avewright/proofwiki-math huggingface dataset:
Theorem: Laplace Transform of Half Wave Rectified Sine Curve

Consider the half wave rectified sine curve:
:$\map f t = \begin {cases} \sin t & : 2 n \pi \le t \le \paren {2 n + 1} \pi \\ 0 & : \paren {2 n + 1} \pi \le t \le \paren {2 n + 2} \pi \end {cases}$
The Laplace transform of $\map f t$ is given by:
:$\laptrans {\map f t} = \dfrac 1 {\paren {1 - e^{-\pi s} } \paren {s^2 + 1} }$

Proof:
We have that $\map f t$ is periodic with period $2 \pi$:
:800px
Hence:
{{begin-eqn}}
{{eqn | l = \laptrans {\map f t}
      | r = \dfrac 1 {1 - e^{-2 \pi s} } \int_0^{2 \pi} e^{-s t} \map f t \rd t
      | c = Laplace Transform of Periodic Function
}}
{{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \paren {\int_0^\pi e^{-s t} \sin t \rd t + \int_\pi^{2 \pi} e^{-s t} \times 0 \rd t}
      | c = 
}}
{{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \int_0^\pi e^{-s t} \sin t \rd t
      | c = 
}}
{{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \intlimits {\dfrac {e^{-s t} \paren {-s \sin t - \cos t} } {s^2 + 1} } 0 \pi
      | c = Primitive of $e^{a x} \sin b x$
}}
{{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} \paren {-s \sin \pi - \cos \pi} } {s^2 + 1} - \dfrac {e^{-s \times 0} \paren {-s \sin 0 - \cos 0} } {s^2 + 1} }
      | c = 
}}
{{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} \paren {-\cos \pi} } {s^2 + 1} - \dfrac {e^{-s \times 0} \paren {- \cos 0} } {s^2 + 1} }
      | c = Sine of Integer Multiple of Pi
}}
{{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} \paren {-\paren {-1} } } {s^2 + 1} - \dfrac {e^{-s \times 0} \paren {-1} } {s^2 + 1} }
      | c = Cosine of Integer Multiple of Pi
}}
{{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} \paren {-\paren {-1} } } {s^2 + 1} - \dfrac {1 \times \paren {-1} } {s^2 + 1} }
      | c = Exponential of Zero
}}
{{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} + 1} {s^2 + 1} }
      | c = simplification
}}
{{eqn | r = \dfrac 1 {\paren {1 + e^{-\pi s} } \paren {1 - e^{-\pi s} } } \paren {\dfrac {e^{-s \pi} + 1} {s^2 + 1} }
      | c = Difference of Two Squares
}}
{{eqn | r = \dfrac 1 {\paren {1 - e^{-\pi s} } \paren {s^2 + 1} }
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Integral to Infinity of Exponential of -t^2

:$\ds \int_0^\infty \map \exp {-t^2} \rd t = \dfrac {\sqrt \pi} 2$

Proof:
Let $\ds I = \int_0^\infty \map \exp {-t^2} \rd t$.
Let $\ds I_P = \int_0^P \map \exp {-x^2} \rd x = \int_0^P \map \exp {-y^2} \rd y$.
Then we have:
:$I = \ds \lim_{P \mathop \to \infty} I_P$
Hence:
{{begin-eqn}}
{{eqn | l = {I_P}^2
      | r = \paren {\int_0^P \map \exp {-x^2} \rd x} \paren {\int_0^P \map \exp {-y^2} \rd y}
      | c = 
}}
{{eqn | r = \int_0^P \int_0^P \map \exp {-\paren {x^2 + y^2} } \rd x \rd y
      | c = 
}}
{{eqn | r = \iint_{\mathscr R_P} \map \exp {-\paren {x^2 + y^2} } \rd x \rd y
      | c = 
}}
{{end-eqn}}
where $\mathscr R_P$ is the square $\Box OACE$ in the figure below:
:400px
Because the integrand is positive:
{{begin-eqn}}
{{eqn | n = 1
      | l = \iint_{\mathscr R_1} \map \exp {-\paren {x^2 + y^2} } \rd x \rd y
      | o = \le
      | r = {I_P}^2
      | c = 
}}
{{eqn | o = \le
      | r = \iint_{\mathscr R_2} \map \exp {-\paren {x^2 + y^2} } \rd x \rd y
      | c = 
}}
{{end-eqn}}
where $\mathscr R_1$ and $\mathscr R_2$ are the regions in the first quadrant bounded by the circles with centers $O$ and radii $P$ and $P \sqrt 2$ respectively.
Using polar coordinates, we can express $(1)$ as:
{{begin-eqn}}
{{eqn | n = 1
      | l = \int_{\theta \mathop = 0}^{\frac \pi 2} \int_{r \mathop = 0}^P \map \exp {-r^2} r \rd r \rd \theta
      | o = \le
      | r = {I_P}^2
      | c = 
}}
{{eqn | o = \le
      | r = \int_{\theta \mathop = 0}^{\frac \pi 2} \int_{r \mathop = 0}^{P \sqrt 2} \map \exp {-r^2} r \rd r \rd \theta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac \pi 4 \paren {1 - \map \exp {-P^2} }
      | o = \le
      | r = {I_P}^2
      | c = 
}}
{{eqn| o = \le
      | r = \dfrac \pi 4 \paren {1 - \map \exp {-2 P^2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \lim_{P \mathop \to \infty} {I_P}^2
      | r = I^2
      | c = 
}}
{{eqn | r = \dfrac \pi 4
      | c = 
}}
{{eqn | ll= \leadsto
      | l = I
      | r = \dfrac {\sqrt \pi} 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Integral to Infinity of Exponential of -t^2

:$\ds \int_0^\infty \map \exp {-t^2} \rd t = \dfrac {\sqrt \pi} 2$

Proof:
Let $\lambda$ be a non-negative real number. 
Then, we have: 
:$\ds \size {\frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} } \le \frac 1 {1 + x^2}$
for each $x \in \R$.
Note that from Definite Integral to Infinity of $\dfrac 1 {x^2 + a^2}$:
:$\ds \int_0^\infty \frac 1 {x^2 + 1} \rd x = \frac \pi 2$ 
So by the Comparison Test for Improper Integral:
:$\ds \int_0^\infty \frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} \rd x$ converges.
We can therefore define a real function $I : \hointr 0 \infty \to \R$ by: 
:$\ds \map I \lambda = \int_0^\infty \frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} \rd x$
for each $\lambda \in \hointr 0 \infty$.
We then have from Definite Integral of Partial Derivative, for $\lambda > 0$:
{{begin-eqn}}
{{eqn	| l = \map {I'} \lambda
	| r = \frac \d {\d \lambda} \paren {\int_0^\infty \frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} \rd x}
}}
{{eqn	| r = \int_0^\infty \frac \partial {\partial \lambda} \frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} \rd x
}}
{{eqn	| r = -2 \lambda \int_0^\infty e^{-\lambda^2 \paren {1 + x^2 } } \rd x
	| c = Derivative of Exponential Function
}}
{{eqn	| r = -2 \lambda e^{-\lambda^2} \int_0^\infty e^{-\paren {\lambda x}^2} \rd x 
}}
{{eqn	| r = -2 e^{-\lambda^2} \int_0^\infty e^{-t^2} \rd t
	| c = substituting $t \mapsto \lambda x$
}}
{{end-eqn}}
We have, using Definite Integral to Infinity of $\dfrac 1 {x^2 + a^2}$ again: 
:$\ds \map I 0 = \int_0^\infty \frac 1 {1 + x^2} \rd x = \frac \pi 2$ 
We also have, for $\lambda > 0$:
{{begin-eqn}}
{{eqn	| l = 0
	| o = \le
	| r = \int_0^\infty \frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} \rd x
}}
{{eqn	| o = \le
	| r = \int_0^\infty e^{-\lambda^2 \paren {1 + x^2} } \rd x
}}
{{eqn	| o = \le
	| r = \int_0^\infty e^{-\lambda x} \rd x
	| c = since $x \le 1 + x^2$
}}
{{eqn	| r = \frac 1 \lambda
	| c = Laplace Transform of Real Power
}}
{{end-eqn}}
so:
:$\ds \lim_{\lambda \to \infty} \int_0^\infty \frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} \rd x = 0$
Now we have: 
{{begin-eqn}}
{{eqn	| l = \int_0^\infty \map {I'} \lambda \rd \lambda
	| r = \lim_{R \to \infty} \int_0^R \map {I'} \lambda \rd \lambda
}}
{{eqn	| r = \lim_{R \to \infty} \paren {\map I R - \map I 0}
}}
{{eqn	| r = -\frac \pi 2
}}
{{end-eqn}}
On the other hand: 
{{begin-eqn}}
{{eqn	| l = \int_0^\infty \map {I'} \lambda \rd \lambda
	| r = -2 \lim_{R \to \infty} \int_0^R \map {I'} \lambda \rd \lambda
}}
{{eqn	| r = -2 \int_0^\infty e^{-t^2} \rd t \lim_{R \to \infty} \int_0^R e^{-\lambda^2} \rd \lambda
}}
{{eqn	| r = -2 \paren {\int_0^\infty e^{-t^2} \rd t}^2
	| c = {{Defof|Improper Integral on Closed Interval Unbounded Above}}
}}
{{end-eqn}}
So:
:$\ds \paren {\int_0^\infty e^{-t^2} \rd t}^2 = \frac \pi 4$
Since:
:$\ds \int_0^\infty e^{-t^2} \rd t \ge 0$
we have: 
:$\ds \int_0^\infty e^{-t^2} \rd t = \frac {\sqrt \pi} 2$
{{qed}}
Theorem: Laplace Transform of Real Power

Let $n$ be a constant real number such that $n > -1$
Let $f: \R \to \R$ be the real function defined as:
:$\map f t = t^n$
Then $f$ has a Laplace transform given by:
{{begin-eqn}}
{{eqn | l = \laptrans {\map f t}
      | r = \int_0^\infty e^{-s t} t^n \rd t
}}
{{eqn | r = \frac {\map \Gamma {n + 1} } {s^{n + 1} }
}}
{{end-eqn}}
where $\Gamma$ denotes the gamma function.

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {t^n}
      | r = \int_0^\infty e^{-s t} t^n \rd t
      | c = {{Defof|Laplace Transform}}
}}
{{eqn | r = \int_0^\infty e^{-u} \paren {\dfrac u s}^n \rd \paren {\dfrac u s}
      | c = Integration by Substitution: $u := s t$ where $s > 0$ is assumed
}}
{{eqn | r = \dfrac 1 {s^{n + 1} } \int_0^\infty u^n e^{-u} \rd u
      | c = 
}}
{{eqn | r = \dfrac {\map \Gamma {n + 1} } {s^{n + 1} }
      | c = {{Defof|Gamma Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Laplace Transform of Reciprocal of Square Root

:$\forall t \in \R_{\ne 0}: \laptrans {\dfrac 1 {\sqrt t} } = \sqrt {\dfrac \pi s}$
where $\laptrans f$ denotes the Laplace transform of the real function $f$.

Proof:
Let $\map f t = \dfrac 1 {\sqrt t}$.
By definition of the Laplace transform with a discontinuity at zero, $\laptrans f$ is the improper integral:
:$\ds \laptrans f := \lim_{\epsilon \mathop \to 0^+} \int_\epsilon^{\to +\infty} \dfrac {e^{-s t} } {\sqrt t} \rd t$
if it exists.
{{explain|Demonstrate that it does exist}}
It follows that this improper integral converges.
So the Laplace transform $\laptrans f$ exists.
Hence:
{{begin-eqn}}
{{eqn | l = \laptrans f
      | r = \dfrac {\map \Gamma {1 / 2} } {s^{1 / 2} }
      | c = Laplace Transform of Real Power
}}
{{eqn | r = \dfrac {\sqrt \pi} {\sqrt s}
      | c = Gamma Function of One Half
}}
{{eqn | r = \sqrt {\dfrac \pi s}
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Gamma Function of 3 over 2

