avewright/proofwiki-math huggingface dataset:
Theorem: Fourier Cosine Coefficients for Even Function over Symmetric Range

Let $\map f x$ be an even real function defined on the interval $\openint {-\lambda} \lambda$.
Let the Fourier series of $\map f x$ be expressed as:
:$\map f x \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
Then for all $n \in \Z_{\ge 0}$:
:$a_n = \dfrac 2 \lambda \ds \int_0^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x$

Proof:
As suggested, let the Fourier series of $\map f x$ be expressed as:
:$\map f x \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
By definition of Fourier series:
:$a_n = \dfrac 1 \lambda \ds \int_{-\lambda}^{-\lambda + 2 \lambda} \map f x \cos \frac {n \pi x} \lambda \rd x$
From Cosine Function is Even:
:$\cos a = \map \cos {-a}$
for all $a$.
By Even Function Times Even Function is Even, $\map f x \cos \dfrac {n \pi x} \lambda$ is even.
Thus:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \frac 1 \lambda \int_{-\lambda}^{-\lambda + 2 \lambda} \map f x \cos \frac {n \pi x} \lambda \rd x
      | c = 
}}
{{eqn | r = \frac 1 \lambda \int_{-\lambda}^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x
      | c = 
}}
{{eqn | r = \frac 2 \lambda \int_0^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x
      | c = {{Corollary|Definite Integral of Even Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Fourier Sine Coefficients for Even Function over Symmetric Range

Let $\map f x$ be an even real function defined on the interval $\openint {-\lambda} \lambda$.
Let the Fourier series of $\map f x$ be expressed as:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
Then for all $n \in \Z_{> 0}$:
:$b_n = 0$

Proof:
As suggested, let the Fourier series of $\map f x$ be expressed as:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
By definition of Fourier series:
{{begin-eqn}}
{{eqn | l = b_n
      | r = \frac 1 \lambda \int_{-\lambda}^{-\lambda + 2 \lambda} \map f x \sin \frac {n \pi x} \lambda \rd x
      | c = 
}}
{{eqn | r = \frac 1 \lambda \int_{-\lambda}^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x
      | c = 
}}
{{end-eqn}}
From Sine Function is Odd:
:$\sin a = -\map \sin {-a}$
for all $a$.
By Odd Function Times Even Function is Odd, $\map f x \sin \dfrac {n \pi x} l$ is odd.
The result follows from {{Corollary|Definite Integral of Odd Function}}:
:$\ds \frac 1 \lambda \int_{-\lambda}^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x = 0$
{{qed}}
Theorem: Fourier Series for Even Function over Symmetric Range

Let $\map f x$ be an even real function defined on the interval $\openint {-\lambda} \lambda$.
Then the Fourier series of $\map f x$ can be expressed as:
:$\map f x \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty a_n \cos \frac {n \pi x} \lambda$
where for all $n \in \Z_{\ge 0}$:
:$a_n = \dfrac 2 \lambda \ds \int_0^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x$

Proof:
By definition of the Fourier series for $f$:
:$\map f x \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
From Fourier Cosine Coefficients for Even Function over Symmetric Range:
:$a_n = \ds \dfrac 2 \lambda \int_0^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x$
for all $n \in \Z_{\ge 0}$.
From Fourier Sine Coefficients for Even Function over Symmetric Range
:$b_n = 0$
for all $n \in \Z_{\ge 0}$:
{{qed}}
Theorem: Fourier Cosine Coefficients for Odd Function over Symmetric Range

Let $\map f x$ be an odd real function defined on the interval $\openint {-\lambda} \lambda$.
Let the Fourier series of $\map f x$ be expressed as:
:$\ds \map f x \sim \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
Then for all $n \in \Z_{\ge 0}$:
:$a_n = 0$

Proof:
As suggested, let the Fourier series of $\map f x$ be expressed as:
:$\ds \map f x \sim \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
By definition of Fourier series:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \frac 1 \lambda \int_{-\lambda}^{-\lambda + 2 \lambda} \map f x \cos \frac {n \pi x} \lambda \rd x
      | c = 
}}
{{eqn | r = \frac 1 \lambda \int_{-\lambda}^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x
      | c = 
}}
{{end-eqn}}
From Cosine Function is Even:
:$\cos a = \map \cos {-a}$
for all $a$.
By Odd Function Times Even Function is Odd, $\map f x \cos \dfrac {n \pi x} \lambda$ is odd.
The result follows from the {{Corollary|Definite Integral of Odd Function}}:
:$\ds \dfrac 1 \lambda \int_{-\lambda}^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x = 0$
{{qed}}
Theorem: Fourier Sine Coefficients for Odd Function over Symmetric Range

