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In these notes (http://www.physics.rutgers.edu/~gmoore/SCGP-FourManifoldsNotes-2017.pdf) the following integral is given on page 9- $$ Z=\int^\infty_{-\infty} \frac{dx}{\sqrt{2\pi}}s'(x) e^{-\frac{1}{2}s(x)^2}, $$ and the solution is said to be $$ Z=\sum_{\mathcal{Z}(s)}\frac{s'(x_l)}{|s'(x_l)|} $$ where $\mathcal{Z}(s)=\{x_l:s(x_l)=0\}$. It is claimed that this solution can be found via change of variables.

My attempt is as follows. $s'(x)dx =s(x)$, so we have $$ Z=\int^\infty_{-\infty} \frac{ds}{\sqrt{2\pi}} e^{-\frac{1}{2}s(x)^2}, $$ which would give us $$ Z=1, $$ which is not correct. Where have I gone wrong, and what is the correct solution?

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  • $\begingroup$ Related: math.stackexchange.com/questions/2389100/… $\endgroup$ – md2perpe Aug 27 '17 at 15:59
  • $\begingroup$ I think that the error in your calculation is that $s'$ might change signs. $\endgroup$ – md2perpe Aug 27 '17 at 16:14
  • $\begingroup$ Thank you for the link and comment, may I know why is it that when $s'$ changes sign, the integration measure transforms as $ds=dx/|s'(x)|$? $\endgroup$ – Mtheorist Aug 28 '17 at 6:58
  • $\begingroup$ $$I := \int_{-\infty}^{+\infty} f(x) \, dx = \{ x = -y \} = \int_{+\infty}^{-\infty} f(-y) \, (-dy) = \int_{-\infty}^{+\infty} f(-y) \, dy$$ since we also get one minus sign when we swap the limits of the integral. $\endgroup$ – md2perpe Aug 28 '17 at 16:30
  • $\begingroup$ By the way, your result $Z=1$ is correct if $s(-\infty) = -\infty$ and $s(+\infty) = +\infty$. If you instead take $s(-\infty) = +\infty$ and $s(+\infty) = -\infty$ (e.g. $s(x) = -x$) you get $Z = -1$. $\endgroup$ – md2perpe Aug 28 '17 at 16:34

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