avewright/proofwiki-math huggingface dataset:
Theorem: Reduction Formula for Definite Integral of Power of Cosine

Let $n \in \Z_{> 0}$ be a positive integer.
Let $I_n$ be defined as:
:$\ds I_n = \int_0^{\frac \pi 2} \cos^n x \rd x$
Then $\sequence {I_n}$ is a decreasing sequence of real numbers which satisfies:
:$n I_n = \paren {n - 1} I_{n - 2}$
Thus:
:$I_n = \dfrac {n - 1} n  I_{n - 2}$
is a reduction formula for $I_n$.

Proof:
From Shape of Cosine Function:
:$\forall x \in \closedint 0 {\dfrac \pi 2}: 0 \le \cos x \le 1$
So, on the same interval:
:$0 \le \cos^{n + 1} x \le \cos^n x$
Therefore:
:$\forall n \in \N: 0 < I_{n + 1} < I_n$
From Reduction Formula for Integral of Power of Cosine:
:$\ds \int \cos^n x \rd x = \dfrac {\cos^{n - 1} x \sin x} n + \dfrac {n - 1} n \int \cos^{n - 2} x \rd x$
Thus:
{{begin-eqn}}
{{eqn | l = I_n
      | r = \int_0^{\frac \pi 2} \cos^n x \rd x
      | c = 
}}
{{eqn | r = \dfrac {n - 1} n \int_0^{\frac \pi 2} \cos^{n - 2} x \rd x - \intlimits {\dfrac {\cos^{n - 1} x \sin x} n} 0 {\pi / 2}
      | c = 
}}
{{eqn | r = \dfrac {n - 1} n I_{n - 2} - \frac {\paren {\cos^{n - 1} \frac \pi 2 \sin \frac \pi 2 - \cos^{n - 1} 0 \sin 0} } n
      | c = 
}}
{{eqn | r = \dfrac {n - 1} n I_{n - 2} - \frac {\paren {0 - \cos^{n - 1} 0 \sin 0} } n
      | c = Cosine of Right Angle
}}
{{eqn | r = \dfrac {n - 1} n I_{n - 2} - \frac {\paren {0 - 0} } n
      | c = Sine of Zero is Zero
}}
{{eqn | ll= \leadsto
      | l = n I_n
      | r = \paren {n - 1} I_{n - 2}
      | c = multiplying both sides by $n$
}}
{{end-eqn}}
{{qed}}
Theorem: Definite Integral from 0 to Half Pi of Even Power of Cosine x

:$\ds \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2$

Proof:
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \int_0^{\frac \pi 2} \cos^2 x \rd x
      | r = \frac \pi 4
      | c = Definite Integral from 0 to Half Pi of Square of Cosine x
}}
{{eqn | r = \dfrac 1 2 \times \dfrac \pi 2
      | c = 
}}
{{eqn | r = \dfrac 2 {\paren {2 \times 1}^2} \dfrac \pi 2
      | c = 
}}
{{eqn | r = \dfrac {\paren {2 \times 1}!} {\paren {2^1 \times 1!}^2} \dfrac \pi 2
      | c = 
}}
{{end-eqn}}
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \int_0^{\frac \pi 2} \cos^{2 k} x \rd x = \dfrac {\paren {2 k}!} {\paren {2^k k!}^2} \dfrac \pi 2$
from which it is to be shown that:
:$\ds \int_0^{\frac \pi 2} \cos^{2 \paren {k + 1} } x \rd x = \dfrac {\paren {2 \paren {k + 1} }!} {\paren {2^{k + 1} \paren {k + 1}!}^2} \dfrac \pi 2$
=== Induction Step ===
This is the induction step:
Let $I_k = \ds \int_0^{\frac \pi 2} \cos^{2 k} x \rd x$.
{{begin-eqn}}
{{eqn | l = I_{k + 1}
      | r = \frac {2 \paren {k + 1} - 1} {2 \paren {k + 1} } I_k
      | c = Reduction Formula for Definite Integral of Power of Cosine
}}
{{eqn | r = \frac {2 \paren {k + 1} - 1} {2 \paren {k + 1} } \dfrac {\paren {2 k}!} {\paren {2^k k!}^2} \dfrac \pi 2
      | c = Induction Hypothesis
}}
{{eqn | r = \frac {2 \paren {k + 1} \paren {2 \paren {k + 1} - 1} } {2^2 \paren {k + 1}^2} \dfrac {\paren {2 k}!} {\paren {2^k k!}^2} \dfrac \pi 2
      | c = 
}}
{{eqn | r = \dfrac {\paren {2 \paren {k + 1} }!} {\paren {2^{k + 1} \paren {k + 1}!}^2} \dfrac \pi 2
      | c = 
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{> 0}: \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2$
{{qed}}
Theorem: Definite Integral from 0 to Half Pi of Even Power of Cosine x

