Can I change the limits in the Fourier transform definition of the Dirac delta function?

Question

The Dirac delta function is often defined as $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{i p x} dp$$ Is there a way in which $$\delta(x)=\int_0^\infty e^{ipx} dp$$ is also correct? For instance if $x$ or $p$ obey some condition.

Answer 1

Not exactly like that. Formally, $$ \delta(x) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{i p x} \, dp = \frac{1}{2\pi} \left( \int_{-\infty}^0 e^{i p x} \, dp + \int_0^\infty e^{i p x} \, dp \right) \\ = \frac{1}{2\pi} \left( \int_0^\infty e^{-i p x} \, dp + \int_0^\infty e^{i p x} \, dp \right) \\ = \frac{1}{2\pi} \int_0^\infty \left( e^{-i p x} + e^{i p x} \right) \, dp \\ = \frac{1}{\pi} \int_0^\infty \cos(p x) \, dp \\ $$

Answer 2

No, you can't. The definition you're giving is the one arising from the Fourier transform. It can be easily proven: take $f\in L^1(\mathbb{R})$ then $$f(x) = {1\over{2\pi}}\int_{-\infty}^{\infty}e^{ikx}\left(\int_{-\infty}^{\infty}e^{-iky}f(y)dy\right)dk$$ here you cannot interchange the integrals, but you can write $$f(x) = \lim_{N\rightarrow\infty}{1\over{2\pi}}\int_{-N}^{N}e^{ikx}\left(\int_{-\infty}^{\infty}e^{-iky}f(y)dy\right)dk\tag{1}$$ By the properties that the Dirac delta should have, there is $$f(x)=\int_{-\infty}^{\infty}\delta(y-x)f(y)dy$$ so from $(1)$ we get that $$\delta_N(x) = {1\over{2\pi}} \int_{-N}^{N}e^{ikx}dk = {1\over{2\pi}}\int_{-N}^0 e^{ikx}dk+{1\over{2\pi}}\int_0^{N}e^{ikx}dk = \\ {1\over{2\pi}}\int_{0}^N e^{-ikx}dk+{1\over{2\pi}}\int_0^{N}e^{ikx}dk = \\ {1\over{2\pi}}\int_0^{N}e^{ikx}+e^{-ikx} = {1\over\pi}\int_0^N \cos(kx)dk = {1\over\pi}\frac{\sin(Nx)}{x} \tag{2}$$ Does this definition of the delta, in the limit, follow the properties that should have? Yes, it does, mainly $$\int_{-\infty}^{\infty}\delta_n(x)dx=1\;\;\;\;\lim_{N\rightarrow\pm\infty}\delta_n(x)=0$$ So we can define the delta function as the limit of the sequence $$\delta(x) = \lim_{N\rightarrow\infty}{1\over\pi}\frac{\sin(Nx)}{x} = \lim_{N\rightarrow\infty} {1\over{2\pi}} \int_{-N}^{N}e^{ikx}dk$$ The Fourier representation of the delta function is as follows $$\delta(x) = \int_{-\infty}^{\infty}e^{ikx}dx$$ As you can see from equation $(2)$ the only integral of $\int_0^{N}e^{ikx}dk$ it's not enough to define a Dirac delta