Calculate the Fourier transform of $b(x) =\frac{1}{x^2 +a^2}$ [closed]

Solution 1:

Consider the function $f(x)=e^{-a|x|}$. Then \begin{align*} \hat{f}(\omega)&=\int_{-\infty}^{\infty}e^{-a|x|}e^{-i\omega x}\, dx= \int_{-\infty}^{0}e^{ax}e^{-i\omega x}\, dx+\int_{0}^{\infty}e^{-ax}e^{-i\omega x}\, dx = \\ &= \left[ \frac{e^{(a-i\omega)x}}{a-i\omega} \right]_{-\infty}^0-\left[ \frac{e^{-(a+i\omega)x}}{a+i\omega} \right]_{0}^{\infty}=\frac{1}{a-i\omega}+\frac{1}{a+i\omega}=\frac{2a}{a^2+\omega^2} \end{align*} Now, by the inversion forumla, we have \begin{equation*} e^{-a|x|}=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2a}{a^2+\omega^2}e^{i\omega x}\, d\omega \end{equation*} Changing the sign on $x$ and multiplying by $\frac{\pi}{a}$, we finally get \begin{equation*} \frac{\pi}{a} e^{-a|-x|}=\frac{\pi}{a} e^{-a|x|}=\int_{-\infty}^{\infty}\frac{e^{-i\omega x}}{a^2+\omega^2}\, d\omega \end{equation*} Thus, \begin{equation*} \hat{b}(\omega)=\frac{\pi}{a}e^{-a|\omega|} \end{equation*}

Solution 2:

One way to attack this is via the residue theorem. Consider

$$\oint_C dz \frac{e^{i k z}}{z^2+a^2}$$

where $C$ is a semicircular contour in the upper half plane of radius $R$. Note that, to use the residue theorem, we expect the integral over the circular arc to vanish as $R \to \infty$; however, this only happens when $k > 0$. (I leave it to the reader to show this.) The residue at the pole $z=i a$ is $e^{-k a}/(i 2 a)$, so, by the residue theorem,

$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{x^2+a^2} = i 2 \pi \frac{e^{-a k}}{i 2 a} = \frac{\pi}{a} e^{-a k}$$

when $k > 0$. When $k < 0$, however, we must use the semicircular contour in the lower half plane, rather than the upper half plane. Thus, we now consider the pole at $z=-i a$, and the integral takes the value $(\pi/a) e^{a k}$ when $k < 0$. Putting this together, we have

$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{x^2+a^2} = \frac{\pi}{a} e^{- a |k|}$$