Suppose we have $f(x) = \int G(x_{0},x_{1},...)\,dx_{0}\,dx_{1}\dotsb dx_{n}$ When can we affirm that $$df = G(x_{0},x_{1},...)? \tag1$$ Basically, I am having trouble to understand how to deal with differentials of functions, intuitively I thought that we can do that:

$$\delta f(x) = \int \sum\left(\frac{\partial G}{\partial dx_{i}}\right)\,dx_{0}\,dx_{1} \dotsb dx_{n} \tag2$$

But I am not sure how $(2)$ reduces to $(1)$.

  • $\begingroup$ Formatting note: you can use \tag1, \tag2 etc. to number equations. $\endgroup$
    – saulspatz
    Feb 19, 2021 at 14:05
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    $\begingroup$ See math.stackexchange.com/questions/704186/…. $\endgroup$ Feb 19, 2021 at 16:13
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    $\begingroup$ You cannot get $(1)$ from the original equation. In fact the original equation is somewhat odd since LHS has a single independent variable but RHS has $n+1.$ Anyway, assuming I understand what you mean, let us consider the two variable case for simplicity. So let $$f(x,y)=\iint G(x,y) dx dy.$$ Then we have that $f_y=\int H(x,y)dx,$ so that $$f_{yx}=F(x,y).$$ $\endgroup$
    – Allawonder
    Feb 19, 2021 at 23:05
  • $\begingroup$ @Allawonder Hello, thank you for the correction. Just to make sure, how is H and F in your equations related to the initial G? $\endgroup$
    – Lac
    Feb 20, 2021 at 4:12
  • $\begingroup$ @LSS Looking at the equations again, I didn't need to introduce the new symbols $H, F.$ It suffices to note that if $$f(x, y)=\iint G(x, y)dx dy,$$ then differentiating twice gives $$f_{yx}=G(x, y).$$ $\endgroup$
    – Allawonder
    Feb 20, 2021 at 8:29

1 Answer 1


In some notations this is true; for example, the volume element can be written $dV=A \ du \ dv$ for some $A$. In the case of your integral, the proper way of dealing with it is probably $df=Gdx_0dx_1\cdots dx_{n-1}dx_n$, for which its usefulness depends on the context.


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