= Solution
For positive standard deviations $a,b$, substitute the two normal densities:
$$
f_V(v)=\frac1{\pi ab}\int_0^\infty u\exp\left[-\frac{u^2}{2}\left(a^{-2}+v^2b^{-2}\right)\right]du.
$$
The integral is $1/(a^{-2}+v^2b^{-2})$, by substituting the exponent's quadratic argument. Thus, with $c=b/a$,
$$
\boxed{f_V(v)=\frac{ab}{\pi(b^2+a^2v^2)}=\frac c{\pi(c^2+v^2)}.}
$$
This is a centered <Cauchy distribution> of scale $c$. Since $P(X=0)=0$, the desired probability is
$$
\boxed{P(|Y|<|X|)=P(|V|<1)
=\frac2\pi\arctan\frac1c=\frac2\pi\arctan\frac ab.}
$$
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