hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \int e^{a x} \sinh b x \rd x       | r = \int e^{a x} \paren {\frac {e^{b x} - e^{-b x} } 2} \rd x       | c = {{Defof|Hyperbolic Sine}} }} {{eqn | r = \frac 1 2 \int e^{a x} \paren {e^{b x} - e^{-b x} } \rd x       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \int \paren {e^{a x} e^{b x} - e^{a x} e^{-b x} } \rd x       | c =  }} {{eqn | r = \frac 1 2 \int \paren {e^{a x + b x} - e^{a x - b x} } \rd x       | c = Exponent Combination Laws: Product of Powers }} {{eqn | r = \frac 1 2 \int e^{a x + b x} \rd x - \frac 1 2 \int e^{a x - b x} \rd x       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 2 \int e^{\paren {a + b} x} \rd x - \frac 1 2 \int e^{\paren {a - b} x} \rd x       | c =  }} {{eqn | r = \frac 1 2 \frac {e^{\paren {a + b} x} } {a + b} - \frac 1 2 \frac {e^{\paren {a - b} x} } {a - b} + C       | c = Primitive of $e^{a x}$ }} {{eqn | r = \frac 1 2 \frac {e^{a x + b x} } {a + b} - \frac 1 2 \frac {e^{a x - b x} } {a - b} + C       | c =  }} {{eqn | r = \frac 1 2 \frac {e^{a x} e^{b x} } {a + b} - \frac 1 2 \frac {e^{a x} e^{-b x} } {a - b} + C       | c = Exponent Combination Laws: Product of Powers }} {{eqn | r = \frac 1 2 \frac {e^{a x} e^{b x} \paren {a - b} } {\paren {a + b} \paren {a - b} } - \frac 1 2 \frac {e^{a x} e^{-b x} \paren {a + b} } {\paren {a - b} \paren {a + b} } + C       | c =  }} {{eqn | r = \frac {e^{a x} e^{b x} \paren {a - b} - e^{a x} e^{-b x} \paren {a + b} } {2 \paren {a + b} \paren {a - b} } + C       | c =  }} {{eqn | r = \frac {e^{a x} e^{b x} \paren {a - b} - e^{a x} e^{-b x} \paren {a + b} } {2 \paren {a^2 - b^2} } + C       | c = Difference of Two Squares }} {{eqn | r = \frac {a e^{a x} e^{b x} - b e^{a x} e^{b x} - a e^{a x} e^{-b x} - b e^{a x} e^{-b x} } {2 \paren {a^2 - b^2} } + C       | c =  }} {{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {\frac {a e^{b x} - b e^{b x} - a e^{-b x} - b e^{-b x} } 2} + C       | c =  }} {{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {a \frac {e^{b x} - e^{-b x} } 2 - b \frac {e^{b x} + e^{-b x} } 2} + C       | c =  }} {{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {a \frac {e^b x - e^{-b} x} 2 - b \cosh b x} + C       | c = {{Defof|Hyperbolic Cosine}} }} {{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {a \sinh b x - b \cosh b x } + C       | c = {{Defof|Hyperbolic Sine}} }} {{eqn | r = \frac {e^{a x} \paren {a \sinh b x - b \cosh b x} } {a^2 - b^2} + C       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int e^{a x} \cosh b x \rd x       | r = \int e^{a x} \paren {\frac {e^{b x} + e^{-b x} } 2} \rd x       | c = {{Defof|Hyperbolic Cosine}} }} {{eqn | r = \frac 1 2 \int e^{a x} \paren {e^{b x} + e^{-b x} } \rd x       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \int \paren {e^{a x} e^{b x} + e^{a x} e^{-b x} } \rd x       | c =  }} {{eqn | r = \frac 1 2 \int \paren {e^{a x + b x} + e^{a x - b x} } \rd x       | c = Exponent Combination Laws: Product of Powers }} {{eqn | r = \frac 1 2 \int e^{a x + b x} \rd x + \frac 1 2 \int e^{a x - b x} \rd x       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 