How can I prove this formula $\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right)\mathrm{d}x=\frac{1}{a}\int_0^\infty f(x^2+2ab)\mathrm{d}x$? While solving this trigonometric integral :
$$\int_0^\infty \sin\left(a^2x^2+\frac{b^2}{x^2}\right)\mathrm{d}x$$
I came across this formula :
$$\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right)\mathrm{d}x=\frac{1}{a}\int_0^\infty f(x^2+2ab)\mathrm{d}x$$
And I'm wondering if I can prove it, but I hadn't any idea, because it seems like it depends on the properties of the function $f$.

With $a,\>b>0$
\begin{align}
\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right){d}x 
\overset{x\to \frac b{ax}}= &
\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right) \frac b{ax^2} dx\\
= & \frac12 \int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right) \left(1+\frac b{ax^2}\right)  dx\\
 = & \frac1{2a}\int_0^\infty f\left( (ax-\frac bx)^2+2ab\right) d\left(ax-\frac b{x}\right)\\
=& \frac1{2a}\int_{-\infty}^\infty f(x^2+2ab)dx\\
=&\frac1{a}\int_{0}^\infty f(x^2+2ab)dx
\end{align}
