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I would like to know how partial derivatives under the integral work, and whether the obvious expected translation of the Leibniz integral rule is correct. I am asking this for the purpose as solving the same question I asked here. But I thought this more basic detail needed asking here in a separate question.

src1 claims an answer in the case with constant integral bounds.

src2 and src3 give answers but somehow in the process of answering, the problem got switched to the regular-d derivative, which defeats the whole point of my question.

src2

src3 (skip to Fredrik's responses)

The Leibniz integral rule states $$ \frac{d}{dx} \int\limits_{a(x)}^{b(x)} f(x,t) dt = f(x,b(x))\frac{d}{dx}b(x) - f(x,a(x))\frac{d}{dx}a(x) + \int\limits_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x,t) dt $$ To me, the obvious translation to a partial derivative would be to simply replace all regular-d derivatives with partial-$\partial$ derivatives. $$ \frac{\partial}{\partial x} \int\limits_{a(x)}^{b(x)} f(x,t) dt =^{(??)} f(x,b(x))\frac{\partial}{\partial x}b(x) - f(x,a(x))\frac{\partial}{\partial x}a(x) + \int\limits_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x,t) dt $$ Is this correct? Is it even necessary? Since partial derivatives keep everything else constant, maybe it is always correct to say? $$ \frac{\partial}{\partial x} \int\limits_{a(x)}^{b(x)} f(x,t) dt =^{(??)} \int\limits_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x,t) dt $$

The $\infty$-dimension wrinkle is,

what if x is actually a continuous differentiable (and perhaps even monotonic) function of another variable y, which $f$ and the integral bounds may depend on? This would come from calculus-of-variations problems. So, $$x = g(y)$$ $$ \frac{\partial}{\partial x} \int\limits_{a(x(y),y)}^{b(x(y),y)} f(x(y),y,t) dt =^{(??)} f(x,b(x(y),y))\frac{\partial}{\partial x}b(x(y),y) - f(x,a(x(y),y))\frac{\partial}{\partial x}a(x(y),y) + \int\limits_{a(x(y),y)}^{b(x(y),y)} \frac{\partial}{\partial x} f(x(y),y,t) dt $$

This source also makes a useful statement, but it also has to do with regular-d derivatives src4

Important follow-up - the constant integrand case (not 'continuous'?)

So, the answer is that this is actually necesarry for the Liebniz Integral rule to be correct. See Abheziko's response to other q. It must be that $$ \frac{\partial}{\partial x} \int\limits_{a(x)}^{b(x)} f(x,t) dt = \int\limits_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x,t) dt $$

However, I think this assumes that f is continuous and that $\frac{df}{dt}$ is continuous?

So, what if $f(x,t) = t$ (or any other constant wrt x)?

Suddenly we have a break down of consistency. For example:

$$ \frac{\partial}{\partial x} \int\limits_{x}^{2x} t dt = \\ = \frac{\partial}{\partial x}(2x^2 - \frac{x}{2}) \\ = 4x- \frac{1}{2} \\ $$ Alternatively, by simply bringing the partial inside the integral... $$ \int\limits_{x}^{2x} \frac{\partial}{\partial x} t dt = \\ \int\limits_{x}^{2x} 0 dt = \\ 0 $$ and $4x-\frac{1}{2}\neq 0$ unless $x=\frac{-1}{8}$.

What assumption is being broken here?

