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Hello everyone,

I am currently reading how to solve Renewal equations with the help of Fourier Transforms. I know some basics about this topic (I had a course in Functional Analysis and one in Spectral theory), however, some things do not make sense to me. What does the notation mean where you subtract $i0$ (see red underlined formulas in the image linked above.)? How does it come that the integrals corressponding to the inverse Fourier transformation converge? I have the suspicion that this has something to do with the Residue theorem, but I am not sure. Unfortunately the author of the paper where this notaion occurs does not explain what it means.

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  • $\begingroup$ Perhaps $i\, 0$ refers to $\Phi'(0)=i\ J$ where $J=\sum\limits_{j=1}^M c_j\,\gamma_j$, so $J=0$ corresponds to $0=\sum\limits_{j=1}^M c_j\,\gamma_j$? $\endgroup$ Commented Aug 16, 2022 at 17:17

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I don't know what the renewal equation is, but $i 0$ almost certainly refers to an infinitesimally small (positive) imaginary part. You might also see this written as $i0^+$ or $i \delta$ or $i\epsilon$ in other places. If you calculate the (inverse) Fourier transform with this, there is a pole in upper half-plane of complex $\zeta$. The inverse Fourier transform leads to a step function in time. Consider the integral

$$\int_{-\infty}^{\infty}\frac{d\zeta}{2\pi}e^{-i\zeta t}\frac{1}{\zeta-i 0}$$.

We will evaluate this by turning it into a contour integral. The only question is whether we should close the contour in the upper or lower half-plane.

First, assume $t < 0$. Then the exponential goes to zero as $\zeta$ goes to infinity in the upper half-plane. Therefore, we should close the contour in the upper half-plane to ensure convergence. There is a pole in the upper half-plane (at $\zeta = i 0$), so the integral yields $2\pi i$ (through application of the residue theorem.)

Next, assume $t>0$. In this case, we must close the contour in the lower half-plane to ensure convergence. However, there is no pole in the lower half-plane, so the integral is zero by the residue theorem.

As such, the integral is equal to $2\pi i \Theta(-t)$ (this result may change based off your Fourier transform conventions). The presence of these $i0$ terms is pretty common (in physics at least) to ensure that causality holds in models.

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  • $\begingroup$ Thank you. I am a mathematician, therefore I have to be a little bit more careful with such formalisms. Still, thanks to your comment I got the right idea. $\endgroup$ Commented Aug 17, 2022 at 11:49

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