lim_{n \mathop \to \infty} \sum_{r \mathop = 0}^n \frac 1 {F_{2^r} }       |c =  }} {{eqn |r = \lim_{n \mathop \to \infty} \left(3 - \frac {F_{2^n - 1} } {F_{2^n } }\right)       |c =  }} {{eqn |r = 3 - \frac 2 {1 + \sqrt 5}       |c = by Ratio of Consecutive Fibonacci Numbers }} {{eqn |r = 3 - \frac {\sqrt 5 - 1} 2       |c =  }} {{eqn |r = \frac {7 - \sqrt 5} 2        |c =  }} {{end-eqn}} as required. {{qed}} {{Namedfor|D.A. Millin|cat=Millin}}
{{refactor|Move Sum of Bernoulli Numbers by Binomial Coefficients Vanishes proof here - more thorough - currently duplicative}} From the generating function definition: {{begin-eqn}} {{eqn | l = \frac x {e^x - 1}       | r = \sum_{n \mathop = 0}^\infty \frac{B_n x^n} {n!}       | c =  }} {{eqn | ll= \leadsto       | l = x        | r = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} \left({ \sum_{k \mathop = 0}^\infty \frac {x^k} {k!} - 1}\right)       | c = {{Defof|Real Exponential Function}} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac{B_n x^n} {n!} \sum_{k \mathop = 0}^\infty \frac {x^{k + 1} } {\left({k + 1}\right)!}        | c = $1 = \dfrac {x^0} {0!}$ }} {{eqn | ll= \leadsto       | l = 1       | r = \sum_{n \mathop = 0}^\infty \frac{B_n x^n} {n!} \sum_{k \mathop = 0}^\infty \frac {x^k } {\left({k + 1}\right)!}       | c =  }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} \sum_{k \mathop = 0}^n \binom n k \frac {B_k} {n - k + 1}       | c = Cauchy Product, Product of Absolutely Convergent Series }} {{end-eqn}} Equating coefficients: For $n = 0$: {{begin-eqn}} {{eqn | l = 1       | r = \sum_{k \mathop = 0}^0 \binom 0 k \frac {B_k} {0 - k + 1}       | c =  }} {{eqn | r = \binom 0 0 \frac {B_0} {0 - 0 + 1}       | c =  }} {{eqn | r = B_0       | c = Binomial Coefficient with Zero }} {{end-eqn}} For $n > 0$: {{begin-eqn}} {{eqn | l = 0       | r = \frac 1 {n!} \sum_{k \mathop = 0}^n \binom n k \frac {B_k} {n - k + 1}       | c =  }} {{eqn | r = \frac 1 {n!} \left({\sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n - k + 1} + \binom n n \frac {B_n} {n - n + 1} }\right)       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n - k + 1} + B_n       | c = Binomial Coefficient with Self and simplifying }} {{eqn | ll= \leadsto       | l = B_n       | r = -\sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n - k + 1}       | c =  }} {{end-eqn}} Hence the result: :$B_n = \begin{cases} 1 & : n = 0 \\ \displaystyle - \sum_{k \mathop = 0}^{n-1} \binom n k \frac {B_k} {n - k + 1} & : n > 0 \end{cases}$ {{qed}} Category:Bernoulli Numbers e56k6sa4xyt1wxofkye9k2toiaw55pp
By definition of the Bernoulli numbers: :$\displaystyle \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac{B_n x^n} {n!}$ or equivalently: :$\displaystyle \frac x {e^x - 1} + \frac x 2 = 1 + \sum_{n \mathop = 2}^\infty \frac{B_n x^n} {n!}$ It remains to be shown that this is an even function. {{explain}} {{begin-eqn}} {{eqn | l = \frac x {e^x - 1} + \frac x 2       | r = \frac {x \left({1 + e^x}\right)} {2\left({e^x - 1}\right)}       | c =  }} {{eqn | r = \frac {-x \left({e^{-x} + 1}\right)} {2\left({e^{-x} - 1}\right)}       | c =  }} {{eqn | r = \frac {-x \left({2 + e^{-x} - 1}\right)} {2\left({e^{-x} - 1}\right)}       | c =  }} {{eqn | r = \frac {-x} {e^{-x} - 1} + \frac {-x} 2       | c =  }} {{end-eqn}} {{qed}}
