How to interpret $\int_0^\infty \exp(ikx) dx$ in distribution theory?

Question

It is well known that $F(k):=\int_{-\infty}^\infty \exp(ikx) dx=2\pi\delta(k)$. Upon expanding out definitions, this is nothing but a peculiar way to state the inversion formula for the Fourier transform. What about $G(k):=\int_0^\infty \exp(ikx) dx$? The only obvious property I see is that that $G(k)+\overline{G}(k)=2\pi\delta(k)$, thus $\mathrm{Re}(G(k))=\pi\delta(k)$. But does it have an imaginary part?

To put it another way, I am asking for a distributional interpretation of $\int_0^\infty \sin(kx) dx$.

Answer 1

Your $G(k)$ is essentially the Fourier Transform of the Heaviside unit step (although the forward transform usually uses $-i$):

$$\begin{align*}F(k) &= \int_0^\infty e^{-ikx}dx \\ &= \int_{-\infty}^\infty H(x)e^{-ikx}dx\\ &= \dfrac{1}{ik} + \pi\delta(k) \end{align*}$$

That result can be found in any decent table of Fourier Transforms.

To derive that result, one can use contour integration with an infinite radius, semi-circular contour, with a vanishing small semi-circular detour around the discontinuity at the origin, in the upper half plane to find the Cauchy Principal Value of the integral along the real axis. Similar to what I did here for the signum function, but without the sign mistake: Fourier Integral Representation of $H(t)-\frac{1}{2}$

Answer 2

I'll offer another motivation for Andy's result, one that doesn't require quite so much theory. The real part is trivial from $G+\bar{G}=2\pi\delta (k)$, but what about the imaginary part? First note that $$\Im\int_0^N\exp (ikx) dx=\Im\frac{\exp (ikN)-1}{ik}=\frac{2}{k}\sin^2 (kN/2).$$For large $N$ and $k\ne 0$, the squared sine averages out to $\frac{1}{2}$. Thus $$\int_0^\infty\exp (ikx) dx=\pi\delta (k)+\frac{i}{k}.$$