2 \int e^{\paren {a + b} x} \rd x + \frac 1 2 \int e^{\paren {a - b} x} \rd x       | c =  }} {{eqn | r = \frac 1 2 \frac {e^{\paren {a + b} x} } {a + b} + \frac 1 2 \frac {e^{\paren {a - b} x} } {a - b} + C       | c = Primitive of $e^{a x}$ }} {{eqn | r = \frac 1 2 \frac {e^{a x + b x} } {a + b} + \frac 1 2 \frac {e^{a x - b x} } {a - b} + C       | c =  }} {{eqn | r = \frac 1 2 \frac {e^{a x} e^{b x} } {a + b} + \frac 1 2 \frac {e^{a x} e^{-b x} } {a - b} + C       | c = Exponent Combination Laws: Product of Powers }} {{eqn | r = \frac 1 2 \frac {e^{a x} e^{b x} \paren {a - b} } {\paren {a + b} \paren {a - b} } + \frac 1 2 \frac {e^{a x} e^{-b x} \paren {a + b} } {\paren {a - b} \paren {a + b} } + C       | c =  }} {{eqn | r = \frac {e^{a x} e^{b x} \paren {a - b} + e^{a x} e^{-b x} \paren {a + b} } {2 \paren {a + b} \paren {a - b} } + C       | c =  }} {{eqn | r = \frac {e^{a x} e^{b x} \paren {a - b} + e^{a x} e^{-b x} \paren {a + b} } {2 \paren {a^2 - b^2} } + C       | c = Difference of Two Squares }} {{eqn | r = \frac {a e^{a x} e^{b x} - b e^{a x} e^{b x} + a e^{a x} e^{-b x} + b e^{a x} e^{-b x} } {2 \paren {a^2 - b^2} } + C       | c =  }} {{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {\frac {a e^{b x} - b e^{b x} + a e^{-b x} + b e^{-b x} } 2} + C       | c =  }} {{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {a \frac {e^{b x} + e^{-b x} } 2 + b \frac {e^{b x} - e^{-b x} } 2} + C       | c =  }} {{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {a \frac {e^b x + e^{-b} x} 2 + b \sinh b x} + C       | c = {{Defof|Hyperbolic Sine}} }} {{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {a \cosh b x + b \sinh b x } + C       | c = {{Defof|Hyperbolic Cosine}} }} {{eqn | r = \frac {e^{a x} \paren {a \cosh b x + b \sinh b x} } {a^2 - b^2} + C       | c =  }} {{end-eqn}} {{qed}}
For all $N \in \Z_{>0}$: :$\displaystyle \int_0^\infty e^{-s t} \map f t \rd t = \int_0^N e^{-s t} \map f t \rd t + \int_N^\infty e^{-s t} \map f t \rd t$ We have that $f$ is piecewise continuous in every closed interval $\closedint 0 N$. Hence the first of the integrals on the {{RHS}} exists. Also, as $\map f t$ is of exponential order $\gamma$ for $t > N$, so does the second integral on the {{RHS}}. Indeed: {{begin-eqn}} {{eqn | l = \size {\int_N^\infty e^{-s t} \map f t \rd t}       | o = \le       | r = \int_N^\infty \size {e^{-s t} \map f t} \rd t       | c =  }} {{eqn | o = \le       | r = \int_0^\infty e^{-s t} \size {\map f t} \rd t       | c =  }} {{eqn | o = \le       | r = \int_0^\infty e^{-s t} M e^{\gamma t} \rd t       | c =  }} {{eqn | r = \dfrac M {s - \gamma}       | c = Laplace Transform of Exponential }} {{end-eqn}} Thus the Laplace transform exists for $s > \gamma$. {{qed}}
Let $\map g t = \displaystyle \int_0^t \map f u \rd u$. Then: :$\map {g'} t = \map f t$ and: :$\map g 0 = 0$ Thus: {{begin-eqn}} {{eqn | l = \laptrans {\map {g'} t}       | r = s \laptrans {\map g t} - \map g 0       | c = Laplace Transform of Derivative }} {{eqn | r = s \laptrans {\map g t}       | c =  }} {{eqn | r = \map F s       | c = as $\map F s = \laptrans {\map f t}$ }} {{eqn | ll= \leadsto       | l = \laptrans {\map g t}       | r = \dfrac {\map F s} s       | c =  }} {{eqn | ll= \leadsto       | l = \laptrans {\int_0^t \map f u \rd u}       | r = \dfrac {\map F s} s       | c =  }} {{end-eqn}} {{qed}}