Let $\map f x$ be an odd real function defined on the interval $\openint {-\lambda} \lambda$.
Let the Fourier series of $\map f x$ be expressed as:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
Then for all $n \in \Z_{> 0}$:
:$b_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x$

Proof:
As suggested, let the Fourier series of $\map f x$ be expressed as:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
By definition of Fourier series:
:$b_n = \ds \frac 1 \lambda \int_{-\lambda}^{-\lambda + 2 \lambda} \map f x \sin \frac {n \pi x} \lambda \rd x$
From Sine Function is Odd:
:$\sin a = -\map \sin {-a}$
for all $a$.
By Odd Function Times Odd Function is Even, $\map f x \sin \dfrac {n \pi x} \lambda$ is even.
Thus:
{{begin-eqn}}
{{eqn | l = b_n
      | r = \frac 1 \lambda \int_{-\lambda}^{-\lambda + 2 \lambda} \map f x \sin \frac {n \pi x} \lambda \rd x
      | c = 
}}
{{eqn | r = \frac 1 \lambda \int_{-\lambda}^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x
      | c = 
}}
{{eqn | r = \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x
      | c = {{Corollary|Definite Integral of Even Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Fourier Series for Odd Function over Symmetric Range

Let $\map f x$ be an odd real function defined on the interval $\openint {-\lambda} \lambda$.
Then the Fourier series of $\map f x$ can be expressed as:
:$\map f x \sim \ds \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} \lambda$
where for all $n \in \Z_{> 0}$:
:$b_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x$

Proof:
By definition of the Fourier series for $f$:
:$\map f x \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
From Fourier Cosine Coefficients for Odd Function over Symmetric Range:
:$a_n = 0$
for all $n \in \Z_{\ge 0}$.
From Fourier Sine Coefficients for Odd Function over Symmetric Range
:$b_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x$
for all $n \in \Z_{> 0}$:
{{qed}}
Theorem: Half-Range Fourier Sine Series over Negative Range

Let $\map f x$ be a real function defined on the open real interval $\openint 0 \lambda$.
Let $f$ be expressed using the half-range Fourier sine series over $\openint 0 \lambda$:
:$\ds \map S x \sim \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} \lambda$
where:
:$b_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x$
for all $n \in \Z_{\ge 0}$.
Then over the interval $\openint {-\lambda} 0$, $\map S x$ takes the values:
:$\map S x = -\map f {-x}$
{{refactor|This following bit depends upon what happens at $x {{=}} 0$ which needs to be carefully considered, so put it here as a corollary|level = basic}}
That is, the real function expressed by the half-range Fourier sine series over $\openint 0 \lambda$ is an odd function over $\openint {-\lambda} \lambda$.

Proof:
From Fourier Series for Odd Function over Symmetric Range, $\map S x$ is the Fourier series of an odd real function over the interval $\openint 0 \lambda$.
We have that $\map S x \sim \map f x$ over $\openint 0 \lambda$.
Thus over $\openint {-\lambda} 0$ it follows that:
:$\map S x = -\map f {-x}$
{{qed}}
Theorem: Half-Range Fourier Cosine Series over Negative Range

Let $\map f x$ be a real function defined on the open real interval $\openint 0 \lambda$.
Let $f$ be expressed using the half-range Fourier cosine series over $\openint 0 \lambda$:
:$\ds \map C x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos \frac {n \pi x} \lambda$
where:
:$a_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x$
for all $n \in \Z_{\ge 0}$.
Then over the closed real interval $\openint {-\lambda} 0$, $\map C x$ takes the values:
:$\map C x = \map f {-x}$
{{refactor|This following bit depends upon what happens at $x {{=}} 0$ which needs to be carefully considered, so put it here as a corollary|level = basic}}
That is, the real function expressed by the half-range Fourier cosine series over $\openint 0 \lambda$ is an even function over $\openint {-\lambda} \lambda$.