:$\ds \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2$

Proof:
{{begin-eqn}}
{{eqn	| l = \int_0^{\frac \pi 2} \cos^{2 n} x \rd x
	| r = \int_0^{\frac \pi 2} \paren {\sin x}^{\frac 2 2 - 1} \paren {\cos x}^{2 \paren {n + \frac 1 2} - 1} \rd x
}}
{{eqn	| r = \frac 1 2 \Beta \paren {\frac 1 2, n + \frac 1 2}
	| c = {{Defof|Beta Function|index = 2}}
}}
{{eqn	| r = \frac 1 2 \cdot \frac {\map \Gamma {n + \frac 1 2} \, \map \Gamma {\frac 1 2} } {\map \Gamma {n + 1} }
	| c = {{Defof|Beta Function|index = 3}}
}}
{{eqn	| r = \frac {\map \Gamma {n + \frac 1 2} \sqrt \pi} {2 \paren {n!} }
	| c = Gamma Function of One Half
}}
{{eqn	| r = \frac {\paren {2 n}! \paren {\sqrt \pi}^2} {2 \cdot 2^{2 n} \paren {n!}^2}
	| c = Gamma Function of Positive Half-Integer
}}
{{eqn	| r = \frac {\paren {2 n}!} {\paren {2^n n!}^2} \frac \pi 2
}}
{{end-eqn}}
{{qed}}
Theorem: Definite Integral from 0 to Half Pi of Odd Power of Cosine x

:$\ds \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\paren {2^n n!}^2} {\paren {2 n + 1}!}$

Proof:
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\paren {2^n n!}^2} {\paren {2 n + 1}!}$
=== Basis for the Induction ===
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \int_0^{\frac \pi 2} \cos x \rd x
      | r = \bigintlimits {\sin x} 0 {\frac \pi 2}
      | c = Primitive of $\cos x$
}}
{{eqn | r = \sin \frac \pi 2 - \sin 0
      | c = 
}}
{{eqn | r = 1
      | c = Sine of Right Angle, Sine of Zero is Zero
}}
{{eqn | r = \frac {\paren {2^0 \times 1!}^2} {\paren {2 \times 0 + 1}!}
      | c = 
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \int_0^{\frac \pi 2} \cos^{2 k + 1} x \rd x = \dfrac {\paren {2^k k!}^2} {\paren {2 k + 1}!}$
from which it is to be shown that:
:$\ds \int_0^{\frac \pi 2} \cos^{2 \paren {k + 1} + 1} x \rd x = \dfrac {\paren {2^{k + 1} \paren {k + 1}!}^2} {\paren {2 \paren {k + 1} + 1}!}$
=== Induction Step ===
This is the induction step:
Let $I_k = \ds \int_0^{\frac \pi 2} \cos^{2 k + 1} x \rd x$.
{{begin-eqn}}
{{eqn | l = I_{k + 1}
      | r = \frac {2 \paren {k + 1} } {2 \paren {k + 1} + 1} I_k
      | c = Reduction Formula for Definite Integral of Power of Cosine
}}
{{eqn | r = \frac {2 \paren {k + 1} } {2 \paren {k + 1} + 1} \dfrac {\paren {2^k k!}^2} {\paren {2 k + 1}!}
      | c = Induction Hypothesis
}}
{{eqn | r = \frac {\paren {2 \paren {k + 1} }^2} {\paren {2 \paren {k + 1} + 1} \paren {2 \paren {k + 1} } } \dfrac {\paren {2^k k!}^2} {\paren {2 k + 1}!}
      | c = multiplying top and bottom by $2 \paren {k + 1}$
}}
{{eqn | r = \dfrac {\paren {2^{k + 1} \paren {k + 1}!}^2} {\paren {2 \paren {k + 1} + 1}!}
      | c = 
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\paren {2^n n!}^2} {\paren {2 n + 1}!}$
{{qed}}
Theorem: Definite Integral from 0 to Half Pi of Odd Power of Cosine x

