hcju/proofwikiqa huggingface dataset:
By definition of Fourier series over the range of integration $\openint \alpha {\alpha + 2 \pi}$: :$(1): \quad \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$ where: {{begin-eqn}} {{eqn | l = a_n       | r = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x }} {{eqn | l = b_n       | r = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \sin n x \rd x }} {{end-eqn}} Let $\xi = \dfrac {\pi x} \lambda$. Then let: :$\map \phi \xi \equiv \map f x$ We have that $\displaystyle \int_{\mathop \to \alpha}^{\mathop \to \alpha + 2 \lambda} \map f x \rd x$ converges absolutely. Thus we are ensured of the existence of: :the limit from the right of $f$ at $\alpha$ :the limit from the left of $f$ at $\alpha + 2 \lambda$ Thus $f$ is defined and bounded on $\openint \alpha {\alpha + 2 \lambda}$. Then: {{begin-eqn}} {{eqn | l = \map f {\alpha^+}       | r = \map \phi {\paren {\frac {\pi \alpha} \lambda}^+}       | c = {{Defof|Limit from Above}} }} {{eqn | l = \map f {\paren {\alpha + 2 \lambda}^-}       | r = \map \phi {\paren {\dfrac {\pi \alpha + \pi 2 \lambda} \lambda}^-}       | c =  }} {{eqn | r = \map \phi {\paren {\frac {\pi \alpha} \lambda + 2 \pi}^-}       | c =  }} {{end-eqn}} Setting $\beta = \dfrac {\pi \alpha} \lambda$, this allows us: {{begin-eqn}} {{eqn | l = \map f {\alpha^+}       | r = \map \phi {\beta^+}       | c =  }} {{eqn | l = \map f {\paren {\alpha + 2 \lambda}^-}       | r = \map \phi {\paren {\beta + 2 \pi}^-}       | c =  }} {{end-eqn}} Thus we have that $\phi$ is defined and bounded on $\openint \beta {\beta + 2 \pi}$. Then: :$\map \phi \xi \sim \displaystyle \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \xi + b_n \sin \xi}$ where: {{begin-eqn}} {{eqn | l = a_n       | r = \dfrac 1 \pi \int_{\mathop \to \pi}^{\mathop \to \alpha + 2 \lambda} \map \phi \xi \cos \xi \rd \xi }} {{eqn | l = b_n       | r = \dfrac 1 \pi \int_{\mathop \to \pi}^{\mathop \to \alpha + 2 \lambda} \map \phi \xi \sin \xi \rd \xi }} {{end-eqn}} We have that: :$\dfrac {\d \xi} {\d x} = \dfrac \pi \lambda$ and so: {{begin-eqn}} {{eqn | l = a_n       | r = \dfrac 1 \lambda \int_{\mathop \to \beta}^{\mathop \to \beta + 2 \lambda} \map f x \cos \frac {n \pi x} \lambda \rd x }} {{eqn | l = b_n       | r = \dfrac 1 \lambda \int_{\mathop \to \beta}^{\mathop \to \beta + 2 \lambda} \map f x \sin \frac {n \pi x} \lambda \rd x }} {{end-eqn}} and so: :$\displaystyle \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$ {{qed}}
{{begin-eqn}} {{eqn	| l = \operatorname {Li_2} \left({z}\right) + \operatorname {Li_2} \left({-z}\right) 	| r = -\left({\int_0^z \frac {\ln \left({1 - t}\right)} t \, \mathrm d t + \int_0^z \frac{\ln \left({1 + t}\right)} t \, \mathrm d t} \right) 	| c = {{Defof|Spence's Function}} }} {{eqn	| r = -\int_0^z \frac{\ln \left({\left({1 - t}\right) \left({1 + t}\right)} \right)} t \, \mathrm d t 	| c = Linear Combination of Integrals, Sum of Logarithms }} {{eqn	| r = -\int_0^z \frac{\ln \left({1 - t^2}\right)} t \, \mathrm d t 	| c = Difference of Two Squares }} {{eqn	| r = -\int_0^{z^2} \frac{\ln \left({1 - t^2}\right)} t \frac {\mathrm d \left({t^2}\right)} {2 t} 	| c = Substituting $t \to t^2$ }} {{eqn	| r = -\frac 1 2 \int_0^{z^2} \frac{\ln \left({1 - t^2}\right)} {t^2} \, \mathrm d \left({t^2}\right) }} {{eqn	| r = \frac 1 2 \operatorname {Li_2} \left({z^2}\right) 	| c = {{Defof|Spence's Function}} }} {{end-eqn}} {{qed}} Category:Spence's Function h2q89pio6nm1e79r7uvw0fabng82568
{{begin-eqn}} {{eqn	| l = \operatorname {Li}_2 \left({z}\right) 	| r = -\int_0^z \frac {\ln \left({1 - t}\right)} t \mathrm d t 	| c = {{Defof|Spence's Function}} }} {{eqn	| r = -\int_0^z \frac 1 t \sum_{n \mathop = 1}^\infty \left({-\frac {t^n} n}\right) \mathrm d t  	| c = Corollary to Power Series Expansion for $\ln \left({1 + x}\right)$ }} {{eqn	| r = \sum_{n \mathop = 1}^\infty \int_0^z \frac {t^{n - 1} } n \mathrm d t 	| c = Fubini's Theorem }} {{eqn	| r = \sum_{n \mathop = 1}^\infty \frac{z^n}{n^2} 	| c = Primitive of Power }} {{end-eqn}} {{qed}} Category:Spence's Function Category:Examples of Power Series dwgz58tqrdvicp1hprzp073rtasqtbl
By definition of odd function: :$\map f {-x} = -\map f x$ and so: {{begin-eqn}} {{eqn | l = \map f {-0}       | r = \map f 0       | c =  }} {{eqn | r = -\map f 0       | c =  }} {{end-eqn}} The only real number $a$ for which $a = -a$ is $0$. Hence the result. {{qed}}