Proof:
From Fourier Series for Even Function over Symmetric Range, $\map C x$ is the Fourier series of an even real function over the interval $\openint 0 \lambda$.
We have that $\map C x \sim \map f x$ over $\openint 0 \lambda$.
Thus over $\openint {-\lambda} 0$ it follows that $\map C x = \map f {-x}$.
{{qed}}
Theorem: Fourier Series over General Range from Specific

Let $a, b \in \R$ be real numbers.
Let $f: \R \to \R$ be a function such that $\ds \int_a^b \map f x \rd x$ converges absolutely.
Then $f$ can be expressed by a Fourier series of the form:
:$\ds \frac {A_0} 2 + \sum_{m \mathop = 1}^\infty \paren {A_m \cos \frac {2 m \pi \paren {x - a} } {b - a} + B_m \sin \frac {2 m \pi \paren {x - a} } {b - a} }$
where:
{{begin-eqn}}
{{eqn | l = A_m
      | r = \dfrac 2 {b - a} \int_a^b \map f x \cos \frac {2 m \pi \paren {x - a} } {b - a} \rd x
}}
{{eqn | l = B_m
      | r = \dfrac 2 {b - a} \int_a^b \map f x \sin \frac {2 m \pi \paren {x - a} } {b - a} \rd x
}}
{{end-eqn}}

Proof:
Consider the Fourier series:
:$(1): \quad \ds \map S x = \frac {A_0} 2 + \sum_{m \mathop = 1}^\infty \paren {A_m \cos \frac {2 m \pi \paren {x - a} } {b - a} + B_m \sin \frac {2 m \pi \paren {x - a} } {b - a} }$
Let $\xi = \dfrac {2 \pi \paren {x - a} } {b - a}$.
Then:
:$\dfrac {\d \xi} {\d x} = \dfrac {2 \pi} {b - a}$
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \xi
      | r = \dfrac {2 \pi \paren {a - a} } {b - a}
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = x
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \xi
      | r = \dfrac {2 \pi \paren {b - a} } {b - a}
      | c = 
}}
{{eqn | r = 2 \pi
      | c = 
}}
{{end-eqn}}
Substituting $\xi$ for $x$ in $(1)$:
:$(2): \quad \map S \xi = \dfrac {A_0} 2 + \ds \sum_{m \mathop = 1}^\infty \paren {A_m \cos m \xi + B_m \sin m \xi}$
which is a Fourier series for a real function $\map g \xi$ such that $\ds \int_0^{2 \pi} \map g \xi \rd x$ converges absolutely.
The Fourier coefficients of $\map g \xi$ are given by:
{{begin-eqn}}
{{eqn | l = A_n
      | r = \dfrac 1 \pi \int_0^{2 \pi} \map g \xi \cos n \xi \rd \xi
}}
{{eqn | l = B_n
      | r = \dfrac 1 \pi \int_0^{2 \pi} \map g \xi \sin n \xi \rd \xi
}}
{{end-eqn}}
Substituting $x$ for $\xi$:
{{begin-eqn}}
{{eqn | l = A_n
      | r = \dfrac 1 \pi \dfrac {2 \pi} {b - a} \int_a^b \map g x \cos n \dfrac {2 \pi \paren {x - a} } {b - a} \rd x
}}
{{eqn | r = \dfrac 1 {b - a} \int_a^b \map g x \cos \dfrac {2 n \pi \paren {x - a} } {b - a} \rd x
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = B_n
      | r = \dfrac 1 \pi \dfrac {2 \pi} {b - a} \int_a^b \map g x \sin n \dfrac {2 \pi \paren {x - a} } {b - a} \rd x
}}
{{eqn | r = \dfrac 1 {b - a} \int_a^b \map g x \sin \dfrac {2 n \pi \paren {x - a} } {b - a} \rd x
}}
{{end-eqn}}
Thus while $A_n$ and $B_n$ are exactly the Fourier coefficients of $\map g \xi$, they are also exactly the Fourier coefficients of $\map S x$.
Thus $\map S x$ is the Fourier series for $\map f x$.
{{qed}}
Theorem: Definite Integral to Infinity of Reciprocal of 1 plus Power of x

:$\ds \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac \pi n \map \csc {\frac \pi n}$

