When evaluating contour integrals, it is often of interest to prove that Fourier-type integrals vanish on large enough semicircles (see the figure). This holds under the following condition:
Theorem. Suppose that $$f(z)=O(|z|^{-a}), \quad a>0$$ for \(z\) in the upper half-plane. Then for any \(\lambda > 0\) we have $$\int_{\gamma_R} f(z)\mathrm{e}^{i\lambda z} \rightarrow 0, \quad R\to+\infty,$$ where \(\gamma_R\) is the upper half-circle of radius \(R\).
This result is stronger than other ways of developing vanishing integration contours in the upper half-plane, compare for instance with the MIT lecture notes by Jeremy Orloff1. The version above can be found in advanced books on Fourier transforms, for example2.
To prove that, parametrize the upper half-circle \(\gamma_R\) by \(z=R\mathrm{e}^{i\theta} = R(\cos\theta + i\sin\theta)\) where \(0<\theta<\pi\). Under this parametrization, the Fourier multiplier becomes \(\mathrm{e}^{i\lambda z} = \mathrm{e}^{-\lambda R \sin \theta}\mathrm{e}^{i R \lambda \cos\theta}\). Thus, the integral can be bounded by $$ \left|\int_{\gamma_R} f(z)\mathrm{e}^{i\lambda z}\right|\leqslant \int_{0}^{\pi} |f(R\mathrm{e}^{i\theta})| R \mathrm{e}^{-R\lambda \sin\theta} \mbox{d}\theta \\
\leqslant C\int_{0}^{\pi} R^{1-a} \mathrm{e}^{-R\lambda \sin\theta} \mbox{d}\theta\\ = 2C\int_{0}^{\frac{\pi}{2}} R^{1-a} \mathrm{e}^{-R\lambda \sin\theta} \mbox{d}\theta \\
\leqslant 2C\int_{0}^{\frac{\pi}{2}} R^{1-a} \mathrm{e}^{-2 R\lambda \theta / \pi} \mbox{d}\theta \\
= C\cdot \frac{\pi R^{- a} \left(1 – e^{- R \lambda}\right)}{\lambda},$$
which tends to zero as long as \(a>0\) and \(R\to \infty\).
- 1.Orloff J. Definite integrals using the residue theorem. Lecture Notes. Accessed 2023. https://math.mit.edu/~jorloff/18.04/notes/topic9.pdf
- 2.Spiegel MR. Laplace Transforms. McGraw Hill; 1965.
