Change of variable in integral. Why is it performed this way in the below example?

27 Views Asked by At

On a book, I read, for a certain $y\in\mathbb{R}$:

given the integral $\displaystyle\int\limits_{-y}^{+\infty}f(x)dx$,
performing the substitution $x=-u$ yields: $$\displaystyle\int\limits_{-\infty}^{y}f(-u)du$$

As to the bounds of integration, substitution is ok to me. What sounds strange is the differential. That is, I would expect:$$\frac{du}{dx}=-1$$ hence $$dx=-du$$ hence, in conclusion $$\displaystyle\int\limits_{-\infty}^{y}-f(-u)du$$ and not $$\displaystyle\int\limits_{-\infty}^{y}f(-u)du$$

Could you please clarify this doubt? I'm getting crazy since I do not know if I am making some silly mistake or if I am not realizing some basic fact or if book is wrong.

1

There are 1 best solutions below

6
On BEST ANSWER

$$\int_{-y}^\infty f(x)dx=\int_\color{blue}y^\color{red}{-\infty}-f(-x)dx\\ =\int_\color{red}{-\infty}^\color{blue}yf(-x)dx$$