Integrating exp(x^2)

  • Thread starter Wishe Deom
  • Start date
  • #1
12
0
Hello everyone,

This is my first post on the forum; I'm pretty sure my question fits in this section.

I am having a lot of difficulty finding the definite integral
[tex]\int^{+\infty}_{-\infty}e^{-2ax^{2}}[/tex] dx
where a is positive and real.
I know the answer is [tex]\\sqrt{\frac{\pi}{2a}}, but I have no idea at all how to get there.
 

Answers and Replies

  • #2
51
0
hint: write another integral of that form, but with y as the variable. multiply them together. convert to polar coordinates, and integrate. then take the square root.
 
  • #3
12
0
I will try this, but why would this work? As far as I understand it, the product of the integrals is not necessarily equal to the integral of the products.

Is this what you are suggesting I try?

[tex]\int{\inf}_{-\inf}e{-ax{2}}e{-ay{2}}dxdy
 
  • #4
51
0
[tex] \int \int f(x) g(y) dx dy = \int f(x) dx \int g(y) dy [/tex]

(most of the time)

and yes, i'm suggesting you look at:

[tex] \sqrt{\int \int e^{-\alpha x^2} e^{-\alpha y^2} dxdy [/tex]
 
  • #5
51
0
As a bonus, once you convince yourself that
[tex] \int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{{\frac{\pi}{\alpha}} [/tex],
you can easily calculate integrals of the form
[tex] \int_{-\infty}^{\infty}x^{2n} e^{-\alpha x^2} dx [/tex]
by differentiating the first expression (under the integral sign, on the left) with respect to alpha n times. This is a very useful fact.
 
  • #6
1,101
3
I will try this, but why would this work? As far as I understand it, the product of the integrals is not necessarily equal to the integral of the products.
To justify the interchange you can use "[URL [Broken] Theorem[/URL].
 
Last edited by a moderator:
  • #7
12
0
Thank you for all the help. I'm well on my way, but I am having trouble with the limits of integration. Since x and y go from -inf to inf, what are the new limits?

Since r = sqrt(x^2+y^2), it seems that r goes from inf to inf, which would result in 0. I know, logically, r should be 0 to infinity, but how do I calculate that?
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
41,833
956
I will try this, but why would this work? As far as I understand it, the product of the integrals is not necessarily equal to the integral of the products.

Is this what you are suggesting I try?

[tex]\int{\inf}_{-\inf}e{-ax{2}}e{-ay{2}}dxdy
[tex] \int \int f(x) g(y) dx dy = \int f(x) dx \int g(y) dy [/tex]

(most of the time)
This is "Fubini's Theorem"
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
41,833
956
1) [tex]\int_{-\infty}^\infty e^{-2ax^2}dx= 2\int_0^\infty e^{-2ax^2}dx [/itex]
by symmetry.

2) If you let [tex]I= \int_{-\infty}^\infty} e^{-2ax^2}dx[/tex]
then [tex]\frac{I}{2}= \int_0^\infty e^{-2ax^2}dx[/tex]
and it is also true that
[tex]\frac{I}{2}= \int_0^\infty e^{-2ay^2}dy[/tex]
since that is the same integral with a different "dummy" variable.

so
3) [tex]I^2/4= \left(\int_0^\infty e^{-2ax^2}dx\right)\left(\int_0^\infty e^{-2ay^2}dy\right)[/tex]
[tex]= \int_0^\infty \int_0^\infty e^{-2a(x^2+ y^2)} dy dx[/tex]
by Fubini's theorem.

You can interpret that last as a double integral over the first quadrant and change to polar coordinates. Remember that the "differential of area" in polar coordinates is [itex]r drd\theta[/itex] and that [itex]\theta[/itex] goes from 0 to [itex]\pi/2[/itex] in the first quadrant.
 

Related Threads on Integrating exp(x^2)

  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
8
Views
113K
Replies
7
Views
17K
Replies
8
Views
9K
Replies
5
Views
980
Replies
17
Views
10K
Replies
8
Views
7K
  • Last Post
Replies
9
Views
15K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
3K
Top