Let $\map g t := \dfrac {\map f t} t$. Then: {{begin-eqn}} {{eqn | l = \map f t       | r = t \, \map g t       | c =  }} {{eqn | ll= \leadsto       | l = \laptrans {\map f t}       | r = \laptrans {t \, \map g t}       | c =  }} {{eqn | ll= \leadsto       | l = \laptrans {\map f t}       | r = -\dfrac \d {\d s} \laptrans {\map g t}       | c = Derivative of Laplace Transform }} {{eqn | ll= \leadsto       | l = \map F s       | r = -\dfrac {\d G} {\d s}       | c = $\map F s := \laptrans {\map f t}$, and so on }} {{eqn | ll= \leadsto       | l = \map G s       | r = -\int_{-\infty}^s \map f u \rd u       | c = }} {{eqn | r = \int_s^\infty \map f u \rd u       | c =  }} {{end-eqn}} The result follows. {{qed}}
Let $x, y \in S$. Then: {{begin-eqn}} {{eqn | l = \map {i_S} x + \map {i_S} y       | r = x + y       | c = {{Defof|Inclusion Mapping}} }} {{eqn | r = x \mathbin{ + {\restriction_S} } y       | c = as $x, y \in S$ }} {{eqn | r = \map {i_S} {x \mathbin{ + {\restriction_S} } y}       | c = as $x \mathbin{ + {\restriction_S} } y \in S$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \map {i_S} x \circ \map {i_S} y       | r = x \circ y       | c = {{Defof|Inclusion Mapping}} }} {{eqn | r = x \mathbin{\circ {\restriction_S} } y       | c = as $x, y \in S$ }} {{eqn | r = \map {i_S} {x \mathbin{ \circ {\restriction_S} } y}       | c = as $x \mathbin{ \circ {\restriction_S} } y \in S$ }} {{end-eqn}} Hence ${i_S}$ is a ring homomorphism by definition. {{qed}} Category:Ring Homomorphisms Category:Inclusion Mappings n1i0aoahsj9suxon5donbspmxfj84sr
By Inclusion Mapping on Subring is Homomorphism, $i_S$ is a ring homomorphism. By Inclusion Mapping is Injection, $i_S$ is an injection. The result follows by definition of (ring) monomorphism. {{qed}} Category:Group Monomorphisms Category:Inclusion Mappings h5ohqc6l5mbb3zoxs5vxcw9butsp61s
Let $i_S: S \to R$ be the inclusion mapping from $S$ to $R$. By Inclusion Mapping on Subring is Homomorphism, $i_S$ is a ring homomorphism. By Ring Homomorphism of Addition is Group Homomorphism, $i_S$ is a group homomorphism on ring addition $+$. Let $x \in S$. Then: {{begin-eqn}} {{eqn | l = \mathbin \sim x       | r = \map {i_S} {\mathbin \sim x}       | c = as $\mathbin \sim x \in S$ }} {{eqn | r = -x       | c = Group Homomorphism Preserves Inverses }} {{end-eqn}} {{qed}} Category:Ring Theory m4kp7cy9zr5f3hq4oksoxoc5bccqsta
Let $x, y \in S$. Let $-x$ denote the ring negative of $x$ in $R$. Let $\mathbin \sim x$ denote the ring negative of $x$ in $S$. Then: {{begin-eqn}} {{eqn | l = x \sim y       | r = x \mathbin {+ {\restriction_S} } \paren {\mathbin \sim y}       | c = {{Defof|Ring Subtraction}} }} {{eqn | r = x + \paren {\mathbin \sim y}       | c = {{Defof|Subring|Addition on Subring}} }} {{eqn | r = x + \paren {-y}       | c = Negative of Subring is Negative of Ring }} {{eqn | r = x - y       | c = {{Defof|Ring Subtraction}} }} {{end-eqn}} {{qed}} Category:Ring Theory qt1dwlw2zy1nwl6tzgkgzzos92fs2na
