Timeline for How to prove an equality involving Laguerre polynomials

11 events

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Sep 21, 2022 at 18:42	comment	added	Pietro Majer		Well, for the $n$-dimensional Lebesgue measure and $f\in \mathcal L^1(\mathcal R^n)$ we do use the notation $\int_{\mathbb R^n}f(x)dx$, where the dimension of the measure is clear from the domain rather than from "$dx$". So $\int_{\mathbb C} f(z)dz$ is not out of this word; I don't see a big danger of confusion with a path integral $\int_{\gamma} f(z)dz$.
Sep 21, 2022 at 18:31	history	edited	zoran Vicovic	CC BY-SA 4.0	added 209 characters in body
Sep 21, 2022 at 18:16	history	edited	zoran Vicovic	CC BY-SA 4.0	added 242 characters in body
Sep 18, 2022 at 15:14	history	edited	LSpice	CC BY-SA 4.0	More informative title; removed "Thank you"
Sep 18, 2022 at 14:37	comment	added	Daniele Tampieri		Since you are performing a volume integration in $\Bbb C$, you should use the notation $\mathrm{d}z\wedge\mathrm{d}\bar{z}$ which is a multiple of $\mathrm{d}x \mathrm{d} y$.
Sep 18, 2022 at 13:17	comment	added	Fedor Petrov		Usually $dz$ is used for contour integration, where it has another sense
Sep 18, 2022 at 11:35	comment	added	zoran Vicovic		$z=(x,y)\in\Bbb R^2$ so $dz=dxdy$ @Fedor Petrov.
Sep 18, 2022 at 11:27	comment	added	Fedor Petrov		By $dz$ you mean $dxdy$, where $z=x+iy$?
Sep 18, 2022 at 10:58	comment	added	zoran Vicovic		it is $z$ not $w$ see again. Thank you
Sep 18, 2022 at 10:56	history	edited	zoran Vicovic	CC BY-SA 4.0	edited body
Sep 18, 2022 at 10:42	history	asked	zoran Vicovic	CC BY-SA 4.0