:$\map \Gamma {\dfrac 3 2} = \dfrac {\sqrt \pi} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \map \Gamma {\dfrac 3 2}
      | r = \map \Gamma {\dfrac 1 2 + 1}
      | c = 
}}
{{eqn | r = \dfrac 1 2 \map \Gamma {\dfrac 1 2}
      | c = Gamma Difference Equation
}}
{{eqn | r = \dfrac {\sqrt \pi} 2
      | c = Gamma Function of One Half
}}
{{end-eqn}}
{{qed}}
Category:Examples of Gamma Function Values
270bagi9c1wqvb1m7ne040jo5gc3oug
Theorem: Gamma Function of Minus 3 over 2

:$\map \Gamma {-\dfrac 3 2} = \dfrac {4 \sqrt \pi} 3$

Proof:
{{begin-eqn}}
{{eqn | l = \map \Gamma {-\dfrac 1 2}
      | r = -\dfrac 3 2 \map \Gamma {-\dfrac 3 2}
      | c = Gamma Difference Equation
}}
{{eqn | ll= \leadsto
      | l = \map \Gamma {-\dfrac 3 2}
      | r = -\dfrac 2 3 \map \Gamma {-\dfrac 1 2}
      | c = 
}}
{{eqn | r = -\dfrac 2 3 \paren {-2 \sqrt \pi}
      | c = Gamma Function of Minus One Half
}}
{{eqn | r = \dfrac {4 \sqrt \pi} 3
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Gamma Function of Minus 5 over 2

:$\map \Gamma {-\dfrac 5 2} = -\dfrac {8 \sqrt \pi} {15}$

Proof:
{{begin-eqn}}
{{eqn | l = \map \Gamma {-\dfrac 3 2}
      | r = -\dfrac 5 2 \map \Gamma {-\dfrac 5 2}
      | c = Gamma Difference Equation
}}
{{eqn | ll= \leadsto
      | l = \map \Gamma {-\dfrac 5 2}
      | r = -\dfrac 2 5 \map \Gamma {-\dfrac 3 2}
      | c = 
}}
{{eqn | r = -\dfrac 2 5 \paren {\dfrac {4 \sqrt \pi} 3}
      | c = Gamma Function of Minus 3 over 2
}}
{{eqn | r = -\dfrac {8 \sqrt \pi} {15}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Gamma Function of Zero

:$\map \Gamma 0$ is not defined.

Proof:
{{begin-eqn}}
{{eqn | l = \map \Gamma 1
      | r = 0 \, \map \Gamma 0
      | c = Gamma Difference Equation
}}
{{eqn | ll= \leadsto
      | l = \map \Gamma 0
      | r = \dfrac {\map \Gamma 1} 0
      | c = 
}}
{{eqn | r = \dfrac 1 0
      | c = Gamma Function Extends Factorial
}}
{{end-eqn}}
But $\dfrac 1 0$ is not defined.
Hence the result.
{{qed}}
Theorem: Gamma Function of Minus 1

:$\map \Gamma {-1}$ is not defined.

Proof:
{{begin-eqn}}
{{eqn | l = \map \Gamma 0
      | r = \paren {-1} \, \map \Gamma {-1}
      | c = Gamma Difference Equation
}}
{{eqn | ll= \leadsto
      | l = \map \Gamma {-1}
      | r = \dfrac {\map \Gamma 0} {-1}
      | c = 
}}
{{end-eqn}}
But from Gamma Function of Zero, $\map \Gamma 0$ is not defined.
Hence the result.
{{qed}}
Theorem: Gamma Function of Minus 2

:$\map \Gamma {-2}$ is not defined.

Proof:
{{begin-eqn}}
{{eqn | l = \map \Gamma {-1}
      | r = \paren {-2} \, \map \Gamma {-2}
      | c = Gamma Difference Equation
}}
{{eqn | ll= \leadsto
      | l = \map \Gamma {-2}
      | r = \dfrac {\map \Gamma {-1} } {-2}
      | c = 
}}
{{end-eqn}}
But from Gamma Function of Minus 1, $\map \Gamma {-1}$ is not defined.
Hence the result.
{{qed}}
Theorem: Integral to Infinity of Bessel Function of First Kind order Zero

:$\ds \int_0^\infty \map {J_0} t \rd t = 1$
where $J_0$ denotes the Bessel function of the first kind of order $0$.

