# Fourier Integral and the Dirac Delta

• November 30th 2006, 08:45 AM
topsquark
Fourier Integral and the Dirac Delta
This is going to end up being a simple question with a simple answer, but I just can't tweak the answer out of my Math Methods text.

I have the following statement:
$\dot{\Delta}(\vec{x},0) = i \int \frac{d^3k}{2(2 \pi )^3} \left [ e^{i \vec{k} \cdot \vec{x}} + e^{-i \vec{k} \cdot \vec{x}} \right ]$ ( $\vec{k}$ and $\vec{x}$ are real 3-vectors and $\vec{k} \cdot \vec{x}$ is the usual dot product.)

I'm supposed to get that
$\dot{\Delta}(\vec{x},0) = -i \delta ^3 (\vec{x})$

But looking at the integral I'm thinking it ought to be:
$\dot{\Delta}(\vec{x},0) = \frac{i}{2} \left ( \delta ^3 (\vec{x}) + \delta ^3 (-\vec{x}) \right )$
which would give me 0, which obviously isn't true. So (sigh) what am I doing wrong?

Thanks!
-Dan
• November 30th 2006, 12:12 PM
CaptainBlack
Quote:

Originally Posted by topsquark
This is going to end up being a simple question with a simple answer, but I just can't tweak the answer out of my Math Methods text.

I have the following statement:
$\dot{\Delta}(\vec{x},0) = i \int \frac{d^3k}{2(2 \pi )^3} \left [ e^{i \vec{k} \cdot \vec{x}} + e^{-i \vec{k} \cdot \vec{x}} \right ]$ ( $\vec{k}$ and $\vec{x}$ are real 3-vectors and $\vec{k} \cdot \vec{x}$ is the usual dot product.)

I'm supposed to get that
$\dot{\Delta}(\vec{x},0) = -i \delta ^3 (\vec{x})$

But looking at the integral I'm thinking it ought to be:
$\dot{\Delta}(\vec{x},0) = \frac{i}{2} \left ( \delta ^3 (\vec{x}) + \delta ^3 (-\vec{x}) \right )$
which would give me 0, which obviously isn't true. So (sigh) what am I doing wrong?

Thanks!
-Dan

Do you think this might be related to:

$
\delta(x)=\delta(-x)\ ?
$

and that you are using the other sense convention for what is the forward FT

RonL
• November 30th 2006, 12:50 PM
topsquark
Quote:

Originally Posted by CaptainBlack
Do you think this might be related to:

$
\delta(x)=\delta(-x)\ ?
$

and that you are using the other sense convention for what is the forward FT

RonL

Okay, I see what you are saying about the convention. And (duh!) I forgot that $\delta$ is even. I knew it was something simple! Thanks.

-Dan