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Based on Wikipedia, the Dirac delta can be loosely thought of as a function on the real line that is defined by $$ \delta (t)=\begin{cases}+\infty ,&t=0\\0,&t\neq 0\end{cases}$$ and which satisfies $$ \int _{-\infty }^{\infty }\delta (t)\,dt=1.$$ This is merely a heuristic characterization. The Dirac delta is not a function in the traditional ...


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I suggest reading about distributions (generalised functions), e.g. https://en.wikipedia.org/wiki/Distribution_(mathematics) . The point is, Dirac's $\delta$-function is not a function at all, at least not in the usual sense. What you have given as a "definition" is merely an intuition for it. What $\delta$-function is is this: it is a linear functional ...


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I'm assuming you are using the physicist definition of the Dirac delta distribution, where: $$\int_{-\infty}^{\infty} f(x)\delta(x)dx = f(0)$$ If this is the case, then the properties follow from: 1) $\int_{-\infty}^{\infty} f(x) \delta(-x)dx = -\int_{\infty}^{-\infty} f(-y)\delta(y)dy =\int_{-\infty}^{\infty}f(y)\delta(y)dy$ 2) $\int_{-\infty}^{\infty}f(...


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Recall that the Green's function $G(x,s)$ for this BVP solves $$y''-2y'+y = \delta(x-s), \ \ \ y(-1)=y(1)=0.$$ Since this equation is linear, we can add solutions and all that will change is the right-hand side. If we choose $y(x) = G(x,1/2)+G(x,-1/2)$, then $y$ solves the equation with $g(x) = \delta(x-1/2)+\delta(x+1/2)$


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Technically yes. The first term is just zero; in fact this always happens when $f$ is a test function and $d$ is a distribution*, so that distributional integration by parts reads $\int f d' = -\int f' d$ (in shorthand). The second term is just evaluated using the definition of $\delta$. * Here I mean either "$f$ is compactly supported smooth and $d$ is a ...


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You should integrate by parts twice ti shift the derivatives off the delta and on to the function f(p, p'). Doing this, and assuming the boundary terms cannot contribute (they should not saturate the delta function) will leave $$(-1)^2\int dp dp' f''(p, p') \delta(p-p'). $$ Can you finish from here?


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In the classical sense, the integrals do not exist for any $n$, at the value $x=0$. For $n\geq 0$, this is because $a^n$ does not decay at infinity. For $n<0$, the singularity at $0$ is too poor. If we instead consider the tempered distributions associated with $a^n$, that is the functional $T_{a^n}\in \mathcal{S}'(\mathbb{R})$ given by $T_{a^n} (\phi) = \...


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The best way to think of the dirac delta function is: $$\delta_n(x)=\begin{cases} 0 & x<0 \\ n & 0<x<\frac1n \\ 0 & x>\frac1n \end{cases}$$ $$\delta(x)=\lim_{n\to\infty}\delta_n(x)$$ Now try and integrate $\delta_n(x)$ first then take the limit: $$\int_{-\infty}^\infty\delta_n(x)dx=\int_{-\infty}^0\delta_n(x)dx+\int_0^{\frac1n}\...


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Formally, the dirac delta is not a function in the same sense as the heaviside function. Rather, it is a distribution.


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Here's an idea for a proof that seems to be semi-rigorous and within the limitations of the class material (I would give full points to it at least :P). Using complex analysis arguments it is easy to prove that $$\theta_{\epsilon}(x)=e^{-\epsilon x}\theta(x)~,~x\neq0$$ If we define $\theta'(x)\equiv\lim_{\epsilon\to 0}\theta'_{\epsilon}(x)$ then it is ...


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