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This question arises in a problem I'm working on in climate economics. Let $X\left( s \right) % MathType!MTEF!2!1!+- % feaahCart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbcvPDwzYbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0x % e9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKk % Fr0xfr-xfr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam % iwamaabmaabaGaam4CaaGaayjkaiaawMcaaaaa!3B70! $ be a real variable. How does one differentiate $\int\limits_0^t {X\left( s \right) ds} % MathType!MTEF!2!1!+- % feaahCart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbcvPDwzYbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0x % e9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKk % Fr0xfr-xfr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8 % qCaeaacaWGybWaaeWaaeaacaWGZbaacaGLOaGaayzkaaacciGae8hi % aaIaamizaiaadohaaSqaaiaaicdaaeaacaWG0baaniabgUIiYdaaaa!423E! $ with respect to $X\left( t \right) % MathType!MTEF!2!1!+- % feaahCart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbcvPDwzYbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0x % e9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKk % Fr0xfr-xfr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam % iwamaabmaabaGaamiDaaGaayjkaiaawMcaaaaa!3B71! $ , the value of $X\left( \bullet \right) % MathType!MTEF!2!1!+- % feaahCart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbcvPDwzYbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0x % e9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKk % Fr0xfr-xfr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam % iwamaabmaabaGaeyOiGClacaGLOaGaayzkaaaaaa!3BFD! $ at the upper limit? I posed this to ChatGPT and it used the Dirac delta function to conclude that the answer is 1. However, I've found it sometimes makes math mistakes, and in any case I didn't understand its logic.

Any advice will be appreciated.

Alan
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  • Functional derivative: $\displaystyle{\delta\operatorname{X}\left(s\right) \over \delta\operatorname{X}\left(t\right)} = \delta\left(s - t\right)$ – Felix Marin May 25 '24 at 20:12
  • Dimensional analysis shows the answer needs the units of $t$, so ChatGPT is wrong. – J.G. May 25 '24 at 23:20
  • Thank you all for your responses. Felix Marin, could you elaborate on how I use this functional derivative to calculate the answer? – Alan May 27 '24 at 16:30
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    Do not use chatGPT for anything other than amusement. It only strings together words based on a probabilistic model, it is not a search engine, it does not do any computation. It doesn't "sometimes make math mistakes". It sometimes just happens to say the right thing, much like a broken clock is right twice a day. – Arturo Magidin May 28 '24 at 18:46
  • J. G.: If the integral happens to be linear in the terminal value of X(.), then the derivative is a constant. Could you explain the dimensional analysis you're referring to? – Alan May 29 '24 at 19:34

1 Answers1

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From this question we can just use the chain rule in a certain way. Let $$f(t) = \int_0 ^t X(s) \ ds.$$ Then

\begin{align*} \frac{df}{dX(t)} &= \frac{df/dt}{dX(t)/dt} \\ &= \frac{X(t)}{X'(t)} \\ \end{align*}

assuming, of course, $X$ is differentiable.

Sean Roberson
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  • Sean Roberson: In their Mar. 18, 2021 response to the question you linked, user49404 provides a formal derivation of the chain-rule cancellation heuristic. If one applies that to my problem, one step is taking the partial of the integral with respect to the function X(.) - i.e., the function itself, not the terminal value. How does one do that? Thanks. – Alan May 29 '24 at 19:41
  • You may need to look at Steiltjes integrals, where you can integrate with respect to a function (of bounded variation). – Sean Roberson Jun 01 '24 at 18:11