First of all note that the integrand is an even function, so:
$$I=\int_{0}^{\infty}e^{\large-\frac{1}{2}\left(y^2+\frac{t^2}{y^2}\right)}dy=\frac12\int_{-\infty}^{\infty}e^{\large-\frac{1}{2}\left(y^2+\frac{t^2}{y^2}\right)}\,dy$$
Now with some algebra: $$y^2 +\frac{t^2}{y^2}=\left(y-\frac{t}{y}\right)^2 +2t$$
One could have very well written $$y^2 +\frac{t^2}{y^2}=\left(y+\frac{t}{y}\right)^2 -2t$$
but the first one, also prepares to use this Glasser's Master Theorem.
$$I=\frac12\int_{-\infty}^{\infty}e^{\large-\frac{1}{2}\left(\color{red}{\left(y-\frac{t}{y}\right)}^2 +2t\right)}\,dy=\frac12\int_{-\infty}^{\infty} e^{\large-\frac12 \left( \color{red}y^2 +2t\right)}dy$$
$$=\frac12 e^{-t} \int_{-\infty}^\infty e^{\large -\frac12 y^2}dy=\sqrt{\frac{\pi}{2}}e^{-t},\ t>0$$
Similarly for $t<0$.
Also note that the last integrand is just $\mathcal{N}(0,1)$ for the Normal Distribution, as seen from here that $$\Phi(x)=\frac{1}{\sqrt {2\pi}} \int_{-\infty }^x e^{\large-\frac12 y^2}dy$$
and well since it's a pdf we must must have that the total area equals to $1$ or $\lim\limits_{x\to \infty } \Phi(x)=1$.
For a proof just substitute $\frac12 y^2 =u^2$ and use this.