We have that $\map {f'} t$ is piecewise continuous with one-sided limits and of exponential order. Hence: :$\displaystyle \lim_{s \mathop \to \infty} \int_0^\infty e^{-s t} \map {f'} t \rd t = 0$ Suppose that $f$ is continuous at $t = 0$. From Laplace Transform of Derivative: :$(1): \quad \laptrans {\map {f'} t} = s \, \map F s - \map f 0$ Taking the limit as $s \to \infty$ in $(1)$, where it is assumed that $\map f t$ is continuous at $t = 0$: :$0 = \displaystyle \lim_{s \mathop \to \infty} s \, \map F s - \map f 0$ or: :$\displaystyle \lim_{s \mathop \to \infty} s \, \map F s = \map f 0 = \lim_{t \mathop \to 0} \map f t$ {{qed|lemma}} Suppose that $f$ is not continuous at $t = 0$. From Laplace Transform of Derivative with Discontinuity at Zero: :$\laptrans {\map {f'} t} = s \, \map F s - \map f {0^+}$ which means: :$(2): \quad \laptrans {\map {f'} t} = s \, \map F s - \displaystyle \lim_{u \mathop \to 0} \map f u$ Similarly taking the limit as $s \to \infty$ in $(2)$, where it is assumed that $\map f t$ is continuous at $t = 0$: :$0 = \displaystyle \lim_{s \mathop \to \infty} s \, \map F s - \lim_{u \mathop \to 0} \map f u$ and so: :$\displaystyle \lim_{s \mathop \to \infty} s \, \map F s = \lim_{u \mathop \to 0} \map f u = \lim_{t \mathop \to 0} \map f t$ {{qed}}
From Laplace Transform of Derivative: :$(1): \quad \laptrans {\map {f'} t} = s \, \map F s - \map f 0$ We have that: {{begin-eqn}} {{eqn | l = \lim_{s \mathop \to 0} \laptrans {\map {f'} t}       | r = \lim_{s \mathop \to 0} \int_0^\infty e^{-s t} \map {f'} t \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \int_0^\infty \map {f'} t \rd t       | c =  }} {{eqn | r = \lim_{L \mathop \to \infty} \int_0^L \map {f'} t \rd t       | c =  }} {{eqn | r = \lim_{L \mathop \to \infty} \paren {\map f L - \map f 0}       | c = Fundamental Theorem of Calculus }} {{eqn | n = 2       | r = \lim_{t \mathop \to \infty} \map f t - \map f 0       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \lim_{s \mathop \to 0} \laptrans {\map {f'} t}       | r = \lim_{s \mathop \to 0} s \, \map F s - \map f 0       | c = from $(1)$ }} {{eqn | ll= \leadsto       | l = \lim_{t \mathop \to \infty} \map f t - \map f 0       | r = \lim_{s \mathop \to 0} s \, \map F s - \map f 0       | c = from $(2)$ }} {{eqn | ll= \leadsto       | l = \lim_{t \mathop \to \infty} \map f t       | r = \lim_{s \mathop \to 0} s \, \map F s       | c =  }} {{end-eqn}} {{qed|lemma}} Suppose that $f$ is not continuous at $t = 0$. From Laplace Transform of Derivative with Discontinuity at Zero: :$\laptrans {\map {f'} t} = s \, \map F s - \map f {0^+}$ which means: :$(3): \quad \laptrans {\map {f'} t} = s \, \map F s - \displaystyle \lim_{u \mathop \to 0} \map f u$ We have that: {{begin-eqn}} {{eqn | l = \lim_{s \mathop \to 0} \laptrans {\map {f'} t}       | r = \lim_{u \mathop \to 0} \paren {\lim_{s \mathop \to 0} \int_u^\infty e^{-s t} \map {f'} t \rd t}       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \lim_{u \mathop \to 0} \paren {\int_u^\infty \map {f'} t \rd t}       | c =  }} {{eqn | r = \lim_{u \mathop \to 0} \paren {\lim_{L \mathop \to \infty} \int_u^L \map {f'} t \rd t}       | c =  }} {{eqn | r = \lim_{u \mathop \to 0} \paren {\lim_{L \mathop \to \infty} \paren {\map f L - \map f u} }       | c = Fundamental Theorem of Calculus }} {{eqn | r = \lim_{u \mathop \to 0} \paren {\lim_{t \mathop \to \infty} \map f t - \map f u}       | c =  }} {{eqn | n = 4       | r = \lim_{t \mathop \to \infty} \map f t - \lim_{u \mathop \to 0} \map f u       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \lim_{s \mathop \to 0} \laptrans {\map {f'} t}       | r = \lim_{s \mathop \to 0} s \, \map F s - \lim_{u \mathop \to 0} \map f u       | c = from $(3)$ }} {{eqn | ll= \leadsto       | l = \lim_{t \mathop \to \infty} \map f t - \lim_{u \mathop \to 0} \map f u       | r = \lim_{s \mathop \to 0} s \, \map F s - \lim_{u \mathop \to 0} \map f u       | c = from $(4)$ }} {{eqn | ll= \leadsto       | l = \lim_{t \mathop \to \infty} \map f t       | r = \lim_{s \mathop \to 0} s \, \map F s       | c =  }} {{end-eqn}} {{qed}}
By definition of Laplace transform: :$\displaystyle \int_0^{\to \infty} e^{-s t} \map f t \rd t = \map F s$ The result follows by taking the limit as $s \to 0$. {{qed}}
The valuation ring $\mathcal O$ Is the open ball ${B_1}^- \paren {0_R}$ by definition. By Open Balls of Non-Archimedean Division Rings are Clopen then $\mathcal O$ is both open and closed in the metric induced by $\norm {\,\cdot\,}$. {{qed}} Category:Normed Division Rings Category:Valuation Ring of Non-Archimedean Division Ring is Clopen 3mq7moxsruz5gxr2j57b2c8kntv45zv
The $p$-adic integers $\Z_p$ is the valuation ring induced by $\norm {\,\cdot\,}_p$ by definition. By Valuation Ring of Non-Archimedean Division Ring is Clopen then the $p$-adic integers $\Z_p$ is both open and closed in the $p$-adic metric. {{qed}} Category:Valuation Ring of Non-Archimedean Division Ring is Clopen qhjehmabhkaqu8whhsdtkl33zp63q8w
{{begin-eqn}} {{eqn | l = \map {J_{-n} } x        | r = \dfrac 1 \pi \int_0^\pi \map \cos {-n \theta - x \sin \theta} \rd \theta       | c = Integral Representation of Bessel Function of the First Kind/Integer Order }} {{eqn | r = \dfrac 1 \pi \int_0^\pi \map \cos {-n \paren {\pi - \theta} - x \sin \paren {\pi - \theta} } \rd \paren {\pi - \theta}       | c = substitution of $\pi - \theta$ }} {{eqn | r = -\dfrac 1 \pi \int_\pi^0 \map \cos {n \theta - x \sin \theta - n \pi} \rd \theta       | c = Sine of Supplementary Angle }} {{eqn | r = \dfrac 1 \pi \int_0^\pi \map \cos {n \theta - x \sin \theta - n \pi} \rd \theta       | c = Reversal of Limits of Definite Integral }} {{eqn | r = \paren {-1}^{-n} \dfrac 1 \pi \int_0^\pi \map \cos {n \theta - x \sin \theta} \rd \theta       | c = Cosine of Angle plus Integer Multiple of Pi }} {{eqn | r = \paren {-1}^n \map {J_n} x       | c = Integral Representation of Bessel Function of the First Kind/Integer Order }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {x^n \map {J_n} x}       | r = n x^{n - 1} \map {J_n} x + x^n \map {J_n'} x       | c = Product Rule for Derivatives }} {{eqn | r = n