2

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 $(1)$.

DMcMor
  • 10,059
Lac
  • 761
  • Formatting note: you can use \tag1, \tag2 etc. to number equations. – saulspatz Feb 19 '21 at 14:05
  • 1
    See https://math.stackexchange.com/questions/704186/indefinite-double-integral. – Martín-Blas Pérez Pinilla Feb 19 '21 at 16:13
  • 1
    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 $$f_{yx}=F(x,y).$$ – Allawonder Feb 19 '21 at 23:05
  • @Allawonder Hello, thank you for the correction. Just to make sure, how is H and F in your equations related to the initial G? – Lac Feb 20 '21 at 04:12
  • @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 $$f_{yx}=G(x, y).$$ – Allawonder Feb 20 '21 at 08:29

1 Answers1

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.

Alexander Conrad
  • 1,798