Applying the fundamental theorem of calculus in double integral

Question

How would one go about calculating $$\frac{d}{dt}\int^t_{-t}f(z,t)dt.$$ And more specifically, $$\frac{d}{dt}\int^t_{-t}\int^t_{-t}f(x,y)dxdy$$ Assuming the necessary conditions, i got to $$lim_{h\to 0}\frac{\int_{(-t-h,t+h)\setminus(-t,t)}f(x,y)}{h}+\int^t_{-t}\frac{d}{dt}(\int^t_{-t}f(x,y)dx)dy$$ Is this correct? And if it is, what would be necesary for the first term to go to $0$.

Answer 1

Define $G(a,b,c,d)$ by

$$ G(a,b,c,d) = \int_{c}^{d} \int_{a}^{b} f(x, y) \, \mathrm{d}x\mathrm{d}y. $$

Assuming that $f$ is continuous near the rectangle $[a, b] \times [c, d]$, by the fundamental theorem of calculus,

\begin{align*} \frac{\partial G}{\partial a} &= -\int_{c}^{d} f(a, y) \, \mathrm{d}y, & \frac{\partial G}{\partial b} &= \int_{c}^{d} f(b, y) \, \mathrm{d}y, \\ \frac{\partial G}{\partial c} &= - \int_{a}^{b} f(x, c) \, \mathrm{d}x, & \frac{\partial G}{\partial d} &= \int_{a}^{b} f(x, d) \, \mathrm{d}x. \end{align*}

So, if $F(t)$ denotes OP's integral, then

$$ F(t) := \int_{-t}^{t} \int_{-t}^{t} f(x, y) \, \mathrm{d}x\mathrm{d}y = G(-t, t, -t, t) $$

and hence we can compute its derivative by invoking the chain rule:

\begin{align*} F'(t) &= - \frac{\partial G}{\partial a} + \frac{\partial G}{\partial b} - \frac{\partial G}{\partial c} + \frac{\partial G}{\partial d} \biggr|_{(a,b,c,d)=(-t,t,-t,t)} \\ &= \int_{-t}^{t} [f(x, t) + f(x, -t)] \, \mathrm{d}x + \int_{-t}^{t} [f(t, y) + f(-t, y)] \, \mathrm{d}t. \tag{*} \end{align*}

This also has an intuitive explanation. If $0 < t < s$, then $F(s) - F(t)$ is equal to the integral of $f(x, y)$ on the rectangular annulus

\begin{align*} [-s, s]^2 \setminus (-t, t)^2 &= \underbrace{([-s, -t] \times [-s, s])}_{\text{left side}} \cup \underbrace{([t, s] \times [-s, s])}_{\text{right side}} \cup \underbrace{([-s, s] \times [-s, -t])}_{\text{bottom side}} \cup \underbrace{([-s, s] \times [t, s])}_{\text{top side}} . \end{align*}

Letting $ s = t + \Delta t$ and assuming $\Delta t$ is small, the integral of $f$ over the above annulus will be approximately $\Delta t$ times the line integral of $f$ along the boundary of $[-t, t]^2$, which is in accordance with the result $\text{(*)}$.