Let f be a non-negative continuous function, and suppose that $\int_1^{\infty} f(x) \,dx$ converges. Show that $\int_1^{\infty} f(y^2) \,dx$ converges.
I substituted $x=y^2$ into the second equation, which led me to $$\int_1^{\infty} f(y^2) \,dx = \int_1^{\infty} \frac{f(x)}{2x} \,dx$$, which means that $\int_1^{\infty} f(y^2) \,dx$ converges.
However, the answer provided was that $$\int_1^{\infty} f(y^2) \,dx = \int_1^{\infty} \frac{f(x)}{2{\sqrt{x}}} \,dx$$ While the integral still converges, I'm concerned over why my substitution was wrong. I had assumed that $\frac{dx}{dy} = 2x$, hence substituting that into the equation, but I'm not sure how I would get ${2\sqrt{x}}$.
Could someone please help?
The goal of substitution is to replace all occurrences of $y$ in the integrand and differential with expressions of $x$.
So you do not want $y$ to occur anywhere in the right hand side. To do this you replace $y:=\surd x$ and $\mathrm d y:=\mathrm d x/(2\surd x)$ .
$$\begin{align}\int\limits_1^\infty f(y^2)\,\mathrm d y\quad&=\quad\int\limits_1^\infty \dfrac{f(x)}{2\surd x}\,\mathrm d x&\qquad&{\begin{split}y&:=\surd x\\\mathrm d y&:=\mathrm d x/(2\surd x)\end{split}}\end{align}$$