- Let \(\textstyle f ( x ) = { \frac { 1 } { x ^ { 2 } + 1 } }\) . Use residues to compute the Fourier transform
\[ { \widehat { f } } ( t ) = \int _ { - \infty } ^ { + \infty } f ( x ) e ^ { - i t x } d x ~ . \]
Solution
Set \[ F_t(z)=\frac{e^{-itz}}{z^2+1}. \] Its poles are simple, at \(z=i\) and \(z=-i\).
1. The case \(t>0\).
Close the contour by the lower semicircle of radius \(R>1\), oriented clockwise. If \(z=x+iy\) lies in the lower half-plane, then \[ |e^{-itz}|=e^{ty}\le1. \] Hence on the semicircular arc, \[ |F_t(z)|\le \frac1{R^2-1}, \] so the arc integral has absolute value at most \[ \frac{\pi R}{R^2-1}\longrightarrow0. \] The only enclosed pole is \(z=-i\), and \[ \operatorname{Res}_{z=-i}F_t(z) =\frac{e^{-it(-i)}}{-2i} =-\frac{e^{-t}}{2i}. \] Because the contour is clockwise, the residue theorem gives \[ \int_{-\infty}^{\infty}\frac{e^{-itx}}{1+x^2}\,dx =-2\pi i\left(-\frac{e^{-t}}{2i}\right) =\pi e^{-t}. \]
2. The case \(t<0\).
Now close the contour by the upper semicircle. Since \(t<0\) and \(y\ge0\), \[ |e^{-itz}|=e^{ty}\le1, \] so the same arc estimate makes the semicircle contribution tend to zero. The enclosed pole is \(z=i\), with residue \[ \operatorname{Res}_{z=i}F_t(z) =\frac{e^{-iti}}{2i} =\frac{e^t}{2i}. \] Thus \[ \int_{-\infty}^{\infty}\frac{e^{-itx}}{1+x^2}\,dx =2\pi i\frac{e^t}{2i} =\pi e^t. \]
3. The case \(t=0\) and the final formula.
At \(t=0\), \[ \widehat f(0)=\int_{-\infty}^{\infty}\frac{dx}{1+x^2}=\pi. \] Combining the three cases, \[ \boxed{\widehat f(t)=\pi e^{-|t|}}. \]