By definition, the Bernoulli numbers are given by: :$\displaystyle \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}$ We have: {{begin-eqn}} {{eqn | l = \frac x {e^x - 1}       | r = \frac x 2 \left({\frac 2 {e^x -  1} }\right)       | c =  }} {{eqn | r = \frac x 2 \left({\frac {e^x - e^x + 2} {e^x -  1} }\right)       | c =  }} {{eqn | r = \frac x 2 \left({\frac {\left({e^x + 1}\right) - \left({e^x - 1}\right)} {e^x -  1} }\right)       | c =  }} {{eqn | r = \frac x 2 \left({\frac {e^x + 1} {e^x - 1} - 1}\right)       | c =  }} {{eqn | r = -\frac x 2 + \frac x 2 \left({\frac {e^x + 1} {e^x - 1} }\right)       | c =  }} {{end-eqn}} Take $f \left({x}\right) := \dfrac x 2 \left({\dfrac {e^x + 1} {e^x - 1} }\right)$, and note that: {{begin-eqn}} {{eqn | l = f \left({-x}\right)       | r = \frac {-x} 2 \left({\dfrac {e^{-x} + 1} {e^{-x} - 1} }\right)       | c =  }} {{eqn | r = -\frac {-x} 2 \left({\dfrac {1 + e^x} {1 - e^x} }\right)       | c = multiplying top and bottom by $e^x$ }} {{eqn | r = -\frac {-x} 2 \left({\dfrac {e^x + 1} {-\left({e^x - 1}\right)} }\right)       | c =  }} {{eqn | r = \frac x 2 \left({\frac {e^x + 1} {e^x - 1} }\right)       | c =  }} {{eqn | r = f \left({x}\right)       | c =  }} {{end-eqn}} and so $f \left({x}\right) := \dfrac x 2 \left({\dfrac {e^x + 1} {e^x - 1} }\right)$ is an even function. Thus $f \left({x}\right)$ can be seen to take the form: :$\displaystyle f \left({x}\right) = 1 + \sum_{n \mathop = 2}^\infty \frac {B_n} {n!} x^n$ and so: :$\displaystyle f \left({x}\right) = 1 + \sum_{n \mathop = 2}^\infty \frac {B_n} {n!} x^{-n}$ That is: :$\forall n \in \N: n > 1: \dfrac {B_n} {n!} x^n = \dfrac {B_n} {n!} x^{-n}$ and so for odd $n$ where $n > 1$ it follows that $B_n = 0$ {{qed}}
{{begin-eqn}} {{eqn | l = \map \Gamma {z + \frac k n}       | r = \paren {z + \frac k n - 1} \map \Gamma {z + \frac k n - 1}       | c = Gamma Difference Equation }} {{eqn | r = \lim_{m \mathop \to \infty} \frac {m! m^{z + k / n - 1} } {\paren {z + \frac k n} \paren {z + \frac k n + 1} \cdots \paren {z + \frac k n - 1 + m} }        | c = {{Defof|Euler Form of Gamma Function}} }} {{eqn | r = \lim_{m \mathop \to \infty} \frac {\sqrt {2 \pi m} \paren {\frac m e}^m m^{z + k / n - 1} } {\paren {z + \frac k n} \paren {z + \frac k n + 1} \cdots \paren {z + \frac k n - 1 + m} }       | c = Stirling's Formula }} {{eqn | r = \lim_{m \mathop \to \infty} \frac {\sqrt {2 \pi} \paren {\frac {m n} e}^m m^{z + k / n - 1/2} } {\paren {n z + k} \paren {n z + k + n} \cdots \paren {n z + k - n + m n} }       | c =  }} {{end-eqn}} Taking the product for $k = 0$ to $n - 1$, we have: {{begin-eqn}} {{eqn | l = \prod_{k \mathop = 0}^{n - 1} \map \Gamma {z + \frac k n}       | r = \lim_{m \mathop \to \infty} \frac {\paren {\sqrt {2 \pi} }^n \paren {\frac {m n} e}^{m n} m^{n z  - n / 2} m^{\sum_{k \mathop = 0}^{n - 1} k / n} } {\paren {n z} \paren {n z + 1} \cdots \paren {n z - 1 + m n} }       | c =  }} {{eqn | r = \lim_{m \mathop \to \infty} \frac {\paren {\sqrt {2 \pi} }^n \paren {\frac {m n} e}^{m n} m^{n z  - 1 / 2} } {\paren {n z} \paren {n z + 1} \cdots \paren {n z - 1 + m n} }       | c = Sum of Arithmetic Sequence }} {{eqn | r = \lim_{m \mathop \to \infty} \frac {\paren {\sqrt {2 \pi} }^n \paren {\frac m e}^m m^{n z  - 1 / 2} n^{1 / 2 - n z} } {\paren {n z} \paren {n z + 1} \cdots \paren {n z - 1 + m} }       | c = by substituting $m n \mapsto m$ }} {{eqn | r = \lim_{m \mathop \to \infty} \frac {\paren {\sqrt {2 \pi} }^{n - 1} m! m^{n z  - 1} n^{1 / 2 - n z} } {\paren {n z} \paren {n z + 1} \cdots \paren {n z - 1 + m} }       | c = Stirling's Formula }} {{eqn | r = \paren {2 \pi}^{\paren {n - 1} / 2} n^{1 / 2 - n z} \paren {n z - 1} \map \Gamma {n z - 1}       | c = {{Defof|Euler Form of Gamma Function}} }} {{eqn | r = \paren {2 \pi}^{\paren {n - 1} / 2} n^{1 / 2 - n z} \map \Gamma {n z}       | c = Gamma Difference Equation }} {{end-eqn}} {{qed}} {{Namedfor|Carl Friedrich Gauss|cat = Gauss}}
By Fubini's Theorem: :$\displaystyle \int_0^\infty \paren {\int_0^\infty e^{- x y} \sin x \rd y} \rd x = \int_0^\infty \paren {\int_0^\infty e^{- x y} \sin x \rd x} \rd y$ Then: {{begin-eqn}} {{eqn | l = \int_0^\infty e^{- x y} \sin x \rd y       | r = \intlimits {-e^{- x y} \frac {\sin x} x} 0 \infty        | c = Primitive of $e^{a x}$ }} {{eqn | r = \frac {\sin x} x       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \int_0^\infty e^{- x y} \sin x \rd x       | r = \intlimits {\frac {-e^{- x y} \paren {y \sin x + \cos x} } {y^2 + 1} } 0 \infty       | c = Primitive of $e^{a x} \sin b x$ }} {{eqn | r = \frac 1 {y^2 + 1}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \int_0^\infty \frac {\sin x} x \rd x       | r = \int_0^\infty \frac 1 {y^2 + 1} \rd y       | c = }} {{eqn | r = \bigintlimits {\arctan y} 0 \infty       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | r = \frac \pi 2       | c = as $\displaystyle \lim_{y \mathop \to \infty} \arctan y = \frac \pi 2$ }} {{end-eqn}} {{qed}}
Define: :$\displaystyle \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x$ Using Definite Integral of Partial Derivative: {{begin-eqn}} {{eqn | l = \map {I'} \alpha       | r = \frac \partial {\partial \alpha} \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x }} {{eqn | r = \int_0^\infty \frac \partial {\partial \alpha} \frac {e^{-\alpha x} \sin x} x \rd x       | c = Leibniz's Integral Rule }} {{eqn | r = -\int_0^\infty e^{-\alpha x} \sin x \rd x       | c = Derivative of Exponential Function }} {{eqn | r = \intlimits {-\frac {e^{-\alpha x} \paren {-\alpha \sin x + \cos x} } {\paren {-\alpha}^2 + 1} } 0 \infty        | c = Primitive of $e^{\alpha x} \sin b x$ }} {{eqn | r = -\frac 1 {\alpha^2 + 1}  }} {{end-eqn}} Integrating with respect to $\alpha$: {{begin-eqn}} {{eqn | l = \map I \alpha       | r = -\int \frac 1 {\alpha^2 + 1} \rd \alpha }} {{eqn | r = C - \arctan \alpha       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{end-eqn}} To determine $C$, take $\alpha \to \infty$: {{begin-eqn}} {{eqn | l = \lim_{\alpha \mathop \to \infty} \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x       | r = \lim_{\alpha \mathop \to \infty} \paren {C - \arctan \alpha} }} {{eqn | ll= \leadsto       | l = \int_0^\infty \frac {0 \times \sin x} x       | r = C - \frac \pi 2 }} {{eqn | ll= \leadsto       | l = C       | r = \frac \pi 2 }}  {{end-eqn}} Hence: :$\displaystyle \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x = \frac \pi 2 - \arctan \alpha$ Setting $\alpha = 0$ yields: :$\displaystyle \int_0^\infty \frac {\sin x} x \rd x = \frac \pi 2$ {{qed}}
Let: :$\map f z = \dfrac {e^{i z} - 1} z$ We have, by Euler's Formula, on the real line: :$\map \Im {\map f z} = \dfrac {\sin z} z$ So:  :$\displaystyle \map \Im {\int_0^\infty \dfrac {e^{i x} - 1} x \rd x} = \int_0^\infty \dfrac {\sin x} x \rd x$ Let $C_R$ be the semicircular contour of radius $R$ situated on the upper half plane, centred at the origin, traversed anti-clockwise. Let $\Gamma_R = C_R \cup \closedint {-R} R$. Then, by Contour Integral of Concatenation of Contours: :$\displaystyle \oint_{\Gamma_R} \frac {e^{i z} - 1} z \rd z = \int_{C_R} \frac {e^{i z} - 1} z \rd z + \int_{-R}^R \frac {e^{i x} - 1} x \rd x$ From Linear Combination of Contour Integrals, we write:  :$\displaystyle \oint_{\Gamma_R} \frac {e^{i z} - 1} z \rd z = \int_{C_R} \frac {e^{i z} } z \rd z - \int_{C_R} \frac {\rd z} z + \int_{-R}^R \frac {e^{i x} - 1} x \rd x$ The only singularity of $f$ is at $z = 0$. However, we can show $\lim_{z \mathop \to 0} \map f z$ to be finite:  {{begin-eqn}} {{eqn	| l = \lim_{z \mathop \to 0} \frac {e^{i z} - 1} z 	| r = \intlimits {\frac {\map \d {e^{i z} } } {\d z} } {z = 0} {} 	| c = {{Defof|Derivative of Complex Function}} }} {{eqn	| r = i e^0 	| c = Derivative of Exponential of a x }} {{eqn	| r = i }} {{end-eqn}} So, $z = 0$ is a removable singularity of $f$. Therefore, $f$ is holomorphic inside our contour.  It then follows from the Cauchy-Goursat Theorem, that: :$\displaystyle \oint_{\Gamma_R} \frac {e^{i z} - 1} z \rd z = 0$ We also have:  {{begin-eqn}} {{eqn	| l = \size {\int_{C_R} \frac { e^{i z} } z \rd z} 	| o = \le 	| r = \pi \max_{0 \mathop \le \theta \mathop \le \pi} \size {\frac 1 {R e^{i \theta} } } 	| c = Jordan's Lemma }} {{eqn	| r = \frac \pi R }} {{eqn	| o = \to 	| r = 0 	| c = as $R \to \infty$ }} {{end-eqn}} Therefore:  :$\displaystyle \lim_{R \mathop \to \infty} \int_{C_R} \frac {\rd z} z = \lim_{R \mathop \to \infty} \int_{-R}^R \frac {e^{i x} - 1} x \rd x = \int_{-\infty}^\infty \frac {e^{i x} - 1} x \rd x$ Evaluating the integral