nyuuzyou/wikis huggingface dataset:
Integral of Reciprocal is Divergent/Unbounded Above/Proof 1

== Theorem ==
== 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.
== Sources ==
* : $\S 13.33$
Category:Integral of Reciprocal is Divergent
Integral of Series of Positive Measurable Functions

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$, $f_n: X \to \overline \R$ be a sequence of positive measurable functions.
Let $\ds \sum_{n \mathop \in \N} f_n: X \to \overline \R$ be the pointwise series of the $f_n$.
Then:
:$\ds \int \sum_{n \mathop \in \N} f_n \rd \mu = \sum_{n \mathop \in \N} \int f_n \rd \mu$
where the integral sign denotes $\mu$-integration.
== Proof ==
Define the sequence $\sequence {g_N}_{n \mathop \in \N}$ of functions $g_N : X \to \overline \R$ by: 
:$\ds \map {g_N} x = \sum_{n \mathop = 1}^N \map {f_n} x$ 
Since $f_n \ge 0$ for each $n$, we have: 
So $\sequence {g_N}_{N \mathop \in \N}$ is increasing. 
From Pointwise Sum of Measurable Functions is Measurable: General Result, we have that: 
:$g_N$ is $\Sigma$-measurable for each $N$. 
So, from the monotone convergence theorem, we have: 
:$\ds \lim_{N \mathop \to \infty} \int g_N \rd \mu = \int \paren {\lim_{N \mathop \to \infty} g_N} \rd \mu$
From the definition of infinite series, we have: 
:$\ds \lim_{N \mathop \to \infty} g_N = \sum_{n \mathop = 1}^\infty f_n$ 
So, we have: 
:$\ds \lim_{N \mathop \to \infty} \int g_N \rd \mu = \int \paren {\sum_{n \mathop = 1}^\infty f_n} \rd \mu$ 
Manipulating the , we have: 
So:
:$\ds \sum_{n \mathop = 1}^\infty \paren {\int f_n \rd \mu} = \int \paren {\sum_{n \mathop = 1}^\infty f_n} \rd \mu$
== Sources ==
* : $9.9$, $\S 9$: Problem $4$
Category:Integral of Positive Measurable Function
Integral of Shifted Dirac Delta Function by Continuous Function over Reals

== Theorem ==
Let $\map \delta x$ denote the Dirac delta function.
Let $g$ be a continuous real function.
Then:
:$\ds \int_{-\infty}^{+ \infty} \map {\delta} {x - s} \map g x \rd x = \map g s$
== Proof ==
We have that:
:$\map \delta x = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
where:
:$\map {F_\epsilon} x = \begin{cases} 0 & : x < -\epsilon + s \\ \dfrac 1 {2 \epsilon} & : -\epsilon + s \le x \le \epsilon + s \\ 0 & : x > \epsilon + s \end{cases}$
We have that:
From Darboux's Theorem:
:$\ds m \paren {\epsilon + s - \paren {-\epsilon + s} } \le \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x \le M \paren {\epsilon + s - \paren {-\epsilon + s} }$
where:
:$M$ is the maximum of $\map g x$
:$m$ is the minimum of $\map g x$
on $\closedint {-\epsilon + s} {\epsilon + s}$.
Hence:
:$\ds 2 m \epsilon \le \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x \le 2 M \epsilon$
and so dividing by $2 \epsilon$:
:$\ds m \le \dfrac 1 {2 \epsilon} \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x \le M$
Then:
:$\ds \lim_{\epsilon \mathop \to 0} M = m = \map g s$
and so by the Squeeze Theorem:
:$\ds \lim_{\epsilon \mathop \to 0} \int_{-\infty}^{+ \infty} \map {F_\epsilon} x \map g x \rd x = \map g s$
But by definition of the Dirac delta function:
:$\ds \lim_{\epsilon \mathop \to 0} \int_{-\infty}^{+ \infty} \map {F_\epsilon} x \map g x \rd x = \int_{- \infty}^{+ \infty} \map \delta x \map g x \rd x$
Hence the result.
== Sources ==
* 
Category:Dirac Delta Function
Integral of Survival Function

