There should be a $\exp(-v^2)$ istead than $\exp(-v)$ in your last integral.
Define $I(y)=\int_0^\infty e^{-v^2} \frac{\sin(yv)}{v} \,dv$. We have $I(0)=0$ and $$ I'(y)=\int_0^\infty e^{-v^2} \cos(yv) \,dv. $$
We prove that $I'(y)\overset{(*)}{=}\frac{\sqrt{\pi}}{2}e^{-y^2/4}$ and we conclude that $I(y)=\frac{\pi}{2} \text{erf}(y/2)$ by the definition of $\text{erf}(x)$ as the integral of such function that vanishes at $x=0$.
We prove $(*)$ using the Feynman's method once again. We have $I'(0)=\int_0^\infty e^{-v^2}\,dv =\frac{\sqrt{\pi}}{2}$ and
$$ I''(y)=-\int_0^\infty e^{-v^2} v \sin(yv) \,dv = \left[ \frac{e^{-v^2}}{2}\sin(yv) \right]_{v=0}^{v=\infty} - \frac{y}{2}\int_0^\infty e^{-v^2}\cos(yv) \,dv = - \frac{y}{2} I'(y). $$ The solution is indeed $I'(y)=\frac{\sqrt{\pi}}{2} e^{-y^2/4} $.
Another way to prove $(*)$. Use Euler's formula for $\cos(yx)$, complete the square and finally change variable linearly (note that the integral will be on an horizontal line different from the real, but this is not a problem since the integrand is holomorphic).