What is the rigor behind U subsitution?

$$\int f(g(x)) dx = \int \frac {f(u)}{u'} du$$

requires that

$$\int f(x) dx = \int f(x) \cdot dx$$

but dx just represents the variable that F(x) +c is a function of. So why is it legal for dx to be treated algebriacally?

I tried investigating this property by using riemann summation:

$$\lim\limits_{n \to \infty}(\sum_{k=1}^n f(\frac{kx}{n} )\frac{x}{n}) = \int_0^x f(t)dt= (F(x)+c)-(F(0)+c)$$

and so you can define

$$\lim\limits_{n \to \infty} (\sum_{k=1}^n f(\frac{kx}{n} )\frac{x}{n})+F(0)+c) = \int f(x) dx$$

you can write $$\frac{x}{n} = dx$$ and $$k\frac{x}{n}=kdx= x_k$$

then you have

$$\lim\limits_{n \to \infty} (\sum_{k=1}^n f(x_k)dx)+F(0)+c) = \int f(x) dx$$

since dx is being written to be multiplied by the series, then you can define dx in different terms to aquire an integral in terms of other variables.

but that exists only in abstraction. I can't quite sufficiently complete the task of doing so.

• Glados.Perhaps of interest: en.m.wikipedia.org/wiki/Integration_by_substitution Mar 29 '19 at 10:49
• You might find this post interesting: math.stackexchange.com/questions/3114746 Mar 29 '19 at 10:52
• Are you talking about definite or indefinite integration? For indefinite integrals (antiderivatives), it nothing but the chain rule. Mar 29 '19 at 10:53
• I appreciate it but I am more looking in to rigor rather than intuition. Mar 29 '19 at 10:54

Let's prove that $$\int_a^b h(g(x)) g^\prime(x) dx=\int_{g(a)}^{g(b)} h(u) du$$ for $$g$$ monotonic on $$[a,\,b]$$ with $$a. Let $$H$$ denote an antiderivative of $$h$$, without loss of generality satisfying $$H(a)=0$$. Then the right-hand side of the putative result is $$H(g(b))-H(g(a))$$. Differentiating this with respect to $$b$$ gives $$h(g(b)) g^\prime(b)$$, so $$H(g(b))-H(g(a))=\int_a^b h(g(x)) g^\prime(x) dx$$ as required.
• @GLaDOS An indefinite integral just means "your favourite antiderivative $+C$", so if substitution works for definite integrals it also does for indefinite ones. For example, $$\int_a^b 3x^2\sin x^3 dx=\int_{a^3}^{b^3}\sin udu\implies\int 3x^2\sin x^3 dx=\int\sin udu.$$
• @GLaDOS That's a special case of the fundamental theorem of calculus, viz. $f(b)=\frac{d}{db}\int_a^b f(x)dx$.