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How can the following integral can be solved $$I(t_1,t_2)=\int_0^{t_1}dt'e^{-a(t_1-t')}\int_0^{t_2}dt''e^{-a(t_2-t'')}\delta(t''-t')$$ where there are no assumptions regarding $t_1$ or $t_2$ (both cases of $t_1>t_2$ or $t_2>t_1$ are valid).

are there any limitations for using the Dirac delta here?

jarhead
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1 Answers1

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First I do the inner integral: $$ \int_0^{t_2} dt'' \, e^{-a(t_2-t'')} \, \delta(t''-t') = \int_{-\infty}^{\infty} dt'' \, \chi_{[0, t_2]}(t'') \, e^{-a(t_2-t'')} \, \delta(t''-t') = \chi_{[0, t_2]}(t') \, e^{-a(t_2-t')} $$ Here, for any set $A$, $\chi_A(t) = 1$ if $t\in A$, $\chi_A(t)=0$ otherwise.

Then the outer integral: $$ I(t_1,t_2) = \int_0^{t_1} dt' \, e^{-a(t_1-t')} \, \chi_{[0, t_2]}(t') \, e^{-a(t_2-t')} \\ = e^{-a(t_1+t_2)} \int_0^{t_1} dt' \, e^{2at'} \, \chi_{[0, t_2]}(t') \\ = e^{-a(t_1+t_2)} \int_{-\infty}^{\infty} dt' \, \chi_{[0, t_1]}(t') \, e^{2at'} \, \chi_{[0, t_2]}(t') \\ = e^{-a(t_1+t_2)} \int_{-\infty}^{\infty} dt' \, \chi_{[0, \min(t_1,t_2)]}(t') \, e^{2at'} \\ = e^{-a(t_1+t_2)} \int_{0}^{\min(t_1,t_2)} dt' \, e^{2at'} \\ = e^{-a(t_1+t_2)} \left[\frac{1}{2a} e^{2at'} \right]_{0}^{\min(t_1,t_2)} \\ = \frac{1}{2a} e^{-a(t_1+t_2)} \left[e^{2a\min(t_1,t_2)} -1 \right] $$

md2perpe
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  • what is the definition of $A$? – jarhead Jul 22 '18 at 12:20
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    $A$ is just some set. I expanded on it somewhat in my post. – md2perpe Jul 22 '18 at 12:23
  • So can it be considered as the Heaviside step function? – jarhead Jul 22 '18 at 12:24
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    It's similar: $\chi_{(0, \infty)}(t) = H(t).$ But $\chi_{(a,b)}(t) = H(t-a) - H(t-b).$ – md2perpe Jul 22 '18 at 12:27
  • Does the formulation still work if t1 and/or t2 is less than zero? - Not implying it doesn't, just that I know you have to worry about it, and I see no discussion of that... – John Polcari Jul 22 '18 at 12:37
  • @md2perpe, can you please explain why have you taken the minimum value of both sets? – jarhead Jul 22 '18 at 12:39
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    Look at the integral before. There we have both $\chi_{[0, t_1]}(t')$ and $\chi_{[0, t_2]}(t')$. For the product of these not vanish, we must have both $t' \in [0, t_1]$ and $t' \in [0, t_2],$ i.e. $t' \in [0, t_1] \cap [0, t_2] = [0, \min(t_1,t_2)].$ – md2perpe Jul 22 '18 at 14:06
  • @JohnPolcari. In the calculations I have assumed $t_1, t_2 > 0.$ I'm looking into what happens in other cases. – md2perpe Jul 22 '18 at 14:37
  • @md2perpe Intuition says min will change to max when both t1 and t2 are negative, but I am sure you will sort it out, just a little tedious. – John Polcari Jul 22 '18 at 15:06
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    In $I(t_1, t_2)$, $e^{-2 a t'}$ in the second line should be $e^{2 a t'}$. Ultimately $I(t_1, t_2)$ should come out as $$\cases{ \frac 1 {2a} e^{-a (t_1 + t_2)} (e^{2 a \min(t_1, t_2)} - 1) & $t_1 > 0 \land t_2 > 0$ \ -\frac 1 {2a} e^{-a (t_1 + t_2)} ( e^{2 a \max(t_1, t_2)} - 1)& $t_1 < 0 \land t_2 < 0$ \ 0 & otherwise}.$$ – Maxim Jul 22 '18 at 15:20
  • @md2perpe, pls correct your solution – jarhead Jul 22 '18 at 15:28
  • @Maxim, thank you for pointing that out – jarhead Jul 22 '18 at 15:29
  • Sign of exponent fixed. – md2perpe Jul 22 '18 at 15:40
  • @Maxim. Do you have a short explanation for the cases when $t_1,t_2$ are not positive? – md2perpe Jul 22 '18 at 15:42
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    Let's make up a signed indicator function, $\chi_{(0,a)}(t) = [0 < t < a] - [a < t < 0]$. Then your derivation goes through unchanged till the point where you take the minimum. For $t_1 t_2 < 0$, $\chi_{(0, t_1)}(t) \chi_{(0, t_2)}(t) = 0$. For negative $t_1$ and $t_2$, the value which is closer to zero will be $\max(t_1, t_2)$ instead of $\min(t_1, t_2)$. – Maxim Jul 22 '18 at 16:09
  • I also got the idea of a signed indicator function, the same one as you, although written differently. But I didn't really sort out exactly what happens in the non-positive cases. Thanks for fixing and completing my calculations! – md2perpe Jul 22 '18 at 19:05