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a $\sigma$-finite measure space.
Let $f: X \to \R_{\ge 0}$ be a positive $\Sigma$-measurable function.
Let $F_f : \R \to \R$ be the survival function of $f$.
Then:
:$\ds \int f \rd \mu = \int_{\openint 0 \to} F_f \rd \lambda$
where $\lambda$ is Lebesgue measure.
== Proof ==
== Sources ==
* : $13.11$
Category:Survival Functions
Category:Integral of Positive Measurable Function
Integral on L-1 Space is Well-Defined

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {L^1} {X, \Sigma, \mu}$ be the $L^1$ space of $\struct {X, \Sigma, \mu}$.
Let $E \in \map {L^1} {X, \Sigma, \mu}$. 
Then the $\mu$-integral of $E$ is well-defined. 
== Proof ==
Let $E = \eqclass f \sim$.
We first establish that:
:$\ds \int f \rd \mu$
is a well-understood real number.
We then will show that the $\mu$-integral of $E$ is independent of the choice of representative for $E$.
Since $E \in \map {L^1} {X, \Sigma, \mu}$, we have that $f \in \map {\LL^1} {X, \Sigma, \mu}$. 
Then: 
:$\ds \int \size f \rd \mu < \infty$
So, from Characterization of Integrable Functions, we have: 
:$\ds \int f \rd \mu$ is well-defined and exists as a finite number. 
Suppose now that $\eqclass f \sim = \eqclass g \sim$. 
From Equivalence Class Equivalent Statements:
:$f = g$ $\mu$-almost everywhere
from the definition of the $\mu$-almost everywhere relation.
From A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 1, we have: 
:$\ds \int f \rd \mu = \int g \rd \mu$
So:
:$\ds \int E \rd \mu$
is independent of the choice of representative for $E$.
Category:Lp Spaces
Integral over 2 pi of Cosine of m x by Cosine of n x

== Theorem ==
Let $m, n \in \Z$ be integers.
Let $\alpha \in \R$ be a real number.
Then:
:$\ds \int_\alpha^{\alpha + 2 \pi} \cos m x \cos n x \rd x = \begin {cases} 0 & : m \ne n \\ \pi & : m = n \end {cases}$
That is:
:$\ds \int_\alpha^{\alpha + 2 \pi} \cos m x \cos n x \rd x = \pi \delta_{m n}$
where $\delta_{m n}$ is the Kronecker delta.
== Proof ==
Let $m \ne n$.
When $m = n$ we have:
== Sources ==
* : Chapter One: $\S 2$. Fourier Series: $(3)$
Category:Definite Integrals involving Cosine Function
Integral over 2 pi of Cosine of n x

== Theorem ==
Let $m \in \Z$ be an integer.
Then:
:$\ds \int_\alpha^{\alpha + 2 \pi} \cos m x \rd x = \begin {cases} 0 & : m \ne 0 \\ 2 \pi & : m = 0 \end {cases}$
== Proof ==
Let $m \ne 0$.
Let $m = 0$.
== Sources ==
* : Chapter One: $\S 2$. Fourier Series: $(4)$
Category:Definite Integrals involving Cosine Function
Integral over 2 pi of Cosine of n x/Mistake

== Source Work ==
* : Chapter One: $\S 2$. Fourier Series: $(4)$
== Mistake ==
:$\ds \int_0^{2 \pi} \cos n x \rd x = 0$
It is implicit that $n$ ranges over all integers.
The author has neglected to specify the case where $n = 0$:
:$\ds \int_0^{2 \pi} \cos n x \rd x = \begin {cases} 0 & : n \ne 0 \\ 2 \pi & : n = 0 \end {cases}$
As it stands, the Fourier coefficient $a_0$ is glossed over.
== Sources ==
* : Chapter One: $\S 2$. Fourier Series: $(4)$
Category:Mistakes/Fourier Series/Sneddon
Integral over 2 pi of Sine of m x by Cosine of n x

