InverseFourierResidue

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:

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

Finding the Poles:

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 $[-1, 1]$.

Calculating the Integral:

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:

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

2. 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:

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

Final Result:

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.