Derivation of Dirac Delta Function

  • Thread starter coki2000
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  • #1
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Hello,
My question is about how dirac-delta function is derived by using this integral,

[tex]\frac{1}{2\pi }\int_{-\infty}^{\infty}e^{ikx}dk=\delta (x)[/tex]

I couldn't solve this integral. Please help me.
Thanks for all of your helps.
 

Answers and Replies

  • #2
Hurkyl
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The trick is that that's not an integral, at least not of the sort you learned in Calc 2.

Unfolding all the definitions involved, what they really mean is that
[tex]
\frac{1}{2\pi }\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{ikx} f(x)\, dx \, dk=f(0)
[/tex]​
for any f which is smooth and rapidly decreasing. (Or some similar condition, depending on the precise details of what they're doing)
 
  • #3
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OK, but how can I solve this integral? Actually, I wonder it.

[tex]\frac{1}{2\pi }\int_{-\infty}^{\infty}e^{ikx}dk[/tex]
 
  • #4
arildno
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OK, but how can I solve this integral? Actually, I wonder it.

[tex]\frac{1}{2\pi }\int_{-\infty}^{\infty}e^{ikx}dk[/tex]
You can't.
Read Hurkyl's response again.
 
  • #5
Hurkyl
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If they are using the [itex]\int[/itex] symbol to denote the sort of integral you learned in your Calc 2 class, then that integral doesn't exist.
 
  • #6
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But when I try to integrate it, I take this answer,

[tex]\lim_{n \to \infty}\frac{sin(nx)}{\pi x}=\delta (x)[/tex]

It should have a meaning.
 
  • #7
arildno
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But when I try to integrate it, I take this answer,

[tex]\lim_{n \to \infty}\frac{sin(nx)}{\pi x}=\delta (x)[/tex]

It should have a meaning.
No, it does not.
 
  • #8
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Okey thanks for your helps.
 
  • #9
arildno
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