on the left hand side:  {{begin-eqn}} {{eqn	| l = \int_{C_R} \frac {\rd z} z 	| r = \int_0^\pi \frac {i R e^{i \theta} } {R e^{i \theta} } \rd \theta 	| c = {{Defof|Complex Contour Integral}} }} {{eqn	| r = i \int_0^\pi \rd \theta }} {{eqn	| r = \pi i }} {{end-eqn}} So: :$\displaystyle \int_{-\infty}^\infty \frac {e^{i x} - 1} x \rd x = \pi i$ Taking the imaginary part: :$\displaystyle \int_{-\infty}^\infty \frac {\sin x} x \rd x = \pi$ From Definite Integral of Even Function: :$\displaystyle \int_{-\infty}^\infty \frac {\sin x} x \rd x = 2 \int_0^\infty \frac {\sin x} x \rd x$ Hence:  :$\displaystyle \int_0^\infty \frac {\sin x} x \rd x = \frac \pi 2$ {{qed}}
From Integral to Infinity of Function over Argument: :$\displaystyle \int_0^\infty {\dfrac {\map f x} x} = \int_0^{\to \infty} \map F u \rd u$ for a real function $f$ and its Laplace transform $\laptrans f = F$, provided they exist. Let $\map f x := \sin x$. Then from Laplace Transform of Sine: :$\laptrans {\map f x} = \dfrac 1 {s^2 + 1}$ Hence: {{begin-eqn}} {{eqn | l = \int_0^\infty \frac {\sin x} x \rd x       | r = \int_0^{\to \infty} \dfrac {\d u} {u^2 + 1}       | c =  }} {{eqn | r = \bigintlimits {\arctan u} 0 \infty       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | r = \dfrac \pi 2       | c =  }} {{end-eqn}} {{qed}}
From the definition of the Riemann Zeta Function:  {{begin-eqn}} {{eqn | l = \map \zeta s       | r = \prod_p \frac 1 {1 - p^{-s} }       | c = }} {{eqn | ll= \leadsto       | l = \log \map \zeta s       | r = \log \prod_p \frac 1 {1 - p^{-s} }        | c =  }} {{eqn | r = \sum_p \map \log {\frac 1 {1 - p^{-s} } }       | c = Sum of Logarithms }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\map \pi n - \map \pi {n - 1} } \map \log {\frac 1 {1 - n^{-s} } }       | c = {{Defof|Prime-Counting Function}} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \map \pi n \paren {\map \log {\frac 1 {1 - n^{-s} } } - \map \log {\frac 1 {1 - \paren {n + 1}^{-s} } } }       | c = as $\map \pi {-1} = 0$ }} {{eqn | r = \sum_{n \mathop = 0}^\infty \map \pi n \paren {\map \log {1 - \paren {n + 1}^{-s} } - \map \log {1 - n^{-s} } }       | c = Logarithm of Power }} {{end-eqn}} By Derivative of Logarithm Function and the Chain Rule for Derivatives: {{begin-eqn}} {{eqn | l = \frac \d {\d x} \map \log {1 - x^{-s} }       | r = \frac {s x^{- s - 1} } {1 - x^{-s} }       | c =  }} {{eqn | r = \frac s {x \paren {x^s - 1} }       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \log \map \zeta s       | r = \sum_{n \mathop = 0}^\infty \map \pi n \int_n^{n + 1} \frac s {x \paren {x^s - 1} } \rd x       | c = Fundamental Theorem of Calculus }} {{eqn | r = \sum_{n \mathop = 0}^\infty s \int_n^{n + 1} \frac {\map \pi x} {x \paren {x^s - 1} } \rd x       | c =  }} {{eqn | r = s \int_0^{\mathop \to \infty} \frac {\map \pi x} {x \paren {x^s - 1} } \rd x       | c =  }} {{end-eqn}} {{qed}} Category:Riemann Zeta Function Category:Number Theory gi6rrfeuq1n3xjizk0oausf9t4gexvi