:$\ds \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\paren {2^n n!}^2} {\paren {2 n + 1}!}$

Proof:
{{begin-eqn}}
{{eqn | l = \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x
      | r = \int_0^{\frac \pi 2} \paren {\sin x}^{\frac 2 2 - 1} \paren {\cos x}^{2 \paren {n + 1} - 1} \rd x
}}
{{eqn | r = \frac 1 2 \map \Beta {\frac 1 2, n + 1}
      | c = {{Defof|Beta Function|index = 2}}
}}
{{eqn | r = \frac 1 2 \cdot \frac {\map \Gamma {n + 1} \map \Gamma {\frac 1 2} } {\map \Gamma {n + 1 + \frac 1 2} }
      | c = {{Defof|Beta Function|index = 3}}
}}
{{eqn | r = \frac 1 2 \cdot \frac {n! \sqrt \pi} {\map \Gamma {n + 1 + \frac 1 2} }
      | c = Gamma Function Extends Factorial, Gamma Function of One Half
}}
{{eqn | r = \frac 1 2 \cdot n! \sqrt \pi \cdot \frac{2^{2 n + 2} \paren {n + 1}!} {\paren {2 n + 2}! \sqrt \pi}
      | c = Gamma Function of Positive Half-Integer
}}
{{eqn | r = \frac {n! \cdot 2^{2 n + 1} \paren {n + 1}!} {\paren {2 n + 2} \paren {2 n + 1}!}
}}
{{eqn | r = \frac 2 2 \cdot \frac {n! \cdot 2^{2 n} \paren {n + 1} n!} {\paren {n + 1} \paren {2 n + 1}!}
}}
{{eqn | r = \frac {\paren {2^n n!}^2} {\paren {2 n + 1}!}
}}
{{end-eqn}}
{{qed}}
Theorem: Arctangent of Zero is Zero

:$\arctan 0 = 0$

Proof:
By definition, $\arctan$ is the inverse of the tangent function's restriction to $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$. 
By Tangent of Zero:
:$\tan 0 = 0$
As $0 \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$, we have $\arctan 0 = 0$ by the definition of an inverse function.
{{qed}}
Category:Arctangent Function
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Theorem: Integral to Infinity of Sine p x Cosine q x over x

:$\ds \int_0^\infty \frac {\sin p x \cos q x} x \rd x = \begin {cases} \dfrac \pi 2 & : p > q > 0 \\ \\ 0 & : 0 < p < q \\ \\ \dfrac \pi 4 & : p = q > 0 \end {cases}$

Proof:
{{begin-eqn}}
{{eqn | l = \int_0^\infty \frac {\sin p x \cos q x} x \rd x
      | r = \int_0^\infty \frac 1 2 \cdot \frac {\sin \paren {\paren {p + q} x} + \sin \paren {\paren {p - q} x} } x \rd x
      | c = Werner Formula for Sine by Cosine
}}
{{eqn | r = \frac 1 2 \int_0^\infty \frac {\sin \paren {\paren {p + q} x} } x \rd x + \frac 1 2 \int_0^\infty \frac {\sin \paren {\paren {p - q} x} } x \rd x
      | c = Linear Combination of Definite Integrals
}}
{{eqn | r = \lim_{t \mathop \to \infty} \frac 1 2 \int_0^t \frac {\sin \paren {\paren {p + q} x} } x \rd x + \lim_{t \mathop \to \infty} \frac 1 2 \int_0^t \frac {\sin \paren {\paren {p - q} x} } x \rd x
      | c = {{Defof|Improper Integral on Closed Interval Unbounded Above}}
}}
{{eqn | r = \lim_{t \mathop \to \infty} \frac 1 2 \int_0^{t \paren {p + q} } \frac {\sin u} u \rd u + \lim_{t \mathop \to \infty} \frac 1 2 \int_0^t \frac {\sin \paren {\paren {p - q} x} } x \rd x
      | c = Integration by Substitution: $u = \paren {p + q} x$
}}
{{eqn | r = \lim_{t \mathop \to \infty} \frac 1 2 \int_0^{t \paren {p + q} } \frac {\sin u} u \rd u + \lim_{t \mathop \to \infty} \frac 1 2 \int_0^{t \paren {p - q} } \frac {\sin v} v \rd v
      | c = Integration by Substitution: $v = \paren {p - q} x$
}}
{{eqn | r = \frac 1 2 \int_0^\infty \frac {\sin u} u \rd u + \frac 1 2 \lim_{t \mathop \to \infty} \int_0^{t \paren {p - q} } \frac {\sin v} v \rd v
      | c = {{Defof|Improper Integral on Closed Interval Unbounded Above}}: {{hypothesis}} $p + q$ is (strictly) positive
}}
{{eqn | r = \frac \pi 4 + \frac 1 2 \lim_{t \mathop \to \infty} \int_0^{t \paren {p - q} } \frac {\sin v} v \rd v
      | c = Dirichlet Integral
}}
{{end-eqn}}
Theorem: Integral to Infinity of Sine p x Sine q x over x Squared

