If we have the double integral
$\displaystyle \int_0^\infty \int_0^{x/\sqrt{t}}\exp\bigg (-\frac{y^2}{2}\bigg)dydt$, where $x > 0$
We get inequalities:
$0 \le t < \infty$ and $0 \le y \le \frac{x}{\sqrt{t}} \implies 0 \le t < \infty $ and $0 \le t \le \frac{x^2}{y^2}$
The inner integral will then be $\int_0^{x^2/y^2}\exp(-y^2/2)dt$.
The answer to the change will be:
$\displaystyle \int_0^\infty \int_0^{x/\sqrt{t}}\exp\bigg (-\frac{y^2}{2}\bigg)dydt = \int_0^\infty \int_0^{x^2/y^2}\exp\bigg (-\frac{y^2}{2}\bigg)dtdy $
My question is why does the limit of integration for $y$ become $0$ to $\infty$?
Using the comment by @ancientmathematician :
WLOG let $x = 1$ as $x > 0$. Using the bounding curves $t$-axis, $y-$axis and $ty^2 = x^2$. And conditions $0 \le t < \infty, 0 \le y \le \frac{x}{\sqrt{t}}$
To get the limits of integration for $y$, let $t$ be fixed. Then
$y^2=\frac{1}{t}$ and letting $t \to \infty$ we get $y^2 \to 0$ and letting $t \to 0$ we get $y^2 \to \infty$
And we know $y \ge 0$ from the conditions above thus $0 \le y < \infty$ in the change-of-order.