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First created in April 2019
The Dirac Delta Function $\delta(x)$ needs to satisfy $\delta(x)=0$ when $x\ne 0$ and $\int_{-\infty}^{+\infty}\delta(x)~dx=1.$
This is an impulse at $x=0$ with $\delta(0)\to+\infty$.
It is useful in $L^2$ Hilbert Space for square integrable functions, but using $\varphi(x)\equiv\delta(\varphi-x)$ as a basis cannot yield the result $\Braket{\varphi\Verti\varphi}=\lvert\varphi\rvert^2=1.$ But yet, our basis is based on $\delta$, a very unsatisfying situation.
We have $\lim_{\epsilon\to 0}\int_{-\epsilon}^{+\epsilon}\delta(x)~dx=1.$ So conceptually, we should consider $\delta(x)~dx~\Big|_{x=0}=1$ instead of $\delta(0)\to +\infty.$
This article is an attempt to reconcile some aspects of $\delta$, first, by understanding more about $\delta$ as a generalised function (a distribution).
https://en.wikipedia.org/wiki/Distribution_(mathematics)
Consider a set of test functions $D(\mathbf R)=\{\phi\},$ where well-behaved functions $\phi:\mathbf R\mapsto\mathbf R$ have these properties
$\phi$ is smooth (infinitely differentiable), and
$\phi$ has compact support (identically zero outside some bounded interval)
Distributions are a classn of linear functionals $T:D(\mathbf R)\mapsto\mathbf R.$
We write $\Braket{T,\varphi}=T(\varphi),$ where $\varphi\in D(\mathbf R).$
e.g. $\displaystyle\Braket{S,\varphi} =\int_\mathbf R\varphi(x)~dx\in\mathbf R.$
Let $\displaystyle\Braket{T_f,\varphi}=\int_\mathbf R f(x)~\varphi(x)~dx.~~$ Sometimes, we simply write $\Braket{f,\varphi}$, and it is understood that $f$ is in fact $T_f$, a distribution based on $f$.
$\displaystyle \Braket{f',\varphi} =\int_\mathbf R f'\varphi~dx =\big[f\varphi\big]_{-\infty}^{+\infty}-\int_\mathbf R f\varphi'~dx =-\int_\mathbf R f\varphi'~dx =-\Braket{f,\varphi'} .$
Formally, $\Braket{T',\varphi}=-\Braket{T,\varphi'}.$
Naturally, $\Braket{f^{(n)},\varphi}=(-1)^n\Braket{f,\varphi^{(n)}}.$
Discrete form: $~~H[n]=\left\{\begin{matrix}1,~~n\ge 0\\0,~~n<0\end{matrix}\right.~,~~~ \delta[n]=H[n]-H[n-1].~~ \text{i.e. } \delta[n]=\left\{\begin{matrix}1,~~n=0\\0,~~n\ne 0\end{matrix}\right.~ .$
$\displaystyle \sum_{k=-\infty}^n\delta[k] =\sum_{k=-\infty}^nH[k]-\sum_{k=-\infty}^nH[k-1] =H[n]+\sum_{k=-\infty}^{n-1}H[k]-\sum_{k=-\infty}^{n-1}H[k] =H[n] .$
Continuous form: $~~H(x)=\left\{\begin{array}{cl}\frac{d}{dx}\max\{x,0\},&x\ne 0\\0,&x=0\end{array}\right.~,~~~ \text{(ramp function)} .$
An analogy from the discrete form: $\displaystyle ~~H(x)=\int_{-\infty}^x\delta(x)~dx.~~~$ The $\delta$ function here turns out requiring a "distribution treatment".
For a Dirac Delta distribution $\delta_\alpha$, we define it as the evaluation of the test function at at $\alpha\in\mathbb R.$
$\displaystyle \Braket{\delta_\alpha~,~\varphi} \equiv\varphi(\alpha) =\int_\mathbf R\delta_\alpha(x)~\varphi(x)~dx.~~~$ If we write $\delta_0$ as $\delta$, by analogy, $\delta_\alpha(x)=\delta(x-\alpha).$
$\boxed{\int_\mathbf R\delta(x-\alpha)~\varphi(x)~dx=\varphi(\alpha).}$
By definition, $\displaystyle\Braket{\delta_\alpha~,~1} =\boxed{\int_\mathbf R\delta(x-\alpha)~dx=1.}$
Let us examine $\displaystyle H(x) =\int_{-\infty}^x\delta(x)~dx,~~ H'(x)=\delta(x) .$
Now that the $\delta$ function is defined as a distribution, we can use the above as a definition of $H$.