As suggested, let the Fourier series of $\map f x$ be expressed as: :$\map f x \sim \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$ By definition of Fourier series: :$a_n = \dfrac 1 \lambda \displaystyle \int_{-\lambda}^{-\lambda + 2 \lambda} \map f x \cos \frac {n \pi x} \lambda \rd x$ 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 = Definite Integral of Odd Function: Corollary }} {{end-eqn}} {{qed}}
As suggested, let the Fourier series of $\map f x$ be expressed as: :$\displaystyle \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$ 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 the corollary to Definite Integral of Odd Function: :$\displaystyle \frac 1 \lambda \int_{-\lambda}^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x = 0$ {{qed}}
By definition of the Fourier series for $f$: :$\map f x \sim \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$ From Fourier Cosine Coefficients for Even Function over Symmetric Range: :$a_n = \displaystyle \dfrac 2 \lambda \int_0^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x$ 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}}
As suggested, let the Fourier series of $\map f x$ be expressed as: :$\map f x \sim \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$ 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 to Definite Integral of Odd Function: :$\displaystyle \frac 1 \lambda \int_{-\lambda}^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x = 0$ {{qed}}
As suggested, let the Fourier series of $\map f x$ be expressed as: :$\displaystyle \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$ By definition of Fourier series: :$b_n = \displaystyle \frac 1 \lambda \int_{-\lambda}^{-\lambda + 2 \lambda} \map f x \sin \frac {n \pi x} \lambda \rd x$ 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 = Definite Integral of Even Function: Corollary }} {{end-eqn}} {{qed}}
By definition of the Fourier series for $f$: :$\map f x \sim \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$ 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 :$b_n = \displaystyle \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x$ for all $n \in \Z_{> 0}$: {{qed}}
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}}
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}}
Consider the Fourier series: :$(1): \quad \displaystyle \map S x = \frac {A_0} 2 + \sum_{m \mathop = 1}^\infty \paren {A_m \cos \frac {2 m \pi \paren {x - a} } {b - a} + B_m \sin \frac {2 m \pi \paren {x - a} } {b - a} }$ 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 $(1)$: :$(2): \quad \map S \xi = \dfrac {A_0} 2 + \displaystyle \sum_{m \mathop = 1}^\infty \paren {A_m \cos m \xi + B_m \sin m \xi}$ which is a Fourier series for a real function $\map g \xi$ such that $\displaystyle \int_0^{2 \pi} \map g \xi \rd x$ 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}}
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 $\theta \nearrow \dfrac \pi 2$, $\tan \theta \to \infty$ and $\tan 0 = 0$, so making a substitution of $x = \paren {\tan \theta}^{\frac 2 n}$ to our original integral: {{explain|$\theta \nearrow \dfrac \pi 2$ needs to be explained, as it does not exist anywhere else on {{ProofWiki}}.}} {{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:  :$\displaystyle \pi \, \map \csc {\frac \pi n} = n \int_0^\infty \frac 1 {1 + x^n} \rd x$ Hence: :$\displaystyle \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac \pi n \, \map \csc {\frac \pi n}$ {{qed}}
Let $R > 1$ be a real number. Let: :$\displaystyle C_R = \set {Re^{i \theta} : 0 \le \theta \le \frac {2 \pi} n}$ 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 anti-clockwise.   {{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|Contour Integral}} }} {{eqn	| r = \left({1 - e^{\frac {2 \pi i} n} }\right) \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 }} {{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:  :$\displaystyle \oint_{\Gamma_\infty} \frac 1 {1 + z^n} \rd z = \paren {1 - e^{\frac {2 \pi i} n} } \int_0^\infty \frac 1 {1 + x^n} \rd x$ As the integrand is meromorphic with a single simple pole at $z = e^{\frac \pi n i}$ contained within the contour, the 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:  :$\displaystyle \paren {e^{\frac {2 \pi i} n} - 1} \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac {2 \pi i e^{i \frac \pi n} } n$ 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 \csc \left({\frac \pi n}\right) 	| c = Cosecant Exponential Formulation }} {{end-eqn}} {{qed}}
