- #1
Emspak
- 243
- 1
I was trying to understand something about this:
[tex]\int_{-\infty}^\infty \mathrm{e}^{-2x^2}\,\mathrm{d}x[/tex]
According to the solution given, this can be set up where you say:
let
[tex]I = \Big(\int_{-\infty}^\infty \mathrm{e}^{-2x}\,\mathrm{d}x\Big)[/tex]
and
[tex]I^2 = I \cdot\ I = \Big(\int_{-\infty}^\infty \mathrm{e}^{-2x}\,\mathrm{d}x\Big)\Big(\int_{-\infty}^\infty \mathrm{e}^{-2y}\,\mathrm{d}x\Big)[/tex]
Well, OK, I am just trying to understand the logic of this step. It seems just completely out of the blue for me and arbitrary. What was done here? The next step says that you have a double integral where e is raised to the (x2+y2) power, and you then convert to polar coordinates. That last bit actually makes sense to me. It's getting there I seem to stumble on.
Homework Statement
[tex]\int_{-\infty}^\infty \mathrm{e}^{-2x^2}\,\mathrm{d}x[/tex]
The Attempt at a Solution
According to the solution given, this can be set up where you say:
let
[tex]I = \Big(\int_{-\infty}^\infty \mathrm{e}^{-2x}\,\mathrm{d}x\Big)[/tex]
and
[tex]I^2 = I \cdot\ I = \Big(\int_{-\infty}^\infty \mathrm{e}^{-2x}\,\mathrm{d}x\Big)\Big(\int_{-\infty}^\infty \mathrm{e}^{-2y}\,\mathrm{d}x\Big)[/tex]
Well, OK, I am just trying to understand the logic of this step. It seems just completely out of the blue for me and arbitrary. What was done here? The next step says that you have a double integral where e is raised to the (x2+y2) power, and you then convert to polar coordinates. That last bit actually makes sense to me. It's getting there I seem to stumble on.