- #1

- 202

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Can anyone show me how the property

[tex]\frac{1}{2\pi} \int ^{\infty} _{-\infty} e^{i\omega x}d\omega= \delta(x)[/tex]

holds.

Thanks,

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- Thread starter Apteronotus
- Start date

- #1

- 202

- 0

Can anyone show me how the property

[tex]\frac{1}{2\pi} \int ^{\infty} _{-\infty} e^{i\omega x}d\omega= \delta(x)[/tex]

holds.

Thanks,

- #2

- 607

- 0

Do you know about Schwartz distribution theory? Generalized functions? Because that is what you are writing about, not traditional calculus functions. Your equation MEANS...

[tex]

\frac{1}{2\pi}\int_{-\infty}^\infty e^{i\omega x} \phi(x)\,d\omega = \phi(x)

[/tex]

for all test functions [itex]\phi[/itex] from an appropriate class.

- #3

- 202

- 0

edgar thanks for your reply.

It seems I've stumbled on something beyond my means.

I dont know anything about Schwartz distribution theory and a quick search on the net didnt help at all.

I thought the integral would be an easy calculus identity of sorts. Can you show me why the integral holds?

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