Justify exp(-x^2)

  • #1
1. Homework Statement [/b
exp(-x^2) = (1/sqrt(pi))*integral[(e^(2*ixy)*e^(-y^2)], y = -infinity .. infinity
how to justify this equality
 

Answers and Replies

  • #2
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0


Expand the square:

[tex]\frac{1}{\sqrt{\pi}} \int_{-\infty}^\infty e^{\textbf{i} 2 x y} e^{-y^2} dy = \frac{1}{\sqrt{\pi}} \int_{-\infty}^\infty e^{(x+\textbf{i}y)^2} e^{-x^2} dy = \frac{e^{-x^2} }{\sqrt{\pi}} \int_{-\infty}^\infty e^{-(y-\textbf{i}x)^2} dy[/tex]

Finish from there.
 
  • #3


Thanks. But my mentor told me to do it a different way for some reason. He says to differentiate the inside of (1/sqrt(pi))*integral[(e^(2*ixy)*e^(-y^2)], y = -infinity .. infinity, with respect to x and then use integration by parts. I get stuck after I do that. Any ideas on what to do after that?
 
  • #4
I like Serena
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Thanks. But my mentor told me to do it a different way for some reason. He says to differentiate the inside of (1/sqrt(pi))*integral[(e^(2*ixy)*e^(-y^2)], y = -infinity .. infinity, with respect to x and then use integration by parts. I get stuck after I do that. Any ideas on what to do after that?
Let's give the integral a name, for instance F(x).

So your mentor asked you to calculate F'(x), which will yield an integral.
Can you show us what you found for F'(x)?

This will be an integral, that is supposed to be integrated by parts.
Can you do a suggestion which part to use?
And integrate it?
 
  • #5


Well since the original integral is with respect to y and he wants me to differentiate the inside just with respect to x I would just differentiate e^2ixy which would give me 2iy*e^2ixy. Putting this all together we have... (1/sqrt(pi))*integral[2iy*(e^(2*ixy)*e^(-y^2)], y = -infinity .. infinity. as far as choosing how to separate the parts for integration by parts it seems like no matter how i separate it I still get an even more complicated answer... For ex. I choose u=e^(2ixy-y^2) and v' to be 2iy... I end up getting (1/sqrt(pi))*[(e^(2ixy-y^2)*iy^2)-integral((2ix-2y)*(iy^2)*(e^(2ixy-y^2)))dy
 
  • #6
I like Serena
Homework Helper
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Well since the original integral is with respect to y and he wants me to differentiate the inside just with respect to x I would just differentiate e^2ixy which would give me 2iy*e^2ixy. Putting this all together we have... (1/sqrt(pi))*integral[2iy*(e^(2*ixy)*e^(-y^2)], y = -infinity .. infinity. as far as choosing how to separate the parts for integration by parts it seems like no matter how i separate it I still get an even more complicated answer... For ex. I choose u=e^(2ixy-y^2) and v' to be 2iy... I end up getting (1/sqrt(pi))*[(e^(2ixy-y^2)*iy^2)-integral((2ix-2y)*(iy^2)*(e^(2ixy-y^2)))dy
Could you try: v'=2y e-y2 ?

(Have a root √ and also ∫∞, and try using x2 just above the reply box. :wink:)
 
Last edited:
  • #7


when i do that i get (1/sqrt(pi))*[-e^(2ixy-y^2) + integral((e^-(y^2))*2ix*e^(2ixy))dy].
Now what?
 
  • #8
I like Serena
Homework Helper
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when i do that i get (1/sqrt(pi))*[-e^(2ixy-y^2) + integral((e^-(y^2))*2ix*e^(2ixy))dy].
Now what?
In the first part you need to substitute the limits -∞ and +∞.

The second part should look familiar. It looks a lot like the original integral F(x), except for a factor -2x, and I think you dropped an "i" somewhere.

In effect you have found the differential equation:
F'(x) = -2x F(x)

Do you know how to solve that?
The solution would yield your proof.

Cheers!
 

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