Proof:
Using the technique of Evaluation of Integral using Laplace Transform:
{{begin-eqn}}
{{eqn | l = \int_0^\infty e^{-s t} \map {J_0} t \rd t
      | r = \dfrac 1 {\sqrt {s^2 + 1} }
      | c = Laplace Transform of Bessel Function of the First Kind of Order Zero
}}
{{eqn | ll= \leadsto
      | l = \int_0^\infty \map {J_0} t \rd t
      | r = 1
      | c = letting $s \to 0^+$
}}
{{end-eqn}}
{{qed}}
Theorem: Integral to Infinity of e^-t by Gauss Error Function of Root t

:$\ds \int_0^\infty e^{-t} \erf \sqrt t \rd t = \dfrac {\sqrt 2} 2$

Proof:
Using the technique of Evaluation of Integral using Laplace Transform:
{{begin-eqn}}
{{eqn | l = \int_0^\infty e^{-s t} \erf \sqrt t \rd t
      | r = \dfrac 1 {s \sqrt {s + 1} }
      | c = Laplace Transform of Gauss Error Function of Root
}}
{{eqn | ll= \leadsto
      | l = \int_0^\infty \erf \sqrt t \rd t
      | r = \dfrac 1 {1 \sqrt {1 + 1} }
      | c = letting $s \to 0^+$
}}
{{eqn | r = \dfrac 1 {\sqrt 2}
      | c = simplifying
}}
{{eqn | r = \dfrac {\sqrt 2} 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Laplace Transform of Natural Logarithm

:$\laptrans {\ln t} = \dfrac {\map {\Gamma'} 1 - \ln s} s = -\dfrac {\gamma + \ln s} s$
where:
:$\laptrans f$ denotes the Laplace transform of the function $f$
:$\Gamma$ denotes the Gamma function
:$\gamma$ denotes the Euler-Mascheroni constant.

Proof:
From Laplace Transform of Power:
:$\ds \int_0^\infty e^{-s t} t^k \rd t = \dfrac {\map \Gamma {k + 1} } {s^{k + 1} }$
for $k > -1$.
Differentiating {{WRT|Differentiation}} $k$:
:$\ds \int_0^\infty e^{-s t} t^k \ln t \rd t = \dfrac {\map {\Gamma'} {k + 1} - \map \Gamma {k + 1} \ln s} {s^{k + 1} }$
Setting $k = 0$:
{{begin-eqn}}
{{eqn | l = \int_0^\infty e^{-s t} \ln t \rd t
      | r = \laptrans {\ln t}
      | c = 
}}
{{eqn | r = \dfrac {\map {\Gamma'} 1 - \ln s}s
      | c = 
}}
{{eqn | r = -\dfrac {\gamma + \ln s} s
      | c = Derivative of Gamma Function at 1
}}
{{end-eqn}}
{{qed}}
Theorem: Second Derivative of Laplace Transform

Let $f: \R \to \R$ or $\R \to \C$ be a continuous function, twice differentiable on any closed interval $\closedint 0 a$.
Let $\laptrans f = F$ denote the Laplace transform of $f$.
Then, everywhere that $\dfrac {\d^2} {\d s^2} \laptrans f$ exists:
:$\dfrac {\d^2} {\d s^2} \laptrans {\map f t} = \laptrans {t^2 \, \map f t}$

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac {\d^2} {\d s^2} \laptrans {\map f t}
      | r = \map {\frac \d {\d s} } {\dfrac \d {\d s} \laptrans {\map f t} }
      | c = {{Defof|Second Derivative}}
}}
{{eqn | r = \map {\frac \d {\d s} } {-\laptrans {t \, \map f t} }
      | c = Derivative of Laplace Transform
}}
{{eqn | r = -\frac \d {\d s} \laptrans {t \, \map f t}
      | c = 
}}
{{eqn | r = -\paren {-\laptrans {t \paren {t \, \map f t} } }
      | c = Derivative of Laplace Transform
}}
{{eqn | r = \laptrans {t^2 \, \map f t}
}}
{{end-eqn}}
{{qed}}
Theorem: Laplace Transform of Multiple Integral

:$\ds \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$n$ times} } \map f u \rd u^n} = \dfrac {\map F s} {s^n}$
wherever $\laptrans f$ exists.