Proof:
From Euler's Reflection Formula:
:$\map \Gamma {\dfrac 1 n} \map \Gamma {1 - \dfrac 1 n} = \pi \map \csc {\dfrac \pi n}$
Then: 
{{begin-eqn}}
{{eqn	| l = \map \Gamma {\frac 1 n} \map \Gamma {1 - \frac 1 n}
	| r = \frac {\map \Gamma {\frac 1 n} \map \Gamma {1 - \frac 1 n} } {\map \Gamma {1 - \frac 1 n + \frac 1 n} }
	| c = $\map \Gamma 1 = 1$
}}
{{eqn	| r = \map \Beta {\frac 1 n, 1 - \frac 1 n}
	| c = {{Defof|Beta Function|index = 3}}
}}
{{eqn	| r = 2 \int_0^{\pi / 2} \paren {\sin \theta}^{\frac 2 n - 1} \paren {\cos \theta}^{1 - \frac 2 n} \rd \theta
	| c = {{Defof|Beta Function|index = 2}}
}}
{{eqn	| r = 2 \int_0^{\pi / 2} \paren {\frac {\sin \theta} {\cos \theta} }^{\frac 2 n - 1} \rd \theta
}}
{{eqn	| r = 2 \int_0^{\pi / 2} \paren {\tan \theta}^{\frac 2 n - 1} \rd \theta 
	| c = {{Defof|Tangent Function}}
}}
{{end-eqn}}
Note we have: 
{{begin-eqn}}
{{eqn	| l = \frac {\map \d {\paren {\tan \theta}^{\frac 2 n} } } {\d \theta}
	| r = \frac {\map \d {\tan \theta} } {\d \theta} \cdot \frac {\map \d {\paren {\tan \theta}^{\frac 2 n} } } {\map \d {\tan \theta} }
	| c = Chain Rule for Derivatives
}}
{{eqn	| r = \sec^2 \theta \cdot \frac 2 n \paren {\tan \theta}^{\frac 2 n - 1}
	| c = Derivative of Tangent Function, Derivative of Power
}}
{{eqn	| r = \frac 2 n \paren {1 + \tan^2 \theta} \paren {\tan \theta}^{\frac 2 n - 1}
	| c = Difference of Squares of Secant and Tangent
}}
{{eqn	| r = \frac 2 n \paren {1 + \paren {\tan^{\frac 2 n} \theta}^n} \cdot \paren {\tan \theta}^{\frac 2 n - 1}
}}
{{end-eqn}}
As $\theta \to \dfrac \pi 2$, $\tan \theta \to \infty$ and $\tan 0 = 0$, so making a substitution of $x = \paren {\tan \theta}^{\frac 2 n}$ to our original integral:
{{begin-eqn}}
{{eqn	| l = 2 \int_0^{\pi / 2} \paren {\tan \theta}^{\frac 2 n - 1} \rd \theta
	| r = 2 \int_0^\infty \paren {\tan \theta}^{\frac 2 n - 1} \frac {\rd x} {\frac 2 n \paren {1 + \paren {\tan^{\frac 2 n} \theta}^n} \cdot \paren {\tan \theta}^{\frac 2 n - 1} }
	| c = Integration by Substitution
}}
{{eqn	| r = \frac {2 n} 2 \int_0^\infty \frac {\paren {\tan\theta}^{\frac 2 n - 1} } {\paren {\tan \theta}^{\frac 2 n - 1} } \frac {\rd x} {1 + x^n}
}}
{{eqn	| r = n \int_0^\infty \frac 1 {1 + x^n} \rd x
}}
{{end-eqn}}
So we have: 
:$\ds \pi \map \csc {\frac \pi n} = n \int_0^\infty \frac 1 {1 + x^n} \rd x$
Hence:
:$\ds \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac \pi n \map \csc {\frac \pi n}$
{{qed}}
Theorem: Definite Integral to Infinity of Reciprocal of 1 plus Power of x

:$\ds \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac \pi n \map \csc {\frac \pi n}$

