Is $\int_{0}^{\infty} e^{-(a^2x^2+\frac{b^2}{x^2})}dx=\int_{0}^{\infty}e^{-(b^{2}{X}^2+\frac{a^2}{X^2})}dX$ for arbitrary $a,b$ and fixed range? The problem says,

If $\int_{0}^{\infty} \mathbb{e^{-(a^2x^2+\frac{b^2}{x^2})dx}=\frac{\sqrt{\pi}}{2a}.e^{-2ab}} \longrightarrow(i)$, then prove that $\mathbb{\int_{0}^{\infty}\frac{1}{x^2}e^{-(a^2x^2+\frac{b^2}{x^2})}dx=\frac{\sqrt{\pi}}{2b}.e^{-2ab}}$


If we take $\mathrm{x=\frac{b}{a\mathscr{X}}}$, where $\mathrm{a}, \mathrm{b}$ are any constant, then we have $$\mathbb{\int_{0}^{\infty}\frac{1}{x^2}e^{-(a^2x^2+\frac{b^2}{x^2})}dx} \longrightarrow  (ii)\\=\mathbb{e^{-2ab}}\mathbb{\int_{0}^{\infty}\frac{1}{x^2}e^{-(ax+\frac{b}{x})^2}dx}\\=\mathbb{\frac{-a\ e^{-2ab}}{b}}\mathbb{\int_{\infty}^{0}e^{-(b\mathscr{X}+\frac{a}{\mathscr{X}})^2}d\mathscr{X}}\\=\mathbb{\frac{a}{b}}\mathbb{\int_{0}^{\infty}e^{-(b^{2}\mathscr{X}^2+\frac{a^2}{\mathscr{X}^2})}d\mathscr{X}}\longrightarrow(iii)\\$$


*

*Since $\mathrm{a}, \mathrm{b}$ are any constant and condition $(i) \ \textrm{&} \ (iii)$ look same and the integration range remains same, therefore should we use the value of $(i)$ in $(iii)$ ? If we plug in the value of $(i)$ in $(iii)$, the desired result comes. But is this a correct aprroach to do that?

Any help, explanation is valuable and highly appreciated.

You can "plug i into iii" via the "integration-by-substitution"
$$
\mathscr{X} = x \\
d \mathscr{X} = dx
$$
(i.e., using the chain rule in the simplest possible way); this shows that the integral you're calling "iii" is the same as the integral you're calling "i".