$\displaystyle\boxed{H(x)\equiv\int_{-\infty}^x\delta(x)~dx.}$
https://www.quora.com/What-is-the-derivative-of-the-Dirac-delta-function
$\displaystyle \int_{-\infty}^{+\infty}\frac{d^n}{dx^n}\delta(x-x_0)~\gamma(x)~dx =(-1)^n\frac{d^n\gamma(x)}{dx^n}\Big|_{x=x_0}~ .$
It results from $\Braket{\delta_\alpha^{(n)},\varphi}=(-1)^n\Braket{\delta_\alpha,\varphi^{(n)}}.$
https://physics.stackexchange.com/questions/419668/derivative-of-delta-function
$\displaystyle \int_{-\infty}^{+\infty}\delta'(x-x_0)~g(x)~dx=-g'(x_0) .$
It results from $\Braket{\delta_\alpha',g}=-\Braket{\delta_\alpha,g'}.$
http://mathworld.wolfram.com/DeltaFunction.html
$\displaystyle \delta(\alpha x)=\frac{\delta(x)}{|\alpha|},\text{ where }\alpha\ne 0 .$
Proof: Equivalent form $\displaystyle \Braket{\delta(\alpha x),g(x)}=\Braket{\frac{\delta(x)}{\lvert\alpha\rvert},g(x)} .$
$\displaystyle \mathrm{LHS}: \int_{x=-\infty}^{x=+\infty}\delta(y)~g\left(\frac{y}{\alpha}\right)\frac{1}{\alpha}~dy=\frac{g(0)}{\lvert\alpha\rvert}, \text{ where }y=\alpha x .$
(Note the boundary swap when $\alpha\le 0$.)
$\displaystyle \mathrm{RHS}: \int_\mathbf R\frac{\delta(x)}{\lvert\alpha\rvert}~g(x)~dx=\frac{g(0)}{\lvert\alpha\rvert} \ldots$ Q.E.D
$x^n\delta^{(n)}(x)=(-1)^n~n!~\delta(x).$
Proof:
$\displaystyle \mathrm{LHS}: \int_\mathbf R x^n\delta^{(n)}(x)~g(x)~dx =(-1)^n\int_\mathbf R\delta(x)~\frac{d^n}{dx^n}\Big(x^ng(x)\Big)~dx =(-1)^n\frac{d^n}{dx^n}\Big(x^ng(x)\Big)\Big|_{x=0} =(-1)^n~n!g(0) .$
Note: Only the $x^{(0)}$ term would survive when $x=0$, and such term is $\displaystyle (-1)^n~g(x)\frac{d^n}{dx^n}x^n\Big|_{x=0}=(-1)^n~n!g(0)~ .$
$\displaystyle \mathrm{RHS}: \int_\mathbf R(-1)^n~n!~\delta(x)~g(x)~dx =(-1)^n~n!g(0)~ \ldots$ Q.E.D
e.g. When $n=1,~x~\delta'(x)=-\delta(x).$
$f(x)~\delta(x-a)=f(a)~\delta(x-a).$
Proof:
$\displaystyle \mathrm{LHS}: \int_\mathbf R f(x)~\delta(x-a)~g(x)~dx =f(a)~g(a) .$
$\displaystyle \mathrm{RHS}: \int_\mathbf R f(a)~\delta(x-a)~g(x)~dx =f(a)\int_\mathbf R \delta(x-a)~g(x)~dx =f(a)~g(a)~ \ldots$ Q.E.D
e.g. $~~x~\delta(x)=0.$
$\displaystyle\delta(x^2-a^2)=\frac{1}{2a}\big(\delta(x+a)+\delta(x-a)\big),~\text{where }a>0 .$
Proof:
Recall the Heaviside Step Function: $\displaystyle H(x) =\int_{-\infty}^x\delta(x)~dx,~~ H'(x)=\delta(x) .$
As $H(x^2-a^2) =\left\{\begin{matrix}1,~x<a,~x>a\\0,~-a<x<+a\end{matrix}\right. =1-H(x+a)+H(x-a).$
Differentiate both sides:
$2x~\delta(x^2-a^2) =-H'(x+a)+H'(x-a)) =-\delta(x+a)+\delta(x-a) .$
$\delta(x^2-a^2) =-\frac{1}{2x}\big(\delta(x+a)-\delta(x-a)\big) .$
$\displaystyle \mathrm{RHS}: \int_\mathbf R-\frac{1}{2x}\big(\delta(x+a)-\delta(x-a)\big)~g(x)~dx =\frac{1}{2a}\big(g(-a)+g(a)\big) .$
This is the same as $\displaystyle \int_\mathbf R\frac{1}{2a}\big(\delta(x+a)+\delta(x-a)\big)~g(x)~dx .$
$\displaystyle \therefore\delta(x^2-a^2) =\frac{1}{2a}\big(\delta(x+a)+\delta(x-a)\big) .$
Generally, suppose $f(x)=c(x-x_1)(x-x_2)\ldots(x-x_n),\text{ with }x_1<x_2<\ldots<x_n,~$ we will have
$H\big(f(x)\big)=\left\{ \begin{array}{rl} \left(\sum_{k=1}^n(-1)^{k-1}H(x-x_k)\right),&c<0\\ 1-\left(\sum_{k=1}^n(-1)^{k-1}H(x-x_k)\right),&c>0\\ \end{array} \right.$
Differentiate both sides: $H'\big(f(x)\big)~f'(x) =-\mathrm{sgn}(c)\sum_{k=1}^n(-1)^{k-1}H'(x-x_k) .$
$\displaystyle \delta\big(f(x)\big) =\frac{\mathrm{sgn}(c)}{f'(x)}\sum_{k=1}^n(-1)^k\delta(x-x_k) .$
$\displaystyle \mathrm{RHS}: \sum_{k=1}^n\int_\mathbf R\frac{\mathrm{sgn}(c)}{f'(x)}(-1)^k\delta(x-x_k)~g(x)~dx =\sum_{k=1}^n\frac{\mathrm{sgn}(c)}{f'(x_k)}(-1)^k~g(x_k) =\sum_{k=1}^n\frac{g(x_k)}{\lvert f'(x_k)\rvert} .$
Same as $\displaystyle\sum_{k=1}^n\int_\mathbf R\frac{1}{\lvert f'(x)\rvert}\delta(x-x_k)~g(x)~dx.$
$\displaystyle \therefore\boxed{\delta\big(f(x)\big)=\sum_{k=1}^n\frac{1}{\lvert f'(x_k)\rvert}\delta(x-x_k)},~$ where $f(x)$ is a polynomial.
