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So I was watching this video by Michael Penn https://www.youtube.com/watch?v=nkaZEI_e2SU&list=LL&index=8&t=430s and at around the 16:30 mark he puts in t = 0 into I'(t) and gets $-\pi$ as the other integral is supposedly the integral of 0, which is 0 if integrated.

But if you were to put in t = 0 at the starting definition of I'(t), just after differentiating it, by that same logic you would get the integral of 0, which again would be 0.

At first I thought it had something to do with the integral not being defined somewhere in the interval or at the bounds. But if you were to take the limits of the integrand at both limits for both integrals, you find that the integrands approach a value and don't diverge for each bound and there are no undefined points in the interval.

Somebody in the comments mentioned something about the Dirichlet integral being some kind of a special integral and that it doesn't work at the start because of that but I'm not convinced.

So I was just hoping somebody could explain to me how this is possible and why exactly this happens.

Vilius
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    I think part of the problem is that $I(t)$ is not differentiable at $t=0$. So it looks like he is computing the right-hand limit of $I'(t)$ as $t\to 0$, but this is different than if you formally differentiate under the integral and set $t=0$. I'm not sure how to justify passing the derivative under the integral for $t>0$ though... – Trevor Norton Dec 07 '21 at 01:14
  • @TrevorNorton You would use uniform convergence to justify it. – Angel Dec 14 '21 at 20:38
  • @Angel The difference quotients don't look like they are uniformly convergent on $(0,\infty)$. You probably need some integrability condition to hold for the difference quotients in addition to uniform convergence. I thought the dominated convergence theorem might apply here, but the absolute values of the finite differences don't look integrable. So that theorem can't be directly applied. – Trevor Norton Dec 14 '21 at 21:04
  • @TrevorNorton No, you are overthinking this. Take a look at https://en.wikipedia.org/wiki/Leibniz_integral_rule. The quotients are obviously Riemann integrable on any open interval $[0,a]$ with $a\geq0,$ and that is the only condition necessary for the interchange on such intervals. As such, for the improper integral, the only condition required for the interchange of the integral symbol and the derivative symbol is the condition for exchanging limits and derivatives, which is given by the uniform convergence of the convergent, which in this case, are the integrals of the quotients... – Angel Dec 14 '21 at 21:16
  • @TrevorNorton ...for additional reference on the above comment, please look at https://math.stackexchange.com/questions/409178/can-i-exchange-limit-and-differentiation-for-a-sequence-of-smooth-functions. Remember that improper integrals are limits of actual Riemann integrals. – Angel Dec 14 '21 at 21:18
  • @Angel I understand that you can interchange the derivative and integral for Riemann integrals on a finite interval, but I'm not sure I completely agree with your argument here. I'm unsure if the exchange of limits here follows that simply (not sure what ``uniform convergence of the convergent'' means). We can see from the post that interchanging derivatives and integrals fails at $t=0$. The standard way to do this on an infinite intervals is with the DCT, which fails here. Regardless, I'm sure there is a way justify it for $t>0$, but I'm not seeing a simple way to do it. – Trevor Norton Dec 14 '21 at 22:20
  • @TrevorNorton I am not sure you are understanding what my argument actually is, since I never said the interchange works for $t=0.$ My analysis is a direct response to you saying that you do not know how to justify the interchange for $t\gt0,$ which is exactly what I am explaining how to do. Your argument is essentially "it does not work for $t=0,$ therefore, it also does not work for $t\gt0.$" But that is just not how that works. I am explaining to you exactly why it works for $t\gt0$ but not $t=0,$ and you are rejecting the explanation for no reason at all. – Angel Dec 15 '21 at 13:05
  • @TrevorNorton Furthermore, the fact that you can interchange between the derivative and the limit (because interchanging the derivative and integral once the limit itself is taken care of is trivial, as pointed out in the Leibniz rule article that I linked) once uniform convergence is established is a basic theorem of real analysis, a theorem so basic it even has no name. You only need uniform convergence to interchange the limit with a derivative operator. This is a well-known fact. If you think I am wrong, then I encourage you to please come up with an example. – Angel Dec 15 '21 at 13:09
  • @TrevorNorton I even provided a source that explains that you can exchange limits and derivatives once you have uniform convergence. Have you taken a look at it? I am not sure this conversation can get anywhere at all if you are unwilling to accept the fact. – Angel Dec 15 '21 at 13:11
  • @Angel Hello. I’m pretty familiar with real analysis actually haha. I did look at what you referenced and thought about it, I just don’t think they apply as easily as you say. I didn’t think I could explain it succinctly in the comments. The reason I mentioned $t=0$ was because your argument seems to apply to that case too, but perhaps I’m misunderstanding you. If you think you can provide a full answer then I’d post a solution to this problem. – Trevor Norton Dec 15 '21 at 13:48

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