Proof:
Let $R > 1$ be a real number.
Let:
:$\ds C_R = \set {R e^{i \theta} : 0 \le \theta \le \frac {2 \pi} n}$
{{explain|the fact that we are working in the complex plane. This is not immediately obvious.}}
Let $L_R$ be the straight line segment from $0$ to $R e^{\frac {2 \pi i} n}$.
Let $\Gamma_R = \closedint 0 R \cup C_R \cup L_R$, traversed anticlockwise. 
{{explain|Diagram}}
Then: 
{{begin-eqn}}
{{eqn | l = \oint_{\Gamma_R} \frac 1 {1 + z^n} \rd z
      | r = \int_0^R \frac 1 {1 + x^n} \rd x + \int_{C_R} \frac 1 {1 + z^n} \rd z + \int_{L_R} \frac 1 {1 + z^n} \rd z
}}
{{eqn | r = \int_0^R \frac 1 {1 + x^n} \rd x + \int_{C_R} \frac 1 {1 + z^n} \rd z + e^{\frac {2 \pi i} n} \int_R^0 \frac 1 {1 + e^{2 \pi i} x^n} \rd x
      | c = {{Defof|Complex Contour Integral}}
}}
{{eqn | r = \paren {1 - e^{\frac {2 \pi i} n} } \int_0^R \frac 1 {1 + x^n} \rd x + \int_{C_R} \frac 1 {1 + z^n} \rd z
}}
{{end-eqn}}
We can show the second integral to vanish as $R \to \infty$: 
{{begin-eqn}}
{{eqn | l = \cmod {\int_{C_R} \frac 1 {1 + z^n} \rd z}
      | o = \le
      | r = \frac {2 \pi R} n \max_{0 \le \theta \le \frac {2 \pi} n} \cmod {\frac 1 {1 + R^n e^{i n \theta} } }
      | c = Estimation Lemma for Contour Integrals
}}
{{eqn | r = \frac {2 \pi R} {n \paren {R^n - 1} }
}}
{{eqn | o = \sim
      | r = \frac {2 \pi} {n R^{n - 1} }
}}
{{eqn | o = \to
      | r = 0 
      | c = as $n > 1$
}}
{{end-eqn}}
So we have: 
:$\ds \oint_{\Gamma_\infty} \frac 1 {1 + z^n} \rd z = \paren {1 - e^{\frac {2 \pi i} n} } \int_0^\infty \frac 1 {1 + x^n} \rd x$
We have that the integrand is meromorphic with a single simple pole at $z = e^{\frac \pi n i}$.
This is contained within the contour
Hence Cauchy's Residue Theorem can be applied to evaluate the {{LHS}}:
{{begin-eqn}}
{{eqn | l = \oint_{\Gamma_\infty} \frac 1 {1 + z^n} \rd z
      | r = 2 \pi i \Res {\frac 1 {1 + z^n} } {e^{\frac \pi n i} }
}}
{{eqn | r = \frac {2 \pi i} {n e^{\frac \pi n \paren {n - 1} i} } 
      | c = Residue at Simple Pole
}}
{{eqn | r = \frac {2 \pi i e^{i \frac \pi n} } {n e^{\pi i} }
}}
{{eqn | r = -\frac {2 \pi i e^{i \frac \pi n} } n
      | c = Euler's Identity
}}
{{end-eqn}}
So: 
:$\ds \paren {e^{\frac {2 \pi i} n} - 1} \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac {2 \pi i e^{i \frac \pi n} } n$
giving: 
{{begin-eqn}}
{{eqn | l = \int_0^\infty \frac 1 {1 + x^n} \rd x
      | r = \frac \pi n \cdot 2 i \cdot \frac {e^{i \frac \pi n} } {e^{\frac {2 \pi i} n} - 1}
}}
{{eqn | r = \frac \pi n \cdot \frac {2 i e^{i \frac \pi n} } {e^{\frac \pi n i} \paren {e^{\frac \pi n i} - e^{-\frac \pi n i} } }
}}
{{eqn | r = \frac \pi n \cdot \frac {2 i} {e^{\frac \pi n i} - e^{-\frac \pi n i} }
}}
{{eqn | r = \frac \pi n \map \csc {\frac \pi n}
      | c = Euler's Cosecant Identity
}}
{{end-eqn}}
{{qed}}
Theorem: Definite Integral to Infinity of Reciprocal of 1 plus Power of x