$\displaystyle\mu(x-x_0;\epsilon)=\frac{1}{\pi(x-x_0)}\sin\left(\frac{x-x_0}{\epsilon}\right) .$
Here is an example: $\displaystyle \delta(x-\alpha) =\frac{1}{2\pi}\int_\mathbf R e^{ik(x-\alpha)}~dk .$
$\displaystyle \frac{d}{dx}\delta(x-x_0) =\frac{1}{2\pi}\int_{-\infty}^{+\infty}\frac{\partial}{\partial x}e^{ik(x-x_0)}~dk =\frac{1}{2\pi}\int_{-\infty}^{+\infty}ik~e^{ik(x-x_0)}~dk ~\ldots(2)~ .$
Integral by parts:
\hspace{3em} $u=ik,~dv=e^{ik(x-x_0)}~dk,~du=i~dk~,v=\frac{1}{i(x-x_0)}e^{ik(x-x_0)} .$
$\displaystyle (2): \frac{d}{dx}\delta(x-x_0) =\frac{1}{2\pi}\int_{-\infty}^{+\infty}u~dv =\frac{1}{2\pi}\left[uv\right]_{-\infty}^{+\infty}-\frac{1}{2\pi}\int_{-\infty}^{+\infty}v~du.$
$\displaystyle =\frac{1}{2\pi}\left[\cancel{i}k\frac{1}{\cancel{i}(x-x_0)}e^{ik(x-x_0)}\right]_{k=-\infty}^{k=+\infty} -\frac{1}{2\pi}\int_{-\infty}^{+\infty}\frac{1}{\cancel{i}(x-x_0)}e^{ik(x-x_0)}~\cancel{i}~dk\\ .$
$\displaystyle =\cancelto{0}{\frac{1}{2\pi}\left[\cancel{i}k\frac{1}{\cancel{i}(x-x_0)}e^{ik(x-x_0)}\right]_{k=-\infty}^{k=+\infty}} -\frac{\delta(x-x_0)}{x-x_0} .$
Trying to do away the first term:
$\lim_{k_0\to\infty}\left[\frac{k}{x-x_0}e^{ik(x-x_0)}\right]_{k=-k_0}^{k=+k_0} =\lim_{k_0\to\infty}\left(\frac{k_0}{x-x_0}e^{ik_0(x-x_0)}-\frac{-k_0}{x-x_0}e^{-ik_0(x-x_0)}\right) =\lim_{k_0\to\infty}\left(\frac{2k_0}{x-x_0}2\cos(k_0(x-x_0))\right)\ldots .$
$\ldots$
Attempt: $\displaystyle \int_{-\infty}^{+\infty}g(x)\frac{d}{dx}\delta(x-x_0)~dx =\Big[g(x)~\delta(x-x_0)\Big]_{-\infty}^{+\infty} -\int_{-\infty}^{+\infty}\delta(x-x_0)~\frac{d}{dx}g(x)~dx =\ldots .$
Attempt: $\displaystyle \frac{1}{2\pi}\int_{-\infty}^{+\infty}ik~e^{ik(x-y)}~dk =\left[ik\left(\frac{1}{2\pi}\int e^{ik(x-y)}dk\right)\right]_{-\infty}^{+\infty} -\int_{-\infty}^{+\infty}i\cancelto{1}{\frac{dk}{dk}}\left(\frac{1}{2\pi}\int e^{ik(x-y)}~dk\right)dk =\ldots .$
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