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 Orloff^{1}. The version above can be found in advanced books on Fourier transforms, for example^{2}.

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.