x^{n - 1} \map {J_n} x - x^n \paren {\frac {\map {J_{n + 1} } x - \map {J_{n - 1} } x} 2}       | c = Recurrence Formula for Bessel Function of the First Kind }} {{eqn | r = n x^{n - 1} \map {J_n} x - x^n \paren {\frac {\frac {2 n} x \map {J_n} x - 2 \map {J_{n - 1} } x} 2}       | c = Recurrence Formula for Bessel Function of the First Kind }} {{eqn | r = n x^{n - 1} \map {J_n} x - x^n \paren {\frac n x \map {J_n} x - \map {J_{n - 1} } x} }} {{eqn | r = n x^{n - 1} \map {J_n} x - n x^{n - 1} \map {J_n} x + x^n \map {J_{n - 1} } x }} {{eqn | r = x^n \map {J_{n - 1} } x }} {{end-eqn}} {{qed}}
From Series Expansion of Bessel Function of the First Kind: {{begin-eqn}} {{eqn | n = 1       | l = \map {J_n} x       | r = \dfrac {x^n} {2^n \, \map \Gamma {n + 1} } \paren {1 - \dfrac {x^2} {2 \paren {2 n + 2} } + \dfrac {x^4} {2 \times 4 \paren {2 n + 2} \paren {2 n + 4} } - \cdots}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac x 2}^{n + 2 k}       | c =  }} {{end-eqn}} The result follows by substituting $-n$ for $n$ in $1$ and simplifying. {{qed}}
We have that: :$\map \delta x = \displaystyle \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$ 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: {{begin-eqn}} {{eqn | l = \int_0^{+ \infty} \map {F_\epsilon} x \rd x       | r = \int_0^\epsilon \dfrac 1 \epsilon \rd x + \int_\epsilon^\infty 0 \rd x       | c = Definition of $\map {F_\epsilon} x$ }} {{eqn | r = \int_0^\epsilon \dfrac 1 \epsilon \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \rd x       | c = {{Defof|Improper Integral}} }} {{eqn | r = \dfrac 1 \epsilon \paren {\epsilon - 0} + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \epsilon} }       | c = Definite Integral of Constant }} {{eqn | r = 1 + \lim_{L \mathop \to \infty} 0       | c = simplification }} {{eqn | r = 1       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int_0^{+ \infty} \map \delta x \rd x       | r = \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \rd x       | c = {{Defof|Dirac Delta Function}} }} {{eqn | r = \lim_{\epsilon \mathop \to 0} 1       | c = from above }} {{eqn | r = 1       | c =  }} {{end-eqn}} Hence the result. {{qed}}
We have that: :$\map \delta x = \displaystyle \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$ 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: {{begin-eqn}} {{eqn | l = \int_0^{+ \infty} \map {F_\epsilon} x \, \map g x \rd x       | r = \int_0^\epsilon \dfrac 1 \epsilon \, \map g x \rd x + \int_\epsilon^\infty 0 \times \map g x \rd x       | c = Definition of $\map {F_\epsilon} x$ }} {{eqn | r = \int_0^\epsilon \dfrac 1 \epsilon \, \map g x \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \times \map g x \rd x       | c = {{Defof|Improper Integral}} }} {{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \rd x       | c =  }} {{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \epsilon} }       | c = Definite Integral of Constant }} {{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} 0       | c = simplification }} {{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x       | c =  }} {{end-eqn}} From Upper