Let $x \ge 0$. {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n - 1} e^{k x}        | r = \sum_{k \mathop = 0}^{n - 1} \sum_{p \mathop = 0}^\infty \frac {\paren {k x}^p}  {p!}       | c = Power Series Expansion for Exponential Function }} {{eqn | r = \sum_{p \mathop = 0}^\infty \paren {\sum_{k \mathop = 0}^{n - 1} k^p} \frac {x^p} {p!}        | c = rearrangement is valid by Tonelli's Theorem }} {{end-eqn}} We also have: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n - 1} e^{k x}        | r = \frac {1 - e^{n x} } {1 - e^x}       | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \frac {e^{n x} - 1} x \frac x {e^x - 1}       | c = }} {{eqn | r = \sum_{p \mathop = 0}^\infty \frac {n^{p + 1} x^p} {\paren {p + 1}!} \sum_{p \mathop = 0}^\infty \frac {B_p x^p} {p!}       | c = {{Defof|Bernoulli Numbers}} }} {{eqn | r = \sum_{p \mathop = 0}^\infty \sum_{i \mathop = 0}^p \frac {n^{p + 1 - i} x^{p - i} } {\paren {p + 1 - i}!} \frac {B_i x^i} {i!}       | c = {{Defof|Cauchy Product}} }} {{eqn | r = \sum_{p \mathop = 0}^\infty \paren {\frac 1 {p + 1} \sum_{i \mathop = 0}^p \binom {p + 1} i B_i n^{p + 1 - i} } \frac {x^p} {p!}       | c =  }} {{end-eqn}} Equating coefficients: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n - 1} k^p        | r = \frac 1 {p + 1} \sum_{i \mathop = 0}^p \binom {p + 1} i B_i n^{p + 1 - i}       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 1}^n k^p        | r = \frac 1 {p + 1} \sum_{i \mathop = 0}^p \paren {-1}^i \binom {p + 1} i B_i n^{p + 1 - i}       | c = as $B_1 = -\dfrac 1 2$ and Odd Bernoulli Numbers Vanish }} {{end-eqn}} {{qed}} {{Namedfor|Johann Faulhaber|cat = Faulhaber}} Category:Number Theory Category:Bernoulli Numbers rct8iugjp1o10ncpbqbea9nl00enaom
Let $z = x + iy$. Let $\epsilon > 0$. By the definition of limits at infinity, we need to show that there is some $M > 0$ such that: :$x > M \implies \size {e^{-z} - 0} < \epsilon$ But: {{begin-eqn}} {{eqn | l = \size {e^{-z} - 0}       | r = \size {e^{-z} } }} {{eqn | r = \size {e^{-x} }       | c = Modulus of Exponential is Modulus of Real Part }} {{eqn | r = \size {e^{-x} - 0} }} {{end-eqn}} so we need an $M$ such that: :$x > M \implies \size {e^{-x} - 0} < \epsilon$ This is the definition of the limits at infinity of the ''real'' exponential. The result follows from Exponential Tends to Zero and Infinity. {{qed}} Category:Exponential Function axmix7l39dppp8rend957tp5aykx8u2
Let $z = x + iy$. {{begin-eqn}} {{eqn | l = \cmod {\exp z}       | r = \cmod {\map \exp {x + iy} } }} {{eqn | r = \cmod {\paren {\exp x} \paren {\exp i y} }       | c = Exponential of Sum }} {{eqn | r = \cmod {\exp x} \cmod {\exp i y}       | c = Modulus of Product }} {{eqn | r = \cmod {\exp x}       | c = Modulus of Exponential of Imaginary Number is One }} {{eqn | r = e^x       | c = Exponential of Real Number is Strictly Positive }} {{eqn | r = \map \exp {\Re z}       | c = {{Defof|Real Part}} }} {{end-eqn}} {{qed}}