Is this identity containing the Gaussian Integral of any use?

  • #1
MevsEinstein
124
36
TL;DR Summary
What the title says
I found this identity: ##x\int e^{-x^2} dx - \int \int e^{-x^2} dx dx = e^{-x^2}/2## by solving the integral of ##x*e^{-x^2}## and then finding its integration-by-parts equivalent. Is this identity useful at all?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
MevsEinstein said:
Summary:: What the title says

I found this identity: ##x\int e^{-x^2} dx - \int \int e^{-x^2} dx dx = e^{-x^2}/2## by solving the integral of ##x*e^{-x^2}## and then finding its integration-by-parts equivalent. Is this identity useful at all?
IMO, no. Unless I'm missing something, this looks similar to the integral ##\int_{-\infty}^\infty e^{-x^2}dx## is evaluated. IOW, by instead looking at the double integral. I'd bet this technique is in most calculus textbooks, although usually as the integral ##\int_{-\infty}^\infty \int_{-\infty}^\infty e^{(-x^2 - y^2)/2}dy dx##.
 
Last edited:
  • Like
Likes vanhees71
  • #3
Fix notation. Using 'x' too much. For example dxdx?
 
  • Like
Likes PhDeezNutz
  • #4
Mark44 said:
IMO, no. Unless I'm missing something, this looks similar to the integral ##\int_{-\infty}^\infty e^{-x^2}dx## is evaluated. IOW, by instead looking at the double integral. I'd bet this technique is in most calculus textbooks, although usually as the integral ##\int_{-\infty}^\infty \int_{-\infty}^\infty e^{(-x^2 - y^2)/2}dy dx##.
There aren't any infinities in the formula.
 
  • #5
mathman said:
Fix notation. Using 'x' too much. For example dxdx?
I don't know how to fix that problem. I'm still 13.
 
  • #6
MevsEinstein said:
I don't know how to fix that problem. I'm still 13.
Use different letters for different things.
 
  • Like
Likes Vanadium 50
  • #7
MevsEinstein said:
I found this identity: ##x\int e^{-x^2} dx - \int \int e^{-x^2} dx dx = e^{-x^2}/2## by solving the integral of ##x*e^{-x^2}## and then finding its integration-by-parts equivalent. Is this identity useful at all?
Should read
##x\int e^{-x^2} dx - \int \int e^{-y^2} dydx = -e^{-x^2}/2## looks like also a sign error.
These "identities" can be generalized to many functions, its just partial integration and change of variables, for instance consider xsin(x2)
 
  • #8
$$\int xe^{-x^2}dx=-e^{-x^2}/2$$.
 
  • Like
Likes malawi_glenn
  • #9
drmalawi said:
looks like also a sign error.
I couldn't edit the OP.
 

Similar threads

Back
Top