== Theorem ==
Let $m, n \in \Z$ be integers.
Let $\alpha \in \R$ be a real number.
Then:
:$\ds \int_\alpha^{\alpha + 2 \pi} \sin m x \cos n x \rd x = 0$
== Proof ==
Let $m \ne n$.
When $m = n$ we have:
== Sources ==
* : Chapter One: $\S 2$. Fourier Series: $(3)$
Category:Definite Integrals involving Sine Function
Category:Definite Integrals involving Cosine Function
Integral over 2 pi of Sine of m x by Sine of n x

== Theorem ==
Let $m, n \in \Z$ be integers.
Let $\alpha \in \R$ be a real number.
Then:
:$\ds \int_\alpha^{\alpha + 2 \pi} \sin m x \sin n x \rd x = \begin {cases} 0 & : m \ne n \\ \pi & : m = n \end {cases}$
That is:
:$\ds \int_\alpha^{\alpha + 2 \pi} \sin m x \sin n x \rd x = \pi \delta_{m n}$
where $\delta_{m n}$ is the Kronecker delta.
== Proof ==
Let $m \ne n$.
When $m = n$ we have:
== Sources ==
* : Chapter One: $\S 2$. Fourier Series: $(2)$
Category:Definite Integrals involving Sine Function
Integral over 2 pi of Sine of n x

== Theorem ==
Let $m \in \Z$ be an integer.
Then:
:$\ds \int_\alpha^{\alpha + 2 \pi} \sin m x \rd x = 0$
== Proof ==
Let $m \ne n$.
Let $m = 0$.
== Sources ==
* : Chapter One: $\S 2$. Fourier Series: $(4)$
Category:Definite Integrals involving Sine Function
Integral to Infinity of Bessel Function of First Kind order Zero

== Theorem ==
:$\ds \int_0^\infty \map {J_0} t \rd t = 1$
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:
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Evaluation of Integrals: $46 \ \text{(a)}$
Category:Bessel Functions
Integral to Infinity of Cosine p x minus Cosine q x over x

== Theorem ==
:$\ds \int_0^\infty \frac {\cos p x - \cos q x} x \rd x = \ln \frac q p$
where $p$ and $q$ are strictly positive real numbers.
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.38$
Category:Definite Integrals involving Cosine Function
Integral to Infinity of Dirac Delta Function

== Theorem ==
Let $\map \delta x$ denote the Dirac delta function.
Then:
:$\ds \int_0^{+ \infty} \map \delta x \rd x = 1$
== Proof ==
We have that:
:$\map \delta x = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
where:
:$\map {F_\epsilon} x = \begin {cases} 0 & : x < 0 \\ \dfrac 1 \epsilon & : 0 \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$
We have that:
Then:
Hence the result.
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {VIII}$. The Unit Impulse function or Dirac delta function: $1$.
Category:Dirac Delta Function
Integral to Infinity of Dirac Delta Function by Continuous Function

== Theorem ==
Let $\map \delta x$ denote the Dirac delta function.
Let $g$ be a continuous real function.
Then:
:$\ds \int_0^{+ \infty} \map \delta x \map g x \rd x = \map g 0$
== Proof ==
We have that:
:$\map \delta x = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
where:
:$\map {F_\epsilon} x = \begin{cases} 0 & : x < 0 \\ \dfrac 1 \epsilon & : 0 \le x \le \epsilon \\ 0 & : x > \epsilon \end{cases}$
We have that:
From Darboux's Theorem:
:$\ds m \paren {\epsilon - 0} \le \int_0^\epsilon \map g x \rd x \le M \paren {\epsilon - 0}$
where:
:$M$ is the maximum of $\map g x$
:$m$ is the minimum of $\map g x$
on $\closedint 0 \epsilon$.
Hence:
:$\ds m \epsilon \le \int_0^\epsilon \map g x \rd x \le M \epsilon$
and so dividing by $\epsilon$:
:$\ds m \le \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x \le M$
Then:
:$\ds \lim_{\epsilon \mathop \to 0} M = m = \map g 0$
and so by the Squeeze Theorem:
:$\ds \lim_{\epsilon \mathop \to 0} \int_0^{+\infty} \map {F_\epsilon} x \map g x \rd x = \map g 0$
But by :
:$\ds \lim_{\epsilon \mathop \to 0} \int_0^{+\infty} \map {F_\epsilon} x \map g x \rd x = \int_0^{+\infty} \map \delta x \map g x \rd x$
Hence the result.
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {VIII}$. The Unit Impulse function or Dirac delta function: $2$.
Category:Dirac Delta Function
Integral to Infinity of Exponential of -t^2