:$\ds \int_0^\infty \frac {\sin p x \sin q x} {x^2} \rd x = \begin {cases} \dfrac {\pi p} 2 & : 0 < p \le q \\ \\ \dfrac {\pi q} 2 & : p \ge q > 0 \end {cases}$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int f g' \rd t = f g - \int f' g \rd t$
let:
{{begin-eqn}}
{{eqn | l = f
      | r = \sin p x \sin q x
}}
{{eqn | ll= \leadsto
      | l = f'
      | r = p \cos p x \sin q x + q \sin p x \cos q x
      | c = Product Rule for Derivatives and Derivative of $\sin a x$
}}
{{eqn | l = g'
      | r = \frac 1 {x^2}
}}
{{eqn | ll= \leadsto
      | l = g
      | r = -\frac 1 x
      | c = Chain Rule for Derivatives
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \int_0^\infty \frac {\sin p x \sin q x} {x^2} \rd x
      | r = \intlimits {-\frac {\sin p x \sin q x} x} 0 \infty + \int_0^\infty \frac {p \cos p x \sin q x + q \sin p x \cos q x} x \rd x
      | c = Integration by Parts
}}
{{eqn | r = 0 + \int_0^\infty \frac {p \cos p x \sin q x + q \sin p x \cos q x} x \rd x
      | c = evaluating limits using L'Hôpital's Rule and Product Rule for Limits of Functions
}}
{{eqn | r = p \cdot \int_0^\infty \frac {\cos p x \sin q x} x \rd x + q \cdot \int_0^\infty \frac {\sin p x \cos q x} x \rd x
      | c = Linear Combination of Definite Integrals
}}
{{end-eqn}}
Theorem: Order of Finite p-Group is Power of p

Let $G$ be a finite group.
Let $p$ be a prime number.
Let all elements of $G$ have order a power of $p$.
Then $G$ is a $p$-group.

Proof:
{{AimForCont}}:
:$\order G = k p^n: p \nmid k$
where $\order G$ denotes the order of $G$.
By Divisors of Power of Prime:
:$k \nmid p^n$
From the First Sylow Theorem:
:$\exists H \le G: \order H = k$
where $H \le G$ denotes that $H$ is a subgroup of $G$.
Thus:
:$\exists h \in H: \order h \divides k \implies \order h \nmid p$
where $\divides$ denotes divisibility.
Thus:
:$\exists h \in G: \order h \ne p^n: n \in \Z$
Thus by Proof by Contradiction, $\order G$ must be a power of $p$.
{{Qed}}
Theorem: Order of Finite p-Group is Power of p

Let $G$ be a finite group.
Let $p$ be a prime number.
Let all elements of $G$ have order a power of $p$.
Then $G$ is a $p$-group.

