So I am working through some Integrals and I've come across transformations I have yet to see in calculus, it involves the Heaviside function $H$.
If someone can please explain how the limits of the integral are changing that would be highly appreciated.
$$\int_{-\infty}^{\infty}H\left[\frac{x-a}{b}\right]g(x)dx$$ letting $y=\frac{x-a}{b}$ (Now this part of the integral makes sense since -b or +b would change the limits)
$$=\int_{-\infty sign(b)}^{\infty sign(b)}H\left[y\right]g(by+a)bdy$$ but from here onwards I am unsure of most of the steps (theorems?) that were used to change between the Integrals (The only one I know is the Heaviside result)
$$=sign(b)\int_{-\infty}^{\infty}H\left[y\right]g(by+a)bdy \quad \text{this result I know}$$ $$=sign(b)\int_{0}^{\infty}g(by+a)bdy \quad \text{this result I know}$$ $$=sign(b)\int_{a}^{\infty sign(b)}g(x)dx \quad \text{this result is foreign}$$
How did the $a$ get into the lower limit and why another $sign(b)$ in the upper limit? I am certain a few results are used to arrive at this final one.
The proof continues further but the rest I can follow. It's just the $a$ and $sign(b)$ that have me stumped. It could be a simple result that I have just simply forgotten?
As usual, many thanks for your feedback and help.
If $y$ ranges from $0$ to $\infty$ this means when we make our substitution back to bounds in terms of $x$ that we have $x$ ranges from $a$ (the value of $x$ must be $a$ in order for $y$ to have a value of $0$ so we have the right bounds), $x$ must range from negative infinity to $0$ which means that $sign(b)=-1$, then if it equals $0$ then $x=0$ and therefore $x=a$ or it ranges from $0$ to infinity if $sign(b)=1$.