Derivative of double integral using Leibniz integral rule

Question

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?

Answer 1

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}

Answer 2

HINT
Set $$\int_h^t f(s)\mathrm{d}s=F(t)$$ Then apply Leibniz twice