Automatic way to derive $∫_0^x f(x,t) \,\mathrm{d} t$

Question

Correct me if I'm wrong, but \begin{align} F(x) = ∫_a^b f(x,t) \,\mathrm{d} t &\implies F'(x) = ∫_a^b \frac{∂f}{∂t} \,\mathrm{d} t, \qquad \text{and} \\ G(x) = ∫_a^x f(t) \,\mathrm{d} t &\implies G'(x) = f(x). \end{align} But is there any general “formula” to derive this function $$ F(x) = ∫_a^x f(x,t) \,\mathrm{d} t, $$ or, in a more general way $$ \frac{\mathrm{d}}{\mathrm{d}x} \biggl( ∫_{a(x)}^{b(x)} f(x,t) \, \mathrm{d}t \biggr). $$

Answer 1

Yes, apply Leibniz' integral rule (a.k.a. differentiation under the integral sign),

$$\frac{\mathrm{d}}{\mathrm{d}x}\int_{a(x)} ^{b(x)} f(x,t)\,\mathrm{d}t =f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int_{a(x)} ^{b(x)}\frac{\partial f}{\partial x}(x,t)\,\mathrm{d}t$$

which in your case gives

$$F'(x)=f(x,x)+\int_a ^x \frac{\partial f}{\partial x}(x,t)\,\mathrm{d}t.$$

Edit: As asked by a user, the precise conditions for Leibniz' integral rule to hold for $t\in[t_1,t_2]$ are

• $a,b$ are $C^1$ on $[t_1,t_2]$
• $f$ is $C^1$ on $\{(x,t): t\in [t_1,t_2], x\in[a(t),b(t)]\}$

(see, e.g., Kaplan: Advanced Calculus, Chpt. 4.9)

Answer 2

This is almost identitical to the question I asked and answered myself here:

Exact smoothness condition necessary for differentiation under integration sign to hold.

You need some additional smoothness condition, e.g. the partial derivative $\partial_xf(x,t)$ to be continuous. The wiki link is actually pretty dreadful as the smooth conditions are not clearly stated. I wonder if someone can give an authentic source.