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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?

Mtheorist
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  • Related: https://math.stackexchange.com/questions/2389100/composition-of-the-derivative-of-dirac-delta-with-a-function/2389360#2389360 – md2perpe Aug 27 '17 at 15:59
  • I think that the error in your calculation is that $s'$ might change signs. – md2perpe Aug 27 '17 at 16:14
  • Thank you for the link and comment, may I know why is it that when $s'$ changes sign, the integration measure transforms as $ds=dx/|s'(x)|$? – Mtheorist Aug 28 '17 at 06:58
  • $$I := \int_{-\infty}^{+\infty} f(x) , dx = { x = -y } = \int_{+\infty}^{-\infty} f(-y) , (-dy) = \int_{-\infty}^{+\infty} f(-y) , dy$$ since we also get one minus sign when we swap the limits of the integral. – md2perpe Aug 28 '17 at 16:30
  • 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 $Z = -1$. – md2perpe Aug 28 '17 at 16:34
  • Also, if $s(-\infty) = s(+\infty) = \pm \infty$ then $Z = 0$. – md2perpe Aug 28 '17 at 18:46
  • In the actual lecture by Moore, which can be seen here - http://scgp.stonybrook.edu/video_portal/video.php?id=3078 - , at 28:39, he draws $s(x)$ as a function of $x$, and it looks like a degree-3 polynomial in $x$, with $s(-\infty)=-\infty$ and $s(\infty)=\infty$. Yet his answer is not $Z=1$. It seems like he is splitting the integration regimes, and somehow only evaluating the Gaussian over each regime. – Mtheorist Aug 29 '17 at 15:04
  • Ths $s(x)$ given in the video is just a sketchy example of a function while the formula he gives is generic. – md2perpe Aug 29 '17 at 18:34
  • I can not see why the 1-dimensional case should count (with sign) specifically the number of zeros of $s$. It could equally well count the number of points where $s$ passes any $c \in \mathbb R$. – md2perpe Aug 29 '17 at 20:22
  • Perhaps it is because the main contribution to the Gaussian function comes from the regions where $s(x)=0$? – Mtheorist Aug 30 '17 at 06:12
  • I thought so, too, but cannot verify it mathematically. – md2perpe Aug 30 '17 at 06:21
  • May I know why you feel that it could equally well count the number of points where $s$ passes any $c\in\mathbb{R}$? – Mtheorist Aug 30 '17 at 10:51
  • Letting $s(x) \to s(x) - c$ makes $Z$ be a sum over the zeros of $s(x) - c$, i.e. the solutions to $s(x) = c$. – md2perpe Aug 30 '17 at 11:33
  • But the result may be clearer in more dimensions. – md2perpe Aug 30 '17 at 11:34
  • Moore does discuss the generalization to higher dimensions, the answer being $Z=\sum_{\mathcal{Z}(s)}\textrm{sign}\bigg(\textrm{det}\big(\frac{\partial s ^i}{\partial x ^j}\big)\bigg)$. – Mtheorist Aug 31 '17 at 11:09
  • Yes, I saw that. Have you tried to perform the integral in several dimensions? – md2perpe Aug 31 '17 at 14:43
  • I have tried it, but it seems more confusing than the 1d version. But in Moore's notes he mentions the saddle-point approximation, so I think perhaps he is using this approximation, so that the Gaussian can be treated as a delta function in the 1d case. – Mtheorist Sep 01 '17 at 18:29
  • An answer is here- https://physics.stackexchange.com/questions/353504/tqft-adding-a-q-exact-term-which-is-equal-to-the-action-itself/353534?noredirect=1#comment794610_353534 – Mtheorist Sep 13 '17 at 05:22

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