Proof:
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$n$ times} } \map f u \rd u^n} = \dfrac {\map F s} {s^n}$
$\map P 0$ is the case:
:$\map f u = \map F s$
which is the statement of the Laplace transform.
Thus $\map P 0$ is seen to hold.
Theorem: Integral of Reciprocal is Divergent/Unbounded Above

:$\ds \int_1^n \frac {\d x} x \to +\infty$ as $n \to + \infty$

Proof:
From Harmonic Series is Divergent, we have that $\ds \sum_{n \mathop = 1}^\infty \frac 1 n$ diverges to $+\infty$.
Thus from the Cauchy Integral Test:
:$\ds \int_1^n \frac {\d x} x \to +\infty$
diverges.
{{qed}}
Theorem: Integral of Reciprocal is Divergent/Unbounded Above

:$\ds \int_1^n \frac {\d x} x \to +\infty$ as $n \to + \infty$

Proof:
From the definition of natural logarithm:
:$\ds \ln x = \int_1^x \dfrac 1 t \rd t$
The result follows from Logarithm Tends to Infinity.
{{qed}}
Theorem: Integral of Reciprocal is Divergent/To Zero

:$\ds \int_\gamma^1 \frac {\d x} x \to -\infty$ as $\gamma \to 0^+$

Proof:
Put $x = \dfrac 1 z$.
Then:
{{begin-eqn}}
{{eqn | l = \int_\gamma^1 \frac {\d x} x
      | r = \int_{1 / \gamma}^1 \frac {-z} {z^2} \rd z
      | c = Integration by Substitution
}}
{{eqn | r = \int_1^{1 / \gamma} \frac {\d z} z
      | c = 
}}
{{end-eqn}}
But as $\gamma \to 0+$, we have that $\dfrac 1 \gamma \to +\infty$.
Hence, from Integral of Reciprocal Unbounded Above is Divergent:
:$\ds \int_1^{1 / \gamma} \frac {\d z} z \to +\infty$
as $\dfrac 1 \gamma \to +\infty$.
The result follows.
{{qed}}
Theorem: Convergence of P-Series/Absolute Convergence if Real Part of p Greater than 1

Let $\map \Re p > 1$.
Then the $p$-series:
:$\ds \sum_{n \mathop = 1}^\infty n^{-p}$
converges absolutely.

Proof:
=== Lemma ===
{{:Convergence of P-Series/Lemma}}
Since $x > 1$ it follows that $1 - x < 0$.
Thus $P^{1 - x} \to 0$ as $P \to \infty$.
Setting $x - 1 = \delta >0$, this limit is:
:$\ds -\frac 1 {\delta} \lim_{t \mathop \to \infty} \frac 1 {t^\delta} = 0$ 
Hence the integral is just $\dfrac 1 {1 - x}$ (that is, convergent) and so the sum converges as well.
Since the terms of the sum were positive everywhere, it is absolutely convergent and hence so is:
:$\ds \sum_{n \mathop = 1}^\infty n^{-p}$
{{qed}}
Theorem: Convergence of P-Series/Divergence if p between 0 and 1

Let $0 < \map \Re p \le 1$.
Then the $p$-series:
:$\ds \sum_{n \mathop = 1}^\infty n^{-p}$
diverges.

