How can we derive $\frac{\partial}{\partial t}\int^t_{t_0}f(t,\tau)d\tau$?

Question

As I read some textbook of engineering, I've been wondering why the partial equation of integration $\frac{\partial}{\partial t}\int^t_{t_0}f(t,\tau)d\tau$ can be derived like below $$\frac{\partial}{\partial t}\int^t_{t_0}f(t,\tau)d\tau=\int^t_{t_0}\left(\frac{\partial}{\partial t}f(t,\tau)\right)d\tau+f(t,\tau)|_{\tau=t} $$ It is intuitive that the partial differential about $t$ is not related by $\tau$ so outside partial differential can be move to inner side of integral but I don't know exactly why the residual component $f(t,\tau)|_{\tau=t}$ in the right side of equation could follow. Thank you.

Answer 1

I think this is the version of the Leibniz Integral rule:http://en.wikipedia.org/wiki/Leibniz_integral_rule. Use $t$ instead of $\theta$

, for $b(t)=t$, and $a(t)=t_0$ , which cancels out, since $t_0$ is a constant. Then $$

$$\frac{\partial}{\partial t}\int^t_{t_0}f(t,\tau)d\tau =\frac{\partial}{\partial t}f(t,\tau)+f(t,t) \partial d(t)/t =\partial\frac{\partial}{\partial t}f(t,\tau)+f(t,t) (1) $$

Answer 2

For intuition, consider $F(t+ \delta)-F(t)$, where $F(t) = \int^t_{t_0}f(t,\tau)d\tau$.

This gives \begin{eqnarray} F(t+ \delta)-F(t) &=& \int^{t+\delta}_{t_0}f(t+\delta,\tau)d\tau - \int^t_{t_0}f(t,\tau)d\tau \\ &=& \int^t_{t_0} ( f(t+\delta,\tau) - f(t,\tau) ) d\tau + \int^{t+\delta}_{t}f(t+\delta,\tau)d\tau \\ &\approx& \int^t_{t_0} {\partial f(t,\tau) \over \partial t } \delta d\tau + f(t,t) \delta \end{eqnarray}