:$\ds \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac \pi n \map \csc {\frac \pi n}$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = x
      | r = \paren {\tan \theta}^{\frac 2 n}
      | c = Integration by Substitution
}}
{{eqn | ll= \leadsto
      | l = \rd x
      | r = \frac 2 n \paren {\tan \theta}^{\frac 2 n - 1} \sec^2 \theta \rd \theta 
      | c = Derivative of Tangent Function, Derivative of Power, Chain Rule for Derivatives
}}
{{end-eqn}}
We also have:
{{begin-eqn}}
{{eqn | l = \tan 0
      | r = 0
      | c = Tangent of Zero
}}
{{eqn | l = \lim_{\theta \mathop \to \frac \pi 2} \map \tan {\theta}
      | o = \to
      | r = \infty
      | c = Tangent of Right Angle
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn	| l = \int_0^\infty \frac 1 {1 + x^n} \rd x
	    | r = \int_0^{\pi / 2} \frac 1 {1 + \paren {\paren {\tan \theta}^{\frac 2 n} }^n} \cdot \frac 2 n \paren {\tan \theta}^{\frac 2 n - 1} \sec^2 \theta \rd \theta 
	    | c = Integration by Substitution
}}
{{eqn	| r = \frac 2 n \int_0^{\pi / 2} \frac 1 {1 + \tan^2 \theta} \cdot \paren {\tan \theta}^{\frac 2 n - 1} \sec^2 \theta \rd \theta 
        | c = 
}}
{{eqn	| r = \frac 2 n \int_0^{\pi / 2} \frac 1 {\sec^2 \theta} \cdot \paren {\tan \theta}^{\frac 2 n - 1} \sec^2 \theta \rd \theta 
        | c = $\sec^2 \theta = 1 + \tan^2 \theta$
}}
{{eqn	| r = \frac 2 n \int_0^{\pi / 2} \paren {\tan \theta}^{\frac 2 n - 1} \rd \theta 
        | c = 
}}
{{eqn	| r = \frac 2 n \int_0^{\pi / 2} \paren {\frac {\sin \theta} {\cos \theta} }^{\frac 2 n - 1} \rd \theta
        | c = {{Defof|Tangent Function}}
}}
{{eqn	| r = \frac 2 n \int_0^{\pi / 2} \paren {\sin \theta}^{\frac 2 n - 1} \paren {\cos \theta}^{1 - \frac 2 n} \rd \theta
        | c = 
}}
{{eqn	| r = \frac 2 n \int_0^{\pi / 2} \paren {\sin \theta}^{2 \paren {\frac 1 n} - 1} \paren {\cos \theta}^{2 \paren {1 - \frac 1 n} -1 } \rd \theta
        | c = 
}}
{{eqn	| r = \frac 1 n \cdot \map \Beta {\frac 1 n, 1 - \frac 1 n}
	    | c = {{Defof|Beta Function|index = 2}}
}}
{{eqn	| r = \frac 1 n \cdot \frac {\map \Gamma {\frac 1 n} \map \Gamma {1 - \frac 1 n} } {\map \Gamma {1 - \frac 1 n + \frac 1 n} }
	    | c =  {{Defof|Beta Function|index = 3}}
}}
{{eqn	| r = \frac 1 n \cdot \dfrac \pi {\map \sin {\frac \pi n} }
	    | c = Euler's Reflection Formula and $\map \Gamma 1 = 1$
}}
{{eqn	| r = \frac \pi n \map \csc {\frac \pi n}
	    | c = {{Defof|Cosecant}}
}}
{{end-eqn}}
{{qed}}
Theorem: Series Expansion of Function over Complete Orthonormal Set

Let $\map f x$ be a real function defined over the interval $\openint a b$.
Let $\map f x$ be able to be expressed in terms of a complete orthonormal set of real functions $S := \family {\map {\phi_i} x}_{i \mathop \in I}$ for some indexing set $I$:
:$\map f x = \ds \sum_{i \mathop \in I} a_i \map {\phi_i} x$
Then the coefficients $\family {a_i}_{i \mathop \in I}$ can be determined as:
:$\forall i \in I: a_i = \ds \int_a^b \map f x \map {\phi_i} x \rd x$

Proof:
{{begin-eqn}}
{{eqn | l = \map f x
      | r = \sum_{i \mathop \in I} a_i \map {\phi_i} x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map f x \map {\phi_n} x
      | r = \sum_{i \mathop \in I} a_i \map {\phi_i} x \map {\phi_n} x
      | c = for arbitrary $n \in I$
}}
{{eqn | ll= \leadsto
      | l = \int_a^b \map f x \map {\phi_n} x \rd x
      | r = \int_a^b \sum_{i \mathop \in I} a_i \map {\phi_i} x \map {\phi_n} x rd x
      | c = 
}}
{{eqn | r = \sum_{i \mathop \in I} a_i \int_a^b \map {\phi_i} x \map {\phi_n} x \rd x
      | c = 
}}
{{eqn | r = \sum_{i \mathop \in I} a_i \delta_{i n}
      | c = {{Defof|Complete Orthonormal Set of Real Functions}}
}}
{{eqn | r = a_n
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Vector as Sum of Orthogonal Base Vectors

Let $\mathbf v$ be a vector quantity in ordinary $3$-space.
Let $\mathbf i, \mathbf j, \mathbf k$ be orthonormal base vectors.
Then:
:$\mathbf v = \paren {\mathbf v \cdot \mathbf i} \mathbf i + \paren {\mathbf v \cdot \mathbf j} \mathbf j + \paren {\mathbf v \cdot \mathbf k} \mathbf k$

Proof:
{{ProofWanted|Background needed}}