Proof:
=== Lemma ===
{{:Convergence of P-Series/Lemma}}{{qed|lemma}}
Hence, the convergence of the $p$-series is dependent on the convergence of: 
:$\ds \lim_{t \mathop \to \infty} \frac {t^{1 - x} } {1 - x}$
Suppose $0 < x < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{t \mathop \to \infty} \frac {t^{1 - x} } {1 - x}
      | r = \frac 1 {1 - x} \lim_{t \mathop \to \infty} t^{1 - x}
}}
{{eqn | o = \to
      | r = +\infty
      | c = Limit at Infinity of $x^n$
}}
{{end-eqn}}
The special case of $x = 1$ is covered by Integral of Reciprocal is Divergent.
Hence the result from the Cauchy Integral Test.
{{qed}}
Category:Convergence of P-Series
lln0bk249rctnmaujmrh2ds719w37yo
Theorem: Integral to Infinity of Reciprocal of Power of x

The improper integral
:$\ds \int_1^\infty \dfrac {\d t} {t^x}$
exists {{iff}} $x > 1$.

Proof:
First let $x \ne 1$.
Then:
{{begin-eqn}}
{{eqn | l = \int_1^\infty \dfrac {\d t} {t^x}
      | r = \lim_{P \mathop \to \infty} \int_1^P t^{-x} \rd t
      | c = {{Defof|Improper Integral}}
}}
{{eqn | r = \lim_{P \mathop \to \infty} \intlimits {\dfrac {t^{-x + 1} } {-x + 1} } 1 P
      | c = Primitive of Power
}}
{{eqn | r = \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} - \dfrac {1^{1 - x} } {1 - x} }
      | c = 
}}
{{eqn | r = \lim_{P \mathop \to \infty} \dfrac 1 {x - 1} \paren {1 - \dfrac 1 {P^{x - 1} } }
      | c = simplifying
}}
{{end-eqn}}
If $x > 1$, then $x - 1 > 0$.
Hence from Sequence of Powers of Reciprocals is Null Sequence, $\dfrac 1 {P^{x - 1} } \to 0$ as $P \to +\infty$.
If $x < 1$, then $x - 1 < 0$.
Hence $P^{x - 1} \to 0$ as $P \to +\infty$.
Then from Reciprocal of Null Sequence it follows that $\dfrac 1 {P^{x - 1} } \to \infty$ as $P \to +\infty$.
Finally we have that from Integral of Reciprocal is Divergent:
:$\ds \lim_{P \mathop \to \infty} \int_1^P \dfrac {\d t} t \to \infty$
All cases are covered, and the result follows.
{{qed}}
Theorem: Convergence of P-Series/Real

Let $p \in \R$ be a real number.
Then the $p$-series:
:$\ds \sum_{n \mathop = 1}^\infty n^{-p}$
is convergent {{iff}} $p > 1$.

Proof:
By the Cauchy Integral Test:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^x}$ converges {{iff}} the improper integral $\ds \int_1^\infty \frac {\d t} {t^x}$ exists.
The result follows from Integral to Infinity of Reciprocal of Power of x.
{{qed}}
Theorem: Convergence of P-Series/Real

Let $p \in \R$ be a real number.
Then the $p$-series:
:$\ds \sum_{n \mathop = 1}^\infty n^{-p}$
is convergent {{iff}} $p > 1$.

