commutativity of differentiation and indefinite integrals

I know the Leibniz rule which states that differentiation and definite integration with respect to independent variables are commutative.

$$\dfrac{\partial}{\partial x} \left( \int^b_a f(x,t) \ dt \right) =\int^b_a \left( \dfrac{\partial f(x,t)}{\partial x} \right) dt$$

Is this commutative property also applicable to indefinite integrals if we ignore the arbitrary constant? For example, will the functions in $LHS$ and $RHS$ of below equation be equal (ignoring the arbitrary constant)?

$$\dfrac{\partial}{\partial x} \left( \int f(x,t) \ dt \right) =\int \left( \dfrac{\partial f(x,t)}{\partial x} \right) dt$$

Edit: In the answer to this question here, it is written:

\begin{align} - \frac{\partial G}{\partial z} &= A \\ \frac{\partial F}{\partial z} &= B \\ \frac{\partial G}{\partial x} - \frac{\partial F}{\partial y} &= C \end{align}

Then integrate the first two equations

\begin{align} G &= - \int A\ dz + g(x,y) \\ F &= \int B dz + f(x,y) \end{align}

At this point, the two arbitrary functions leftover from integration should be set so that the third equation is satisfied

$$\frac{\partial G}{\partial x} - \frac{\partial F}{\partial y} = -\int \left(\frac{\partial A}{\partial x} + \frac{\partial B}{\partial y} \right)dz + \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} = C$$

Here we are dealing with indefinite integrals. So how can we take differentiation with respect to $x$ and $y$ inside antiderivative with respect to $z$?

• Let $F(x,t)$ be the family of functions that satisfy $\frac{\partial F(x,t)}{\partial t}=f(x,t)$. Assuming that the sufficient conditions are met for equality of mixed partals, then $$\frac{\partial }{\partial t}\frac{\partial F(x,t)}{\partial x}=\frac{\partial f(x,t)}{\partial x}$$Can you finish now? – Mark Viola Jul 3 '18 at 16:12
• $$f(x,t)=\dfrac{\partial}{\partial t} \left[ \int f \ dt \right]$$ $$\dfrac{\partial f}{\partial x}=\dfrac{\partial ^2}{\partial x \ \partial t}\left[ \int f \ dt \right]$$ $$\int \dfrac{\partial f}{\partial x} \ dt=\dfrac{\partial}{\partial x} \left[ \int f \ dt \right] + g(x)$$ where $g(x)$ is constant of integration..... Am I right? – Joe Jul 3 '18 at 16:55
• You have it now. Well done. – Mark Viola Jul 3 '18 at 17:50
• thanks.............. – Joe Jul 3 '18 at 18:16

Every indefinite integral can be written in a form of definite integral $$\int f(x,t)dt=\int^t_{c(x)} f(x,u)du$$

Let $c(x)$ be a constant(w.r.t. $u,t$) such that both $f(x,u)$ and $f_x(x,u)$ is continuous for $c(x)\le u\le t$ when $s_0\le x\le s_1$. Also suppose $c(x)$ has continuous derivative when $s_0\le x\le s_1$.

Then, by Leibniz’s integral rule, for $s_0\le x\le s_1$,
$$\color{red}{\frac{\partial}{\partial x}\int^t_{c(x)} f(x,u)du=-f(x,c(x))\cdot c’(x)+\int^t_{c(x)}\left(\frac{\partial}{\partial x} f(x,u)\right)du}$$

Please note that the choice of $c(x)$, apart from the conditions for continuity, is quite arbitrary; this is analogous to the $+C$ in the indefinite integral of single variable functions.

You can infer that the constant of integration for indefinite integral of multivariable functions is not as simple as the one-variable case. And, of course, differentiating the indefinite integral makes things even more complicated.

• Thanks... In one of my questions here, the answerer (after writing down the constant of integration) uses indefinite integrals with respect to variable $z$ instead of definite integral. Isn't it better to write definite integrals from $c$ (constant) to $z$; after we have written down the constant of integration??? – Joe Jul 3 '18 at 12:40
• @Joe You pointed out an important mistake. I will try to correct it. – Szeto Jul 3 '18 at 13:09

I think this link might help: Leibniz_integral_rule-Measure_theory_statement