Consider the following integral
\begin{equation*}
I_1\triangleq\int_{-a}^{a} \exp\left[-\left(\frac{\nu-y}{\sqrt{2}\sigma_v}\right)^2\right]\text{ d}\nu
\end{equation*}
with the change of variable
\begin{equation*}
t\triangleq\frac{\nu-y}{\sqrt{2}\sigma_v}
\end{equation*}
follows
\begin{equation*}
I_1=
\sqrt{2}\sigma_v \int_{\frac{-a-y}{\sqrt{2}\sigma_v}}^{\frac{a-y}{\sqrt{2}\sigma_v}} \exp(-t^2)\text{ d}t= \sqrt{\frac{\pi}{2}}\sigma_v \left[\text{erf}\left(\frac{a-y}{\sqrt{2}\sigma_v}\right)-\text{erf}\left(\frac{-a-y}{\sqrt{2}\sigma_v}\right)\right]
\end{equation*}
now consider the following integral
\begin{equation*}
I_2\triangleq\int_{-1}^{1} \exp\left[-\left(\frac{a\nu_{\text{b}}-y}{\sqrt{2}\sigma_v}\right)^2\right]\text{ d}\nu_{\text{b}}
\end{equation*}
with the change of variable
\begin{equation*}
t\triangleq\frac{a\nu_{\text{b}}-y}{\sqrt{2}\sigma_v}
\end{equation*}
follows
\begin{equation*}
I_2=
\frac{\sqrt{2}\sigma_v}{a} \int_{\frac{-a-y}{\sqrt{2}\sigma_v}}^{\frac{a-y}{\sqrt{2}\sigma_v}} \exp(-t^2)\text{ d}t= \sqrt{\frac{\pi}{2}}\frac{\sigma_v}{a} \left[\text{erf}\left(\frac{a-y}{\sqrt{2}\sigma_v}\right)-\text{erf}\left(\frac{-a-y}{\sqrt{2}\sigma_v}\right)\right]
\end{equation*}
and so, $I_1\neq I_2$. The problem is that I cannot see if I've made some mistake somewhere or if actually $I_1\neq I_2$ is true. For me $I_1$ and $I_2$ are the same integral.
Best Answer
$$ \begin{array}{rcl} I_2 &=& \displaystyle \int_{-1}^{1} \exp\left(-\left(\frac{a\nu_b-y}{\sqrt{2}\,\sigma_v}\right)^2\right) \mathrm{d}\nu_b \\ &=& \displaystyle \int_{-a}^{a} \exp\left(-\left(\frac{\nu-y}{\sqrt{2}\,\sigma_v}\right)^2\right) \frac{\mathrm{d}\nu}{a} \\ &=& \displaystyle \frac{I_1}{a} \end{array} $$ via the change of variable $\nu = a\nu_b$.