Proof:
Let $p = 1$.
Then from Harmonic Series is Divergent the $p$-series diverges.
So let $p > 1$.
We note that the sequence of partial sums is increasing.
Hence it is sufficient to show that they are bounded above.
Let:
:$s_{2^N} := 1 + \dfrac 1 {2^p} + \dfrac 1 {3^p} + \dotsb + \dfrac 1 {N^p}$
Then:
{{begin-eqn}}
{{eqn | l = s_N
      | o = \le
      | r = S_{2^N - 1}
      | c = 
}}
{{eqn | r = 1 + \dfrac 1 {2^p} + \dfrac 1 {3^p} + \dfrac 1 {4^p} + \dotsb + \dfrac 1 {\paren {2^N - 1}^p}
      | c = 
}}
{{eqn | r = 1 + \paren {\dfrac 1 {2^p} + \dfrac 1 {3^p} } + \paren {\dfrac 1 {4^p} + \dfrac 1 {5^p} + \dfrac 1 {6^p} + \dfrac 1 {7^p} } + \dotsb + \paren {\dfrac 1 {2^{\paren {N - 1} p} } + \dotsb + \dfrac 1 {\paren {2^N - 1}^p} }
      | c = 
}}
{{eqn | o = \le
      | r = 1 + \paren {\dfrac 1 {2^p} + \dfrac 1 {2^p} } + \paren {\dfrac 1 {4^p} + \dfrac 1 {4^p} + \dfrac 1 {4^p} + \dfrac 1 {4^p} } + \dotsb + \paren {\dfrac 1 {2^{\paren {N - 1} p} } + \dotsb + \dfrac 1 {2^{\paren {N - 1} p} } }
      | c = 
}}
{{eqn | r = 1 + \dfrac 2 {2^p} + \dfrac 4 {4^p} + \dotsb + \dfrac {2^{N - 1} } {2^{\paren {N - 1} p} }
      | c = 
}}
{{eqn | r = 1 + \dfrac 1 {2^{p - 1} } + \paren {\dfrac 1 {2^{p - 1} } }^2 + \dotsb + \paren {\dfrac 1 {2^{p - 1} } }^{n - 1}
      | c = 
}}
{{eqn | r = \dfrac {1 - \paren {1 / 2^{p - 1} }^N} {1 - \paren {1 / 2^{p - 1} } }
      | c = 
}}
{{eqn | o = \le
      | r = \dfrac 1 {1 - \paren {1 / 2^{p - 1} } }
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Transitive Law

Let $a, b, c \in \R$ such that $a > b$ and $b > c$.
Then:
:$a > c$

Proof:
From Real Numbers form Ordered Integral Domain, $\struct {\R, +, \times, \le}$ forms an ordered integral domain.
From Ordered Integral Domain is Totally Ordered Ring, the usual ordering $\le$ is a total ordering.
From Relation Induced by Strict Positivity Property is Transitive it follows that $<$ is transitive.
{{qed}}
Theorem: Real Number Ordering is Compatible with Multiplication/Positive Factor

:$\forall a, b, c \in \R: a < b \land c > 0 \implies a c < b c$

Proof:
From Real Numbers form Ordered Integral Domain, $\struct {\R, +, \times, \le}$ forms an ordered integral domain.
Thus:
{{begin-eqn}}
{{eqn | l = a
      | o = <
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b - a
      | o = >
      | r = 0
      | c = {{Defof|Positivity Property}}
}}
{{eqn | ll= \leadsto
      | l = c \times \paren {b - a}
      | o = >
      | r = 0
      | c = {{Defof|Ordered Integral Domain|index = 1}}: $(\text P 1)$: Closure under Ring Addition
}}
{{eqn | ll= \leadsto
      | l = b \times c - a \times c
      | o = >
      | r = 0
      | c = {{Ring-axiom|D}}
}}
{{eqn | ll= \leadsto
      | l = a \times c
      | o = <
      | r = b \times c
      | c = {{Defof|Positivity Property}}
}}
{{end-eqn}}
{{Qed}}
Theorem: Real Number Ordering is Compatible with Multiplication/Negative Factor

:$\forall a, b, c \in \R: a < b \land c < 0 \implies a c > b c$

Proof:
From Real Numbers form Ordered Integral Domain, $\struct {\R, +, \times, \le}$ forms an ordered integral domain.
Thus:
{{begin-eqn}}
{{eqn | l = a
      | o = <
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b - a
      | o = >
      | r = 0
      | c = {{Defof|Positivity Property}}
}}
{{eqn | ll= \leadsto
      | l = c \times \paren {b - a}
      | o = <
      | r = 0
      | c = Product of Strictly Negative Element with Strictly Positive Element is Strictly Negative
}}
{{eqn | ll= \leadsto
      | l = b \times c - a \times c
      | o = <
      | r = 0
      | c = {{Ring-axiom|D}}
}}
{{eqn | ll= \leadsto
      | l = a \times c
      | o = >
      | r = b \times c
      | c = {{Defof|Positivity Property}}
}}
{{end-eqn}}
{{Qed}}