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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.

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1 Answer 1

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$$ \begin{array}{rcl} I_2 &=& \displaystyle \int_{-1}^{1} \exp\left(-\left(\frac{a\nu_b-y}{\sqrt{2}\,\sigma_v}\right)^2\right) \mathrm{d}\nu_b \\ &=& \displaystyle \int_{-a}^{a} \exp\left(-\left(\frac{\nu-y}{\sqrt{2}\,\sigma_v}\right)^2\right) \frac{\mathrm{d}\nu}{a} \\ &=& \displaystyle \frac{I_1}{a} \end{array} $$ via the change of variable $\nu = a\nu_b$.

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  • $\begingroup$ 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 $\nu_{\text{b}}\mapsto \nu$. $\endgroup$
    – matteogost
    May 13, 2023 at 14:48

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