Show that if \(\xi\in \RR\), then \[ \int_{-\infty}^{\infty} \frac{e^{-2 \pi i x \xi}}{\left(1+x^{2}\right)^{2}} d x=\frac{\pi}{2}(1+2 \pi|\xi|) e^{-2 \pi|\xi|} .\]
Solution
Write \(I(\xi) = \int_{-\infty}^\infty \frac{e^{-2\pi i x \xi}}{(1+x^2)^2}\,dx\) and \(a = 2\pi \abs{\xi} \ge 0\). Let \(f(z) = \frac{e^{i a z}}{(z^2 + 1)^2}\), and for \(R > 1\) let \(\Gamma_R\) consist of \([-R, R]\) and the arc \(C_R\): \(z=Re^{i\theta}\), \(\theta \in [0, \pi]\).
1. \(I(\xi) = \int_{-\infty}^\infty \frac{e^{i a x}}{(1+x^2)^2}\,dx\).
1.1. \(I\) is an even function of \(\xi\).
The substitution \(u = -x\) gives \[I(-\xi) = \int_{-\infty}^\infty \frac{e^{2\pi i x \xi}}{(1+x^2)^2}\,dx = \int_{-\infty}^\infty \frac{e^{-2\pi i u \xi}}{(1+u^2)^2}\,du = I(\xi).\]
1.2. Q.E.D.
By step 1 we may assume \(\xi \le 0\). Then \(-2\pi\xi = a\), so \(e^{-2\pi i x\xi} = e^{iax}\).
2. In the upper half-plane, \(f\) has exactly one singularity, a pole of order \(2\) at \(z = i\).
\(f(z) = \frac{e^{i a z}}{(z - i)^2 (z + i)^2}\), the numerator is entire and nonvanishing, and \(z=-i\) lies in the lower half-plane.
3. \(\operatorname{Res}(f, i) = \dfrac{(1 + a) e^{-a}}{4i}\).
For a pole of order \(2\), \[\operatorname{Res}(f, i) = \lim_{z \to i} \frac{d}{dz}\left[\frac{e^{i a z}}{(z + i)^2}\right] = \lim_{z \to i}\frac{i a (z + i) - 2}{(z + i)^3} e^{i a z}.\] At \(z=i\): \(z + i = 2i\), \((z + i)^3 = -8i\), \(ia(2i) - 2 = -2(1 + a)\), and \(e^{i a i} = e^{-a}\). Hence \(\operatorname{Res}(f, i) = \frac{-2(1 + a) e^{-a}}{-8i}\).
4. \(\displaystyle\int_{C_R} f(z)\,dz \to 0\) as \(R\to\infty\).
For \(z = x + i y \in C_R\), \(y \ge 0\) and \(a\ge0\), so \(\abs{e^{i a z}} = e^{-a y} \le 1\). By the reverse triangle inequality, \(\abs{z^2 + 1} \ge R^2 - 1 > 0\), so \(\abs{f(z)} \le (R^2 - 1)^{-2}\). The arc has length \(\pi R\), so \[\left|\int_{C_R} f(z)\,dz\right| \le \frac{\pi R}{(R^2 - 1)^2} \longrightarrow 0.\]
5. \(I(\xi) = \boxed{\dfrac{\pi}{2}(1 + 2\pi \abs{\xi}) e^{-2\pi \abs{\xi}}}\).
1. Substitute \(a = 2\pi\abs{\xi}\).
6. Q.E.D.
Step 5 is the required identity.