Consider the definite integral: :$I_{m n} = \displaystyle \int_0^\lambda \map {\phi_m} x \map {\phi_n} x \rd x$ From Sine Function is Odd, each of $\map {\phi_n} x$ is an odd function. From Odd Function Times Odd Function is Even, $\map {\phi_m} x \map {\phi_n} x$ is even. That is: :$\paren {\sqrt {\dfrac 2 \lambda} \sin \dfrac {m \pi x} \lambda} \paren {\sqrt {\dfrac 2 \lambda} \sin \dfrac {n \pi x} \lambda}$ is an even function. Let $u = \dfrac {\pi x} \lambda$. We have: {{begin-eqn}} {{eqn | l = \dfrac {\d u} {\d x}       | r = \dfrac \pi \lambda       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\d x} {\d u}       | r = \dfrac \lambda \pi       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = x       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = u       | r = \dfrac {\pi \times 0} \lambda       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = x       | r = \lambda       | c =  }} {{eqn | ll= \leadsto       | l = u       | r = \dfrac {\pi \times \lambda} \lambda       | c =  }} {{eqn | r = \pi       | c =  }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = I_{m n}       | r = \int_0^\lambda \map {\phi_m} x \map {\phi_n} x \rd x       | c =  }} {{eqn | r = \int_0^\lambda \paren {\sqrt {\dfrac 2 \lambda} \sin \frac {m \pi x} \lambda} \paren {\sqrt {\dfrac 2 \lambda} \sin \frac {n \pi x} \lambda} \rd x       | c =  }} {{eqn | r = \frac 2 \lambda \int_0^\lambda \sin \frac {m \pi x} \lambda \sin \frac {n \pi x} \lambda \rd x       | c = Linear Combination of Integrals }} {{eqn | r = \frac 2 \lambda \int_0^\pi \frac \lambda \pi \sin m u \sin n u \rd u       | c = Integration by Substitution }} {{eqn | r = \frac 2 \lambda \frac \lambda \pi \int_0^\pi \sin m u \sin n u \rd u       | c = Linear Combination of Integrals }} {{eqn | r = \frac 2 \pi \int_0^\pi \sin m u \sin n u \rd u       | c =  }} {{eqn | r = \frac 1 \pi \int_{-\pi}^\pi \sin m u \sin n u \rd u       | c = Definite Integral of Even Function }} {{eqn | r = \frac 1 \pi \pi \delta_{m n}       | c = Integral over $2 \pi$ of $\sin m u \sin n u$ }} {{eqn | r = \delta_{m n}       | c =  }} {{end-eqn}} Hence the result by definition of orthonormal set. {{qed}}
Follows from Correspondence Theorem for Ring Epimorphisms/Bijection {{qed}} Category:Quotient Rings 3w1z2puogcl7deb6rpox8rno635czn6
The first statements follow from: * Preimage of Ideal under Ring Homomorphism is Ideal * Image of Ideal under Ring Epimorphism is Ideal The last statement follows from: * Image of Preimage of Ideal under Ring Epimorphism * Preimage of Image of Ideal under Ring Homomorphism {{qed}} Category:Ring Homomorphisms 6brohngslt7sxuhimqju0i9m8a0acx8
By Prime Ideal iff Quotient Ring is Integral Domain, $0$ is prime {{iff}} the quotient ring $A / 0$ is an integral domain. By Quotient Ring by Null Ideal, $A \cong A / 0$. {{qed}} Category:Integral Domains Category:Prime Ideals of Rings 4jdfrnqgjrz9fwwlgfghtws97gnx8jb
By definition, the kernel of a ring homomorphism is its kernel when considered as a group homomorphism between the additive groups. The result follows from Kernel of Composition of Group Homomorphisms. {{qed}} Category:Ring Homomorphisms sdjqic2fcbk30acpe13xsgurfciujpe
By definition, $f$ is a semigroup homomorphism between multiplicative semigroups. A unity of a ring is by definition an identity element of its multiplicative semigroup. Thus the result follows from Epimorphism Preserves Identity. {{qed}} Category:Ring Epimorphisms 9nofbcnttzpsssgf79g3ug472j6fwcu
{{AimForCont}} $R \setminus P$ is not multiplicatively closed. That is: :$\exists a, b \in R \setminus P: a b \notin R \setminus P$ This means: :$a b \in P$ This contradicts the assertion that $P$ is a prime ideal of $R$. {{qed}} Category:Prime Ideals of Rings qnn2pb7ia8cmy0049a3850n1nxqpgf6
{{begin-eqn}} {{eqn	| l = \map {\Gamma'} x 	| r = \frac \d {\d x} \int_0^\infty t^{x - 1} e^{-t} \rd t 	| c = {{Defof|Gamma Function}} }} {{eqn	| r = \int_0^\infty \frac {\partial} {\partial x} t^{x - 1} e^{-t} \rd t 	| c = Leibniz's Integral Rule }} {{eqn	| r = \int_0^\infty t^{x-1} \ln t \, e^{-t} \rd t 	| c = Derivative of Power of Constant }} {{end-eqn}} {{qed}} Category:Gamma Function Category:Derivatives lzqf3v058gnfq24aztg36vkj6zedv76