\begin{equation} \text{Let $\hspace{3mm}$ }f(t) = 2\int_{b}^{\infty} \sqrt{\frac{1}{2\pi t}}e^{-x^2/2t}dx. \end{equation} I found that this integral can be written with change of variables can be written as, \begin{equation} f(t) = \sqrt{\frac{2}{\pi}}\int_{|b|t^{-1/2}}^{\infty} e^{-x^2/2}dx. \end{equation} I guess the transformation used is $u(x)^2 = \frac{x^2}{t}$ so that $|u(x) | = \frac{|x|}{\sqrt{t}}$ . Is $u(x)$ 1-1? how we calculate the derivative of $u$ with respect to $x$. Can anyone explain analytically?
2026-04-05 01:58:54.1775354334
Change of variables - integrals
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Use the transformation $u=x/\sqrt{t}$ instead. What you have, by the way, is just a normal (cumulative) distribution function.