== Theorem ==
:$\ds \int_0^\infty \map \exp {-t^2} \rd t = \dfrac {\sqrt \pi} 2$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $29$
Category:Error Function
Category:Definite Integrals involving Exponential Function
Category:Integral to Infinity of Exponential of -t^2
Integral to Infinity of Exponential of -t^2/Proof 1

== Theorem ==
== 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:
where $\mathscr R_P$ is the square $\Box OACE$ in the figure below:
:
Because the integrand is positive:
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:
Category:Integral to Infinity of Exponential of -t^2
Integral to Infinity of Exponential of -t^2/Proof 2

== Theorem ==
== 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 $\lambda > 0$:
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 $\lambda > 0$:
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: 
On the other hand: 
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: 
:$\ds \int_0^\infty e^{-t^2} \rd t = \frac {\sqrt \pi} 2$
Category:Integral to Infinity of Exponential of -t^2
Integral to Infinity of Function over Argument

== Theorem ==
Let $f: \R \to \R$ or $\R \to \C$ be a continuous function on any interval of the form $0 \le t \le A$.
Let $\laptrans f = F$ denote the Laplace transform of $f$.
Then:
:$\ds \int_0^\infty {\dfrac {\map f t} t} = \int_0^{\to \infty} \map F u \rd u$
provided the integrals converge.
== Proof ==
== Also see ==
* Laplace Transform of Integral
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Division by $t$: $22 \ \text{(a)}$
Category:Laplace Transforms
Integral to Infinity of One minus Cosine p x over x Squared

== Theorem ==
:$\ds \int_0^\infty \frac {1 - \cos p x} {x^2} \rd x = \frac {\pi \size p} 2$
where $p$ is a real number.
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.37$
Category:Definite Integrals involving Cosine Function
Integral to Infinity of Reciprocal of Power of x

== Theorem ==
The improper integral
:$\ds \int_1^\infty \dfrac {\d t} {t^x}$
exists $x > 1$.
== Proof ==
First let $x \ne 1$.
Then:
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.
== Sources ==
* : $\S 13.34 \ (3)$
* : $\S 1.2.2$: Summary of convergence tests
Category:Reciprocals
Category:Examples of Definite Integrals
Integral to Infinity of Shifted Dirac Delta Function by Continuous Function

== Theorem ==
Let $\map \delta x$ denote the Dirac delta function.
Let $g$ be a continuous real function.
Let $a \in \R_{\ge 0}$ be a positive real number.
Then:
:$\ds \int_0^{+ \infty} \map \delta {x - a} \, \map g x \rd x = \map g a$
== Proof ==
We have that:
:$\map \delta {x - a} = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
where:
:$\map {F_\epsilon} x = \begin {cases} 0 & : x < a \\ \dfrac 1 \epsilon & : a \le x \le a + \epsilon \\ 0 & : x > a + \epsilon \end {cases}$
We have that:
From Darboux's Theorem:
:$\ds m \paren {\paren {a + \epsilon} - a} \le \int_a^{a + \epsilon} \map g x \rd x \le M \paren {\paren {a + \epsilon} - 0}$
where:
:$M$ is the maximum of $\map g x$
:$m$ is the minimum of $\map g x$
on $\closedint a {a + \epsilon}$.
Hence:
:$\ds m \epsilon \le \int_a^{a + \epsilon} \map g x \rd x \le M \epsilon$
and so dividing by $\epsilon$:
:$\ds m \le \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x \le M$
Then:
:$\ds \lim_{\epsilon \mathop \to 0} M = m = \map g a$
and so by the Squeeze Theorem:
:$\ds \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \map g x \rd x = \map g a$
But by definition of the Dirac delta function:
:$\ds \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \map g x \rd x = \int_0^{+ \infty} \map \delta {x - a} \map g x \rd x$
Hence the result.
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {VIII}$. The Unit Impulse function or Dirac delta function: $3$.
Category:Dirac Delta Function
Integral to Infinity of Sine p x Cosine q x over x