Tejovan
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  • No, partial derivatives commute with integration, while the boundary terms of Leibniz integral rule come from the chain rule due to total differentiation (see this answer of mine for a brief example). – Abezhiko Jan 17 '25 at 17:45
  • @Abezhiko, perhaps you forgot the link? But, from what you wrote here, I interpret that you are saying that the following is correct? $ \frac{\partial}{\partial x} \int\limits_{a(x)}^{b(x)} f(x,t) dt =^{(??)} \int\limits_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x,t) dt $ ? And this is true even when x is a function of an additional variable that the integral and $f$ depend on? – Tejovan Jan 17 '25 at 18:12
  • That's so stupid of me... Here is the link : https://math.stackexchange.com/questions/4951615/partial-term-in-leibniz-rule/4951634#4951634. And yes, it is. – Abezhiko Jan 17 '25 at 18:33
  • excellent, thank you, I see now. I'll add that as the official answer to this later today. If you'd like, perhaps you will also have such an easy answer to my other related question about partial derivatives ? – Tejovan Jan 17 '25 at 19:03
  • @Abezhiko, I have added a follow about the special case when the integrand is a constant (at least with respect to the derivative-variable). As it turns out this is the case in the problem I am trying to solve. How can I handle this? – Tejovan Jan 28 '25 at 21:35
  • I am not sure about this simplification you propose at the end which seems to drop the $\partial_x b(x)$ and $\partial_x a(x)$ terms entirely. Indeed, your example shows that the simplified formula you propose cannot work. Unless I am misunderstanding, it seems the Leibniz formula should be the "obvious translation" as you propose. Ie. Shouldn't $\partial_x b(x)=b'(x)$? To belabor the point, if you believe $\partial_x b(x) g(y) = b'(x) g(y)$, then set $g(y)=1$. – Matt Jan 29 '25 at 02:24
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    The notation $$ \frac{\partial}{\partial x} \int_{a(\color{red}{\boldsymbol x})}^{b(\color{lightgreen}{\boldsymbol{x}})} f(\color{lightblue}{\boldsymbol{x}},t), dt $$ is more than ambiguous because you are not saying w.r.t. which of the three $x$-es ($\color{lightgreen}{\boldsymbol{x}},\color{red}{\boldsymbol{x}},\color{lightblue}{\boldsymbol{x}}$) you are differentiating. Starting with a correct expression for $\frac d{dx}$ and replacing every$\frac d{dx}$ by $\frac\partial{\partial x}$ is naive symbol pushing but not math. Know your weapon. – Kurt G. Jan 29 '25 at 06:32
  • @Matt, See Abheziko's response to this q. – Tejovan Jan 29 '25 at 21:17
  • @KurtG. , so you are saying that each of the three x's, in each bound and in the integrand, need to actually be considered as different variables? If it were a total derivative, then those three x's would be the same variable/parameter? Perhaps that is actually getting right to the point of how partial derivs are different than total derivs, and why this Abhezikos answer is correct? But then, why is it that the correctness breaks down when there is even one term in the integrand that does not depend on the derivative-parameter-x? – Tejovan Jan 29 '25 at 21:24
  • This is not exactly what I am saying and I don't have trouble with Abezhiko's answer per se. I am saying that if a notation is ambiguous you have to find ways of telling your readers what you are doing. – Kurt G. Jan 29 '25 at 21:52
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    An acceptable unambiguous notation would be to define $$ F(x,y,z)=\int_{a(y)}^{b(z)} f(x,t), dt $$ and write $$ \frac{\partial}{\partial x}F(x,y,z)\Bigg|_{y=x,z=x},. $$ Or: $F(x_1,x_2,x_3)$ for the former $F(x,y,z)$ and write $\partial_1F(x,x,x),.$ – Kurt G. Jan 30 '25 at 09:27

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I will develop Kurt G.'s comments. Initially, you face a trivariate function, namely $$ F(u,v,w) = \int_u^v f(w,t) \,\mathrm{d}t, $$ whose partial derivatives are given by $$ \frac{\partial F}{\partial u} = -f(w,u), \quad \frac{\partial F}{\partial v} = f(w,v) \quad\&\quad \frac{\partial F}{\partial w} = \int_u^v \frac{\partial f}{\partial w}(w,t) \,\mathrm{d}t, \tag{$*$} $$ hence $$ \mathrm{d}F(u,v,w) = -f(w,u)\,\mathrm{d}u + f(w,v)\,\mathrm{d}v + \left(\int_u^v \frac{\partial f}{\partial w}(w,t) \,\mathrm{d}t\right)\mathrm{d}w. $$ Now, the central point in your question and the crux of your confusion lies in the role of $x$ in the partial derivative $\partial_x$ you want to consider. Indeed, three cases are possible.

1° If the said variable $x$ refers to one of the previous variables above, then $\partial_xF$ is given by one of the expressions in $(*)$.

2° If $x$ is a new variable and $(u,v,w)$ are all independent of $x$, then $\partial_xF = 0$.

3° If $x$ is a fourth variable, but at least one of the other variables depends on it, then you still have a vanishing partial derivative, i.e. $\partial_xF = 0$, while the associated total derivative need to be computed with the help of the chain rule, as it follows : $$ \frac{\mathrm{d}F}{\mathrm{d}x} = -f(w,u)\frac{\mathrm{d}u}{\mathrm{d}x} + f(w,v)\frac{\mathrm{d}v}{\mathrm{d}x} + \left(\int_u^v \frac{\partial f}{\partial w}(w,t) \,\mathrm{d}t\right)\frac{\mathrm{d}w}{\mathrm{d}x} $$

Abezhiko
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