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).

There should be a $\exp(-v^2)$ instead of $\exp(-v)$ 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 $I'(y)\overset{(*)}{=}\frac{\sqrt{\pi}}{2}e^{-y^2/4}$ and we conclude $I(y)=\frac{\pi}{2} \text{erf}(y/2)$ by the definition of $\text{erf}(x)$ as the integral of such function that vanishes at $x=0$.

We prove $(*)$ with 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 allowed since the integrand is holomorphic).