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How can I perform derivative of double integral

$$\frac{\mathrm d}{\mathrm dt}\int_{t-\mathrm d1}^t \int_h^t f(s) \,\mathrm ds\,\mathrm dh$$

Can I apply a Leibniz rule of some form? How?

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    $\begingroup$ Do you know what the answer is supposed to be? What have you tried? $\endgroup$ Jan 18, 2017 at 6:17

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The Leibniz integral rule can be applied.

Let $\int_{h}^{t}\!f \left( s \right) \,{\rm d}s = F(t,h)$.

\begin{align} \frac{\mathrm d}{\mathrm dt}\int_{t-\mathrm d1}^t \int_h^t f(s) \,\mathrm ds\,\mathrm dh &= {\frac {\mathrm d}{\mathrm dt}}\int_{t-\mathrm d_{{1}}}^{t}F \left( t,h \right) \,{\rm d}h\\ &= F(t,t)\frac {\mathrm d}{\mathrm dt}(t) - F(t,t-\mathrm d_1)\frac {\mathrm d}{\mathrm dt}(t - \mathrm d_1) + \int_{t-\mathrm d_{{1}}}^{t} \frac {\partial}{\partial t} F \left( t,h \right) \,{\rm d}h\\ &= 0 - F(t,t-\mathrm d_1) + \int_{t-\mathrm d_{{1}}}^{t} f(t)\,{\rm d}h\\ &= \boxed{-\int_{t-\mathrm d_{{1}}}^{t} f(s)\,{\rm d}s + \mathrm d_1\,f(t)}\\ \end{align}

Note that this is the same as what Mathematica and Maple give.

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    $\begingroup$ Where you have $F(t,t)$, @math-fun has $F(t,h)$ in effect; likewise for $F(t,t-d_1)$ the other has $F(t-d_1,h)$. That's wrong (that is, math-fun's answer is wrong). Perhaps the highlighting, instead of revealing the error, masks it. $\endgroup$
    – Michael E2
    May 27, 2020 at 22:49
  • $\begingroup$ @MichaelE2 Just came back to this and fixed it: thank you very much for the comment. $\endgroup$
    – Math-fun
    Mar 5, 2021 at 18:07
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Let $\int_h^tf(x)ds = F(t)-F(h)$ with $F'(t)=f(t)$. Then

\begin{align} \frac{\mathrm d}{\mathrm dt}\int_{\color{red}{t-\mathrm d_1}}^{\color{blue}{t}} \int_h^t f(s) \,\mathrm ds\,\mathrm dh&=\frac{\mathrm d}{\mathrm dt}\int_{\color{red}{t-\mathrm d_1}}^{\color{blue}{t}} (F(t)-F(h))\,\mathrm dh\\ &=\int_{t-\mathrm d_1}^{t} \frac{\mathrm d}{\mathrm dt}(F(t)-F(h))\,\mathrm dh\\ &+\frac{\mathrm d}{\mathrm dt}(t)\times (F(t)-F(t))\\ &-\frac{\mathrm d}{\mathrm dt}(t-\mathrm d_1)\times (F(t)-F(t-\mathrm d_1))\\ &=\int_{t-\mathrm d_1}^{t} f(t)\,\mathrm dh-F(t)+F(t-\mathrm d_1)\\ &=(t-t+d_1)f(t)+F(t)-F(t-\mathrm d_1)\\ &=d_1f(t)-\left(F(t)-F(t-\mathrm d_1)\right)\\ \end{align}

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  • $\begingroup$ Why do both Maple and Mathematica answer $$\mathrm d_{{1}}f \left( t \right) -\int_{t-\mathrm d_{{1}}}^{t}\!f \left( s \right) \,{\rm d}s=\mathrm d_{{1}}f \left( t \right) -F \left( t \right) +F \left( t-\mathrm d _{{1}} \right) $$ though? $\endgroup$
    – Leponzo
    May 27, 2020 at 4:28
  • $\begingroup$ @Leponzo Just came back to this and fixed it: thank you very much for the comment. $\endgroup$
    – Math-fun
    Mar 5, 2021 at 18:07
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HINT
Set $$\int_h^t f(s)\mathrm{d}s=F(t)$$ Then apply Leibniz twice

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