== Theorem ==
:$\ds \int_0^\infty \frac {\sin p x \cos q x} x \rd x = \begin {cases} \dfrac \pi 2 & : p > q > 0 \\ \\ 0 & : 0 < p < q \\ \\ \dfrac \pi 4 & : p = q > 0 \end {cases}$
== Proof ==
=== Case $p > q > 0$: ===
=== Case $0 < p < q$: ===
=== Case $p = q > 0$: ===
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.34$
* : $\S 18$: Definite Integrals: Definite Integrals involving Trigonometric Functions: $18.34$
Category:Definite Integrals involving Sine Function
Category:Definite Integrals involving Cosine Function
Integral to Infinity of Sine p x Cosine q x over x/Mistake

== Source Work ==
:Chapter $15$: Definite Integrals
::Definite Integrals involving Trigonometric Functions: $15.34$
:Chapter $18$: Definite Integrals
::Definite Integrals involving Trigonometric Functions: $18.34$
== Mistake ==
:$\ds \int_0^\infty \frac {\sin p x \cos q x} x \rd x = \begin {cases} 0 & : p > q > 0 \\ \\
\dfrac \pi 2 & : 0 < p < q \\ \\
\dfrac \pi 4 & : p = q > 0 \end {cases}$
== Correction ==
As demonstrated in Integral to Infinity of $\dfrac {\sin p x \cos q x} x$, this is incorrect.
It should be:
:$\ds \int_0^\infty \frac {\sin p x \cos q x} x \rd x = \begin {cases} \dfrac \pi 2 & : p > q > 0 \\ \\
0 & : 0 < p < q \\ \\
\dfrac \pi 4 & : p = q > 0 \\
\end {cases}$
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.34$
* : $\S 18$: Definite Integrals: Definite Integrals involving Trigonometric Functions: $18.34$
Category:Mistakes/Mathematical Handbook of Formulas and Tables
Integral to Infinity of Sine p x Sine q x over x Squared

== Theorem ==
:$\ds \int_0^\infty \frac {\sin p x \sin q x} {x^2} \rd x = \begin {cases} \dfrac {\pi p} 2 & : 0 < p \le q \\ \\ \dfrac {\pi q} 2 & : p \ge q > 0 \end {cases}$
== 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:
So:
=== Case $0 < p = q$ ===
=== Case $0 < p \le q$: ===
Suppose $0 < p < q$.
Then:
Adjoin to the case where $p = q$.
=== Case $p \ge q > 0$: ===
Suppose $p > q > 0$.
Then:
Adjoin to the case where $p = q$.
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.35$
* : $\S 18$: Definite Integrals: Definite Integrals involving Trigonometric Functions: $18.35$
Category:Definite Integrals involving Sine Function
Integral to Infinity of Sine p x over x

== Theorem ==
:$\ds \int_0^\infty \frac {\sin p x} x \rd x = \begin {cases} \dfrac \pi 2 & : p > 0 \\ \\ 0 & : p = 0 \\ \\ -\dfrac \pi 2 & : p < 0 \end {cases}$
== Proof 1 ==
 
