$de^{-x^2}$ as an integration variable

Question

I found this expression:

$$\int_0^\infty e^{-x^2}dx=-\int_0^\infty x de^{-x^2}$$

It's the first time I see an integration variable like $de^{-x^2}$. It is even legit to use it? What rule I have to follow to transform the first integral to the second?

Answer 1

Observe that by the chain rule : $$\frac {de^{-x^2}}{dx}=e^{-x^2}(-2x)$$ Multiply by $dx$ $$de^{-x^2}=-2xe^{-x^2}dx$$ $$\int de^{-x^2}=-2\int xe^{-x^2}dx$$ Therefore : $$\int e^{-x^2}dx=- \frac 12\int \frac {d(e^{-x^2})}{x}$$ Not what you have

Answer 2

You could consider this to be an almost chain rule like analog to integration. Consider,

$\,\frac{de^{-x^2}}{dx} = -2xe^{-x^2}$

so , $\,de^{-x^2} = -2xe^{-x^2} {dx} $

So the integral becomes, $\,-\large{\int_0^\infty} x.-2xe^{-x^2} {dx} $

It can now be integrated easily.

Answer 3

This can be viewed at least a couple of different ways.

1. As a Riemann-Stieltjes integral.

2. As the substitution $u=e^{-x^2}$ giving $-\int_0^\infty x\,\mathrm{d}e^{-x^2}=-\int_1^0\sqrt{-\log(u)}\,\mathrm{d}u$

The rule to get from the first to the second is Integration by Parts.