Why does this trick make the oscillating exponential integral converge?

Question

I have the following integral: $$\int_0^{+\infty} e^{iEt} dt,$$ where $E$ is a real constant. I know this integral does not converge. However, I have seen the following trick which makes it converge: \begin{align} \int_0^{+\infty} e^{iEt} dt &= \lim_{\epsilon \rightarrow 0} \int_0^{+\infty} e^{i(E+i\epsilon) t} dt\\ &= \lim_{\epsilon \rightarrow 0}\left[ \frac{1}{i(E+i\epsilon)} e^{i(E+i\epsilon) t} \right] _0^{+\infty}\\ &=\frac{1}{iE} \end{align} If it was just for this, I would have thought that this is completely nonsense, and that it is like thinking that the limit of a discontinuous function at a point actually equals the function at that point. However, if one uses this trick when calculating the solution of the non-homogeneous Klein-Gordon equation $$(\square-m^2)\phi(\vec{x})=g\delta^3(\vec{x})$$ where $g$ is a real constant, using the Green function $G(x)$, where $x$ is a 4d vector, such that $$-(\square-m^2)G(x)=\delta^4(x)$$ the result is actually the Yukawa potential, which can obviously be calculated using other methods. I suspect that the delta is playing some role here, but I do not know distribution theory so I am asking here. What is really going on mathematically?

Answer 1

I guess you encountered this result in the context of a physics topic, where the argument for such a trick is usually omitted or implicit. In fact, two "formalisms" are here merged together without warning, in that the first expression has to be understood within the scope of Fourier theory, when the actual following computations turn out to be a Laplace transform.

Concretely, it is based on the fact that $\mathscr{F}[f(t)\theta(t)](\omega) = \displaystyle \lim_{s\rightarrow i\omega} \mathscr{L}[f(t)](s)$, which is somewhat equivalent to set $s = i(\omega + i\epsilon)$, with $\epsilon \rightarrow 0^+$. Some authors even use the notation $\omega + i0$, so that they write $\int_0^\infty e^{iEt} \,\mathrm{d}t = \frac{-i}{E + i0}$. Note that $\theta(t)$ represents here the Heaviside function, hence $f(t) = 1$ in the present case.

This kind of trick plays the role of a "regularization". Indeed, the Fourier transform suffers from convergence problems, basically because its kernel is non-decreasing (since $|e^{i\omega t}| = 1$), which restricts the class of functions possessing a Fourier transform. This problem may be partially circumvented by working with distributions, but it has the disadvantage to involve nasty expressions (see e.g. the Fourier transform of the Heaviside function here).

On the contrary, the kernel of the Laplace transform involves a complex variable with an auxiliary real/imaginary part (depending on whether the prefactor $i$ is present or not), which makes it decreasing and allows easier convergence. In consequence, "not-fast-decreasing" functions admit nicer Laplace transforms, such as the Heaviside function, whose Laplace transform is the inverse function you derived.

Finally, it is to be noted that this formal change could be considered as a mere new point of view on the same object. Indeed, the quantity $\int_0^\infty e^{iEt} \,\mathrm{d}t$ is an integral representation of a kind of truncated Dirac delta, when the function $\frac{1}{iE}$ may be seen as a Dirac delta in the complex plane, since the Cauchy kernel $\frac{1}{2\pi iz}$ acts as such with respect to analytic test functions.