Proof:
Let every element of $G$ be a $p$-element.
Let $q$ be a prime number which is a divisor of the order $\order G$ of $G$.
By Cauchy's Lemma (Group Theory), there exists an element of $G$ whose order is a divisor of $q$.
But as the order of all elements of $G$ divide $p^n$ it follows that $q = p$.
Thus $G$ is a group whose order is $p^n$ for some $n \in \Z_{>0}$.
{{qed}}
Theorem: Sum of Sequence of Binomial Coefficients by Sum of Powers of Integers

Let $n, k \in \Z_{\ge 0}$ be positive integers.
Let $S_k = \ds \sum_{i \mathop = 1}^n i^k$.
Then:
:$\ds \sum_{i \mathop = 1}^k \binom {k + 1} i S_i = \paren {n + 1}^{k + 1} - \paren {n + 1}$

Proof:
Let $N$ be a positive integer. 
Then: 
{{begin-eqn}}
{{eqn | l = \paren {N + 1}^{k + 1} - N^{k + 1}
      | r = \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i N^i - N^{k + 1}
      | c = Binomial Theorem for Integral Index
}}
{{eqn | r = \binom {k + 1} 0 + \sum_{i \mathop = 1}^k \binom {k + 1} i N^i + \binom {k + 1} {k + 1} N^{k + 1} - N^{k + 1}
}}
{{eqn | r = 1 + \sum_{i \mathop = 1}^k \binom {k + 1} i N^i
      | c = {{Defof|Binomial Coefficient}}
}}
{{end-eqn}}
Summing from $N = 1$ to $N = n$, we have on the {{LHS}}:
{{begin-eqn}}
{{eqn | l = \sum_{N \mathop = 1}^n \paren {\paren {N + 1}^{k + 1} - N^{k + 1} } 
      | r = \paren {n + 1}^{k + 1} - 1^{k + 1}
      | c = Telescoping Series: Example 1
}}
{{eqn | r = \paren {n + 1}^{k + 1} - 1
}}
{{end-eqn}}
So, we have: 
{{begin-eqn}}
{{eqn | l = \paren {n + 1}^{k + 1} - 1
      | r = \sum_{N \mathop = 1}^n \paren {1 + \sum_{i \mathop = 1}^k \binom {k + 1} i N^i}
}}
{{eqn | r = \sum_{N \mathop = 1}^n 1 + \sum_{i \mathop = 1}^k \binom {k + 1} i \sum_{N \mathop = 1}^n N^i
}}
{{eqn | r = n + \sum_{i \mathop = 1}^k \binom {k + 1} i S_i
}}
{{end-eqn}}
giving: 
:$\ds \sum_{i \mathop = 1}^k \binom {k + 1} i S_i = \paren {n + 1}^{k + 1} - \paren {n + 1}$
{{qed}}
Theorem: Sum of Reciprocals of Powers of Odd Integers Alternating in Sign

:$\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s} = \frac 1 {2 \map \Gamma s} \int_0^\infty x^{s - 1} \map \sech x \rd x$
where:
:$\map \Re s > 0$
:$\Gamma$ is the gamma function
:$\sech$ is the hyperbolic secant function.

Proof:
{{begin-eqn}}
{{eqn	| l = \int_0^\infty x^{s - 1} \map \sech x \rd x
	| r = 2 \int_0^\infty \frac { x^{s - 1} } {e^x + e^{-x} } \rd x
	| c = {{Defof|Hyperbolic Secant}} 
}}
{{eqn	| r = 2 \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 - \paren {- e^{-2 x} } } \rd x
}}
{{eqn	| r = 2 \int_0^\infty x^{s - 1} e^{-x} \sum_{n \mathop = 0}^\infty \paren {-1}^n e^{-2 n x} \rd x
	| c = Sum of Infinite Geometric Sequence
}}
{{eqn	| r = 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^\infty x^{s - 1} e^{-\paren {2 n + 1} x} \rd x
	| c = Fubini's Theorem
}}
{{eqn	| r = 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^\infty \paren {\frac t {2 n + 1} }^{s - 1} e^{-t} \frac {\rd t} {2 n + 1}
	| c = substituting $t = \paren {2 n + 1} x$
}}
{{eqn	| r = 2 \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s} \int_0^\infty t^{s - 1} e^{-t} \rd t
}}
{{eqn	| r = 2 \map \Gamma s \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s}
	| c = {{Defof|Gamma Function}}
}} 
{{end-eqn}}
So: 
:$\ds \frac 1 {2 \map \Gamma s} \int_0^\infty x^{s - 1} \map \sech x \rd x = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s}$
{{qed}}
Category:Dirichlet Beta Function
Category:Hyperbolic Secant Function
Category:Definite Integrals
Category:Sums of Sequences
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Theorem: Equivalence of Definitions of Local Ring Homomorphism

