- #1

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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.

- Thread starter coki2000
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- #1

- 91

- 0

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.

- #2

Hurkyl

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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 \frac{1}{2\pi }\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{ikx} f(x)\, dx \, dk=f(0)

[/tex]

- #3

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[tex]\frac{1}{2\pi }\int_{-\infty}^{\infty}e^{ikx}dk[/tex]

- #4

arildno

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You can't.

[tex]\frac{1}{2\pi }\int_{-\infty}^{\infty}e^{ikx}dk[/tex]

Read Hurkyl's response again.

- #5

Hurkyl

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

It should have a meaning.

- #7

arildno

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No, it does not.

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

It should have a meaning.

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Okey thanks for your helps.

- #9

arildno

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You're welcome.Okey thanks for your helps.

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