Derivation of Dirac Delta Function

[coki2000]

Hello,
My question is about how dirac-delta function is derived by using this integral,

$$\frac{1}{2\pi }\int_{-\infty}^{\infty}e^{ikx}dk=\delta (x)$$

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

 

[Hurkyl]

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

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

for any f which is smooth and rapidly decreasing. (Or some similar condition, depending on the precise details of what they're doing)

 

[coki2000]

OK, but how can I solve this integral? Actually, I wonder it.

$$\frac{1}{2\pi }\int_{-\infty}^{\infty}e^{ikx}dk$$

 

[arildno]

OK, but how can I solve this integral? Actually, I wonder it.

$$\frac{1}{2\pi }\int_{-\infty}^{\infty}e^{ikx}dk$$

You can't.
Read Hurkyl's response again.

 

[Hurkyl]

If they are using the $\int$ symbol to denote the sort of integral you learned in your Calc 2 class, then that integral doesn't exist.

 

[coki2000]

But when I try to integrate it, I take this answer,

$$\lim_{n \to \infty}\frac{sin(nx)}{\pi x}=\delta (x)$$

It should have a meaning.

 

[arildno]

But when I try to integrate it, I take this answer,

$$\lim_{n \to \infty}\frac{sin(nx)}{\pi x}=\delta (x)$$

It should have a meaning.

No, it does not.

 

[coki2000]

Okey thanks for your helps.

 

[arildno]

Okey thanks for your helps.

You're welcome.