Let $\struct {A, \mathfrak m}$ and $\struct {B, \mathfrak n}$ be commutative local rings.
Let $f : A \to B$ be a unital ring homomorphism.
{{TFAE|def = Local Ring Homomorphism}}

Proof:
=== 1 iff 2 ===
Follows from Image is Subset iff Subset of Preimage.
{{qed|lemma}}
Theorem: Definite Integral from 0 to 1 of Power of u over 1 + Power of u

{{begin-eqn}}
{{eqn | l = \int_0^1 \dfrac {u^{a - 1} \rd u} {1 + u^d}
      | r = \sum_{j \mathop = 0}^\infty \frac {\paren {-1}^j} {a + j d}
      | c = 
}}
{{eqn | r = \frac 1 a - \frac 1 {a + d} + \frac 1 {a + 2 d} - \frac 1 {a + 3 d} + \cdots
      | c = 
}}
{{end-eqn}}
where $a, d > 0$.

Proof:
{{begin-eqn}}
{{eqn | l = \int_0^1 \frac {u^{a - 1} \rd u} {1 + u^d}
      | r = \int_0^1 \frac {u^{a - 1} \rd u} {1 - \paren {-u^d} }
}}
{{eqn | r = \int_0^1 u^{a - 1} \sum_{j \mathop = 0}^\infty \paren {-1}^j u^{j d} \rd u
      | c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = \int_0^1 \sum_{j \mathop = 0}^\infty \paren {-1}^j u^{a - 1 + j d} \rd u
}}
{{eqn | r = \sum_{j \mathop = 0}^\infty \paren {-1}^j \int_0^1 u^{a - 1 + j d} \rd u
      | c = Fubini's Theorem
}}
{{eqn | r = \sum_{j \mathop = 0}^\infty \paren {-1}^j \intlimits {\frac {u^{a + j d} } {a + j d} } 0 1
      | c = Primitive of Power, Fundamental Theorem of Calculus
}}
{{eqn | r = \sum_{j \mathop = 0}^\infty \frac {\paren {-1}^j} {a + j d}
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Reciprocals of Even Powers of Odd Integers

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} }
      | r = \dfrac 1 {1^{2 n} } + \dfrac 1 {3^{2 n} } + \dfrac 1 {5^{2 n} } + \dfrac 1 {7^{2 n} } + \cdots
      | c = 
}}
{{eqn | r = \paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n} - 1} \pi^{2 n} } {2 \paren {2 n}!}
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^\infty \frac 1 {j^{2 n} }
      | r = \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j}^{2 n} } + \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} }
      | c = 
}}
{{eqn | r = \frac 1 {2^{2 n} } \sum_{j \mathop = 1}^\infty \frac 1 {j^{2 n} } + \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!}
      | r = \frac 1 {2^{2 n} } \times \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!} + \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} }
      | c = Riemann Zeta Function at Even Integers
}}
{{eqn | ll= \leadsto
      | l = \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} }
      | r = \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!} - \frac 1 {2^{2 n} } \times \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!}
}} 
{{eqn | r = \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n} \pi^{2 n} } {2 \paren {2 n}!} - \paren {-1}^{n + 1} \dfrac {B_{2 n} \pi^{2 n} } {2 \paren {2 n}!}
}}
{{eqn | r = \paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n} - 1} \pi^{2 n} } {2 \paren {2 n}!}
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Reciprocals of Even Powers of Integers Alternating in Sign