== Proof 2 ==
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.33$
Category:Definite Integrals involving Sine Function
Category:Integral to Infinity of Sine p x over x
Integral to Infinity of Square of Sine p x over x Squared

== Theorem ==
:$\ds \int_0^\infty \paren {\frac {\sin p x} x}^2 \rd x = \frac {\pi \size p} 2$
where $p$ is a real number.
== Proof ==
We have: 
We also have, by Primitive of Power: 
:$\ds \int \frac {\d x} {x^2} = -\frac 1 x + C$
So: 
We have, from Real Sine Function is Bounded, that: 
:$0 \le \sin^2 p x \le 1$
for all real $x$. 
Therefore: 
:$0 \le \dfrac {\sin^2 p x} x \le \dfrac 1 x$
for all strictly positive real $x$.
We have: 
:$\ds \lim_{x \mathop \to \infty} \frac 1 x = 0$
Therefore, by the Squeeze Theorem, we have: 
:$\ds \lim_{x \mathop \to \infty} \paren {\frac {\sin^2 p x} x} = 0$
As to the other limit, note that: 
So:
So: 
:$\ds \int_0^\infty \paren {\frac {\sin p x} x}^2 \rd x = p \int_0^\infty \frac {\sin 2 p x} x \rd x$
By Integral to Infinity of Sine p x over x, we have: 
:$\ds \int_0^\infty \frac {\sin 2 p x} x \rd x = \begin {cases} \dfrac \pi 2 & : p > 0 \\ \\ 0 & : p = 0 \\ \\ -\dfrac \pi 2 & : p < 0 \end{cases}$
So:
:$\ds p \int_0^\infty \frac {\sin 2 p x} x \rd x = \begin{cases} \dfrac \pi 2 p & : p > 0 \\ \\ 0 & : p = 0 \\ \\ -\dfrac \pi 2 p & : p < 0 \end{cases}$
Hence:
:$\ds p \int_0^\infty \frac {\sin 2 p x} x \rd x = \frac {\pi \size p} 2$
Hence the result.
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.36$
Category:Definite Integrals involving Sine Function
Integral to Infinity of e^-t by Error Function of Root t

== Theorem ==
:$\ds \int_0^\infty e^{-t} \erf \sqrt t \rd t = \dfrac {\sqrt 2} 2$
where $\erf$ denotes the error function.
== Proof ==
Using the technique of Evaluation of Integral using Laplace Transform:
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Evaluation of Integrals: $46 \ \text{(b)}$
Category:Error Function
Category:Definite Integrals involving Exponential Function
Integral with respect to Dirac Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $x \in X$, and let $\delta_x$ be the Dirac measure at $x$.
Let $f \in \MM _{\overline \R}, f: X \to \overline \R$ be a measurable function.
Then:
:$\ds \int f \rd \delta_x = \map f x$
where the integral sign denotes the $\delta_x$-integral.
== Proof 1 ==
 
== Proof 2 ==
== Sources ==
* : $9.10 \ \text{(i)}$
Category:Dirac Measure
Category:Integral of Integrable Function
Category:Integral with respect to Dirac Measure
Integral with respect to Dirac Measure/Proof 1

== Theorem ==
== Proof ==
Define the constant function $g : X \to \overline \R$ by: 
:$\map g {x'} = \map f x$
for each $x' \in X$. 
From Constant Function is Measurable, we have: 
:$g$ is $\Sigma$-measurable. 
From Measurable Functions Determine Measurable Sets:
:$\set {x' \in X : \map g {x'} \ne \map f {x'} } \in \Sigma$
Further: 
:$x \not \in \set {x' \in X : \map g {x'} \ne \map f {x'} }$
So from the definition of the Dirac measure, we have:
:$\map {\delta_x} {\set {x' \in X : \map g {x'} \ne \map f {x'} } } = 0$
So:
:$g = f$ $\delta_x$-almost everywhere.
From A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 1, we have: 
:$\ds \int g \rd \delta_x = \int f \rd \delta_x$ 
We finally have:
Category:Integral with respect to Dirac Measure