and Lower Bounds of Integral: :$\displaystyle m \paren {\epsilon - 0} \le \int_0^\epsilon \map g x \rd x \le M \paren {\epsilon - 0}$ where: :$M$ is the maximum of $\map g x$ :$m$ is the minimum of $\map g x$ on $\closedint 0 \epsilon$. Hence: :$\displaystyle m \epsilon \le \int_0^\epsilon \map g x \rd x \le M \epsilon$ and so dividing by $\epsilon$: :$\displaystyle m \le \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x \le M$ Then: :$\displaystyle \lim_{\epsilon \mathop \to 0} M = m = \map g 0$ and so by the Squeeze Theorem: :$\displaystyle \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \, \map g x \rd x = \map g 0$ But by definition of the Dirac delta function: :$\displaystyle \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \, \map g x \rd x = \displaystyle \int_0^{+ \infty} \map \delta x \, \map g x \rd x$ Hence the result. {{qed}}
We have that: :$\map \delta {x - a} = \displaystyle \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$ 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: {{begin-eqn}} {{eqn | l = \int_0^{+ \infty} \map {F_\epsilon} x \, \map g x \rd x       | r = \int_0^a 0 \times \map g x + \int_a^{a + \epsilon} \dfrac 1 \epsilon \, \map g x \rd x + \int_{a + \epsilon}^\infty 0 \times \map g x \rd x       | c = Definition of $\map {F_\epsilon} x$ }} {{eqn | r = \int_0^a 0 \times \map g x + \int_a^{a + \epsilon} \dfrac 1 \epsilon \, \map g x \rd x + \lim_{L \mathop \to \infty} \int_{a + \epsilon}^L 0 \times \map g x \rd x       | c = {{Defof|Improper Integral}} }} {{eqn | r = \int_0^a 0 + \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x + \lim_{L \mathop \to \infty} \int_{a + \epsilon}^L 0 \rd x       | c =  }} {{eqn | r = \paren {0 \times \paren {a - 0} } + \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \paren {a + \epsilon} } }       | c = Definite Integral of Constant }} {{eqn | r = 0 + \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x + \lim_{L \mathop \to \infty} 0       | c = simplification }} {{eqn | r = \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x       | c =  }} {{end-eqn}} From Upper and Lower Bounds of Integral: :$\displaystyle m \paren {\paren {a + \epsilon} - a} \le \int_a^{a + \epsilon} \map g x \rd x \le M \paren {\paren {a + \epsilon} - 0}$ where: :$M$ is the maximum of $\map g x$ :$m$ is the minimum of $\map g x$ on $\closedint a {a + \epsilon}$. Hence: :$\displaystyle m \epsilon \le \int_a^{a + \epsilon} \map g x \rd x \le M \epsilon$ and so dividing by $\epsilon$: :$\displaystyle m \le \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x \le M$ Then: :$\displaystyle \lim_{\epsilon \mathop \to 0} M = m = \map g a$ and so by the Squeeze Theorem: :$\displaystyle \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \, \map g x \rd x = \map g a$ But by definition of the Dirac delta function: :$\displaystyle \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \, \map g x \rd x = \displaystyle \int_0^{+ \infty} \map \delta {x - a} \, \map g x \rd x$ Hence the result. {{qed}} {{Proofread}}
This is an instance of Integrable Function Zero A.E. iff Absolute Value has Zero Integral. {{qed}}