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^\infty \paren {-1}^{j + 1} \frac 1 {j^{2 n} }
      | r = \dfrac 1 {1^{2 n} } - \dfrac 1 {2^{2 n} } + \dfrac 1 {3^{2 n} } - \dfrac 1 {4^{2 n} } + \cdots
      | c = 
}}
{{eqn | r = \paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n - 1} - 1} \pi^{2 n} } {\paren {2 n}!}
      | c = 
}}
{{end-eqn}}
where $B_{2 n}$ is the $2 n$th Bernoulli number.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^\infty \paren {-1}^{j + 1} \frac 1 {j^{2 n} }
      | r = \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} } - \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j}^{2 n} }
      | c = separating odd positive terms from even negative terms
}}
{{eqn | r = \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} } - \frac 1 {2^{2 n} } \sum_{j \mathop = 1}^\infty \frac 1 {j^{2 n} }
}}
{{eqn | r = \paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n} - 1} \pi^{2 n} } {2 \paren {2 n}!} - \frac 1 {2^{2 n} } \sum_{j \mathop = 1}^\infty \frac 1 {j^{2 n} }
      | c = Sum of Reciprocals of Even Powers of Odd Integers
}}
{{eqn | r = \paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n} - 1} \pi^{2 n} } {2 \paren {2 n}!} - \frac 1 {2^{2 n} } \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!}
      | c = Riemann Zeta Function at Even Integers
}}
{{eqn | r = \paren {-1}^{n + 1} \dfrac {2^{2 n - 1} B_{2 n} \paren {2^{2 n} - 1} \pi^{2 n} - B_{2 n} 2^{2 n - 1} \pi^{2 n} } {2^{2 n} \paren {2 n}!}
}}
{{eqn | r = \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \paren {2^{2 n} - 2} \pi^{2 n} } {2^{2 n} \paren {2 n}!}
}}
{{eqn | r = \paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n - 1} - 1} \pi^{2 n} } {\paren {2 n}!}
}}
{{end-eqn}}
{{qed}}
Theorem: Lagrange's Trigonometric Identities/Sine/Sine Form

{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \sin k x
      | r = \sin 0 + \sin x + \sin 2 x + \sin 3 x + \cdots + \sin n x
      | c = 
}}
{{eqn | r = \frac {\sin \frac {\paren {n + 1} x} 2 \sin \frac {n x} 2} {\sin \frac x 2}
      | c = 
}}
{{end-eqn}}
where $x$ is not an integer multiple of $2 \pi$.

Proof:
By Werner Formula for Sine by Sine:
:$2 \sin \alpha \sin \beta = \map \cos {\alpha - \beta} - \map \cos {\alpha + \beta}$
Thus we establish the following sequence of identities:
{{begin-eqn}}
{{eqn | l = 2 \sin 0 \sin \frac x 2
      | r = 0
      | c = 
}}
{{eqn | l = 2 \sin x \sin \frac x 2
      | r = \cos \frac x 2 - \cos \frac {3 x} 2
      | c = 
}}
{{eqn | l = 2 \sin 2 x \sin \frac x 2
      | r = \cos \frac {3 x} 2 - \cos \frac {5 x} 2
      | c = 
}}
{{eqn | o = \cdots
      | c = 
}}
{{eqn | l = 2 \sin n x \sin \frac x 2
      | r = \cos \frac {\paren {2 n - 1} x} 2 - \cos \frac {\paren {2 n + 1} x} 2
      | c = 
}}
{{end-eqn}}
Summing the above:
{{begin-eqn}}
{{eqn | l = 2 \sin \frac x 2 \paren {\sum_{k \mathop = 0}^n \sin k x}
      | r = \cos \frac x 2 - \cos \frac {\paren {2 n + 1} x} 2
      | c = Sums on {{RHS}} form Telescoping Series
}}
{{eqn | r = -2 \map \sin {\dfrac {\frac x 2 + \frac {\paren {2 n + 1} x} 2} 2} \map \sin {\dfrac {\frac x 2 - \frac {\paren {2 n + 1} x} 2} 2}
      | c = Cosine minus Cosine
}}
{{eqn | r = -2 \sin \dfrac {\paren {n + 1} x} 2 \sin \dfrac {-n x} 2
      | c = 
}}
{{eqn | r = 2 \sin \dfrac {\paren {n + 1} x} 2 \sin \dfrac {n x} 2
      | c = Sine Function is Odd
}}
{{end-eqn}}
The result follows by dividing both sides by $2 \sin \dfrac x 2$.
It is noted that when $x$ is a multiple of $2 \pi$ then:
:$\sin \dfrac x 2 = 0$
leaving the {{RHS}} undefined.
{{finish|Explain what happens in that case}}