Q:

Why does substitution work in integrals

Let's say I have this integral:
$$\int_0^\infty e^{-t} \, dt$$
And I make the substitution:
$$t = nu$$
Then why I can say that:
$$dt = n\,du$$
and then put this into my integral like this:
$$\int_0^\infty e^{-nu}n\,du$$
What's happening in the background that allow this to be done?
I'm asking this because I don't feel confortable threating diferencial operators as fractions and I don't know why this can be done.

A:

It's just a way of looking at the chain rule.
The chain rule is differentiation by substitution.
One can write $$\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x),$$ or one can look at
$$
\frac{d}{dx} f(g(x))
$$
and then do this substitution:
$$
u = g(x),\qquad \frac{du}{dx} = g'(x).
$$
Then one writes
$$
\frac{d}{dx} f(g(x)) = \frac{d}{dx} f(u) = \frac{df(u)}{dx} = \frac{df(u)}{du}\cdot\frac{du}{dx} = f'(u)\cdot g'(x) = f'(g(x))\, g'(x).
$$
In the same way, when one sees
$$
\int f'(g(x)) g'(x) \,dx,
$$
one does the substitution
$$
u=g(x),\qquad \frac{du}{dx} = g'(x),\qquad du = g'(x)\,dx.
$$
Then one has
$$
\int f'(g(x)) g'(x) \,dx = \int f'(u)\,du = f(u)+C = f(g(x))+C.
$$
So integration by substitution is the chain rule in reverse, just as integration by parts is the product rule in reverse.

