InverseFourierResidue - crowlogic/arb4j GitHub Wiki

Given the function to be transformed:

f(x) = \frac{1}{1 - \frac{\lambda}{\sqrt{1 - x^2}}}

To compute the Fourier Transform using contour integration, we represent our function as:

g(x) = \frac{e^{-ixy}}{1 - \frac{\lambda}{\sqrt{1 - x^2}}}

The poles are obtained where the denominator is zero:

1 - \frac{\lambda}{\sqrt{1 - x^2}} = 0

\sqrt{1 - x^2} = \lambda

1 - x^2 = \lambda^2

x^2 = 1 - \lambda^2

x = \pm \sqrt{1 - \lambda^2}

Since $0 < \lambda < 1$, the poles $x = \pm \sqrt{1 - \lambda^2}$ are real and located within the interval $[-1, 1]$.

Upper Semi-Circle Contour: For the pole at $x = \sqrt{1 - \lambda^2}$, enclose it within a semi-circular contour in the upper half-plane. Using the Residue Theorem:

\int_{\text{Contour}} g(x) dx = 2\pi i \times \text{Residue at } x = \sqrt{1 - \lambda^2}

So, the integral becomes:

-2\pi i \frac{\lambda^2}{\sqrt{1 - \lambda^2}} e^{-i\sqrt{1 - \lambda^2}y}

Lower Semi-Circle Contour: For the pole at $x = -\sqrt{1 - \lambda^2}$, enclose it within a semi-circular contour in the lower half-plane. Using the Residue Theorem:

\int_{\text{Contour}} g(x) dx = 2\pi i \times \text{Residue at } x = -\sqrt{1 - \lambda^2}

So, the integral becomes:

2\pi i \frac{\lambda^2}{\sqrt{1 - \lambda^2}} e^{i\sqrt{1 - \lambda^2}y}

Adding both integrals provides the Fourier Transform as:

F(y) = \int_{-1}^{1} g(x) dx = 2\pi i \frac{\lambda^2}{\sqrt{1 - \lambda^2}} \left( e^{i\sqrt{1 - \lambda^2}y} - e^{-i\sqrt{1 - \lambda^2}y} \right)

Using the identity $2i \sin(a) = e^{ia} - e^{-ia}$, this simplifies to:

F(y) = 4\pi \frac{\lambda^2}{\sqrt{1 - \lambda^2}} \sin\left(\sqrt{1 - \lambda^2}y\right)

This is the Fourier transform of the given function over the specific interval $[-1,1]$, considering the real poles within this interval.