- Let \(f ( x ) = { \frac { 1 } { x ^ { 2 } + 1 } }\) . Use a contour integral consisting of the interval \([ - R , R ] \subset \mathbb { R }\) and a semicircle of radius R to compute the Fourier transform
\[ { \widehat { f } } ( 1 ) = \int _ { \mathbb { R } } f ( x ) e ^ { - i x } d x ~ . \]
Show that the contour integral converges to your answer as \(R \to + \infty\)
Solution
The value is \(\boxed{\widehat f(1)=\pi/e}\).
1. The real integral exists absolutely.
For real \(x\), \(|e^{-ix}|=1\). The function \((1+x^2)^{-1}\) is bounded on \([-1,1]\) and is at most \(x^{-2}\) for \(|x|\geq1\). Since the latter is integrable on both tails, the integral of \(e^{-ix}/(1+x^2)\) is absolutely convergent. Thus limits of its truncated integrals determine its ordinary integral, not merely a principal value.
2. The clockwise lower semicircle encloses one pole.
Set \(F(z)=e^{-iz}/(1+z^2)\). For \(R>1\), traverse \([-R,R]\) from left to right, then return from \(R\) to \(-R\) along the lower semicircle \(\Gamma_R\). This contour is clockwise. Its only enclosed pole is \(-i\), and that pole is simple, with residue \[ \operatorname{Res}_{z=-i}F =\frac{e^{-i(-i)}}{-2i}=\frac{i}{2e}. \] The residue theorem therefore gives [@SS03] \[ \int_{-R}^R\frac{e^{-ix}}{1+x^2}\,dx +\int_{\Gamma_R}F(z)\,dz =-2\pi i\frac{i}{2e}=\frac\pi e. \] The minus sign is the contour orientation; the lower half-plane is chosen because it gives exponential decay for the prescribed sign \(e^{-iz}\).
3. The semicircle contribution tends to zero.
On \(\Gamma_R\), one has \(\operatorname{Im}z\leq0\) and \[ |e^{-iz}|=e^{\operatorname{Im}z}\leq1, \qquad |1+z^2|\geq R^2-1. \] The arc length is \(\pi R\), so \[ \left|\int_{\Gamma_R}F(z)\,dz\right| \leq\frac{\pi R}{R^2-1}\longrightarrow0. \] Letting \(R\to\infty\) in the identity of step 2, using this estimate and the absolute convergence in step 1, gives \(\widehat f(1)=\pi/e\) as asserted.