Timeline for Change of variables in a Gaussian integral in matrix form
Current License: CC BY-SA 4.0
14 events
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| Jan 16, 2021 at 22:41 | vote | accept | Rafael | ||
| Jan 16, 2021 at 22:41 | |||||
| Jan 16, 2021 at 22:40 | comment | added | Rafael | Thank you very much, I guess this is the integral I was looking for | |
| Jan 16, 2021 at 10:38 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 16, 2021 at 9:47 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 15, 2021 at 20:33 | comment | added | Carlo Beenakker | @IosifPinelis --- the same identity applies there as well, you can use it to integrate out one of the $y_j$'s, for example $y_1$: $$\mathop{\idotsint}\delta\left(x-{\textstyle{\sum_{j=1}^{k}}y_j}\right) e^{-N^2r\sum_{j=1}^{k}y_j^2} dy_1\dots dy_{k}= $$ $$\mathop{\idotsint}e^{-N^2r(x-\sum_{j=2}^k y_j)^2}e^{-N^2r\sum_{j=2}^{k}y_j^2}\,dy_2\dots dy_{k}.$$ | |
| Jan 15, 2021 at 18:50 | comment | added | Iosif Pinelis | @CarloBeenakker : I understand that the ordinary integral $\int_{-\infty}^\infty f(x)\delta(x-u)dx$ is defined as $f(u)$. But how is your first $k$-fold integral defined? | |
| Jan 15, 2021 at 8:05 | comment | added | Carlo Beenakker | @Rafael --- you asked "how do you define the delta function measure?" --- you can take a look on Wikipedia --- the only property of the delta function is need is the identity $\int_{-\infty}^\infty f(x)\delta(x-u)dx=f(u)$. | |
| Jan 14, 2021 at 21:58 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 14, 2021 at 14:50 | comment | added | Rafael | I'm going to check my iterations for this mistake! Thanks! It is good to know, that method 1 works. Do you have any idea what went wrong with method 2? | |
| Jan 14, 2021 at 11:52 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 14, 2021 at 11:00 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 14, 2021 at 10:48 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 14, 2021 at 8:16 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 14, 2021 at 8:09 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |