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Consider a typical detector equivalent circuit, where the detector can be seen as an ideal current generator $I(t)$. Since $I = I_C + I_R$ (currents through the capacitor and resistor), $I_R R=V$ and $\frac{I_C}{C}=\frac{dV}{dt}$, the equation of the circuit is:

$$\frac{dV}{dt}+\frac{V}{RC}=\frac{I}{C}.$$

When $C$ goes to $0$, the equation should read $V=IR$ as it happens in detectors with a low $RC$ constant. But if I wanted to send $C$ to zero in the solution of the differential equation, which is $V(t)=\int_0^t \frac{I}{C}e^{\frac{t'-t}{RC}}dt'$, to obtain the same exact result, I would be brought to think that $e^{\frac{t'-t}{RC}}$ is kind of double a Dirac delta function. Substituting in the integral: $$V(t)=\int_0^t \frac{I}{C}2\delta(t'-t)RCdt'=I(t)R.$$

Is it actually a delta? Is it correct to reason this way?

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    $\begingroup$ Whether it is correct or not, it's much easier to rewrite your equation as $CR\dfrac{dV}{dt} + V= IR$ and not play complicated mathematical games at all. $\endgroup$
    – alephzero
    Jul 16 at 18:54
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  1. Yes, OP has essentially constructed a one-sided Dirac delta distribution $$\delta_{[0,\infty[}(x)~=~\lim_{\varepsilon\searrow 0} \frac{1}{\varepsilon} e^{-x/\varepsilon}$$ on the positive half line $[0,\infty[$ via a generalized function, so that $$\int_{[0,\infty[} \mathrm{d}x~\delta_{[0,\infty[}(x)~f(x)~=~f(0) $$ for all test functions $f:[0,\infty[\to \mathbb{R}$.

  2. Notice that OP's distribution is different from the usual two-sided Dirac delta distribution $\delta_{\mathbb{R}}(x)$ on the real line $\mathbb{R}$, which satisfies $$\int_{\mathbb{R}} \mathrm{d}x~\delta_{\mathbb{R}}(x)~f(x)~=~f(0) $$ for all test functions $f:\mathbb{R}\to \mathbb{R}$.

  3. This also explains why OP's last formula has a correction factor of 2.

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  • $\begingroup$ Would you have a reference for a more detailled analysis about the "one sided delta function" ? Thanks ! $\endgroup$ Jul 16 at 17:06
  • $\begingroup$ @VincentFraticelli. Do you have a proper distribution-theoretical definition of "one sided delta function"? $\endgroup$
    – md2perpe
    Jul 16 at 17:19
  • $\begingroup$ I found out that it exists on this post. But, I had already asked myself the question about the currents induced on the surface of a conductor when the conductivity tends towards infinity. There is the same problem because the conductor exist only for $z>0$ $\endgroup$ Jul 16 at 17:26
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A more mathematical treatment

We have an ordinary differential equation (ODE): $$\alpha^{-1}y'(t) + y(t) = f(t).$$ In our case, $y(t)=V(t),$ $\alpha=\frac{1}{RC},$ $f(t)=RI(t).$

The ODE can be solved by finding a Green's function $G_\alpha(t)$ satisfying $$\alpha^{-1}G_\alpha'(t) + G_\alpha(t) = \delta(t)$$ and then get the solution $y(t)$ as $$y(t) = (G_\alpha*f)(t) = \int_{-\infty}^{\infty} G_\alpha(t-t')\,f(t')\,dt'.$$

A Green's function that only gives a response for $t>0$ (so that it's causal) is given by $$G_\alpha(t) = \alpha e^{-\alpha t} H(t),$$ where $H(t)$ is the Heaviside step function.

Distributions and convergence as such are defined by how they "act" on, or work in an integral when multiplied with, a test function $\varphi\in C^\infty_c(\mathbb{R}).$ Therefore, to show that $G_\alpha(t) \to \delta(t)$ in the sense of distributions, when $\alpha\to\infty,$ we need to show that $$ \lim_{\alpha\to\infty} \int_{-\infty}^{\infty} G_\alpha(t)\,\varphi(t)\,dt = \int_{-\infty}^{\infty} \delta(t)\,\varphi(t)\,dt = \varphi(0) $$ for all $\varphi\in C^\infty_c(\mathbb{R}).$

That is easy. Using the variable change $t=s/\alpha$ we get $$ \int_{-\infty}^{\infty} G_\alpha(t) \, \varphi(t) \, dt = \int_{-\infty}^{\infty} \alpha e^{-\alpha t} H(t) \, \varphi(t) \, dt = \int_{0}^{\infty} \alpha e^{-\alpha t} \, \varphi(t) \, dt = \int_{0}^{\infty} \alpha e^{-s} \, \varphi(s/\alpha) \, ds/\alpha \\ = \int_{0}^{\infty} e^{-s} \, \varphi(s/\alpha) \, ds \to \int_{0}^{\infty} e^{-s} \, \varphi(0) \, ds = \int_{0}^{\infty} e^{-s} \, ds \, \varphi(0) = 1 \cdot \varphi(0). $$

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    $\begingroup$ Might be worth mentioning it's a linear ODE, which is necessary for solving it with a Green's function. $\endgroup$ Jul 16 at 22:38
  • $\begingroup$ +1, Hopefully there is a correct mathematical answer. $\endgroup$
    – LL 3.14
    Jul 24 at 0:52
  • $\begingroup$ @LL3.14. I would say that this is a correct mathematical answer. There are just a few things to fix to make it fully rigorous. $\endgroup$
    – md2perpe
    Jul 24 at 8:35
  • $\begingroup$ Oh yes, sorry if I was not clear, I was saying that your answer was a rigorous. $\endgroup$
    – LL 3.14
    Jul 24 at 21:49

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