Derivation of Dirac Delta Function

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

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

Thanks for all of your helps.

Hurkyl
Staff Emeritus
Gold Member
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)

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

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

arildno
Homework Helper
Gold Member
Dearly Missed
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.

Hurkyl
Staff Emeritus
Gold Member
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.

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
Homework Helper
Gold Member
Dearly Missed
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.