# Troubles with a change of integration variable

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.

$$\begin{array}{rcl} I_2 &=& \displaystyle \int_{-1}^{1} \exp\left(-\left(\frac{a\nu_b-y}{\sqrt{2}\,\sigma_v}\right)^2\right) \mathrm{d}\nu_b \\ &=& \displaystyle \int_{-a}^{a} \exp\left(-\left(\frac{\nu-y}{\sqrt{2}\,\sigma_v}\right)^2\right) \frac{\mathrm{d}\nu}{a} \\ &=& \displaystyle \frac{I_1}{a} \end{array}$$ via the change of variable $$\nu = a\nu_b$$.
• Yes, the two integrals are not the same. $I_2$ becomes equal to $I_1$ only by introducing the absolute value of the determinant of the Jacobian of the map $\nu_{\text{b}}\mapsto \nu$. May 13, 2023 at 14:48