Lösningsförslag. Fouriertransformen av $f(x)$ ges av

\begin{aligned} F(\omega) &= \mathcal{F}(f)(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x) e^{-i\omega x} \, dx \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-x} e^{-i\omega x} \, dx - \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{x} e^{-i\omega x} \, dx \\ &= \frac{1}{2\pi} \int_{0}^{\infty} e^{-(1 + i\omega)x} \, dx - \frac{1}{2\pi} \int_{-\infty}^{0} e^{(1 - i\omega)x} \, dx \\ &= \frac{1}{2\pi} \left[ -\frac{1}{1 + i\omega} e^{-(1 + i\omega)x} \right]_{0}^{\infty} - \frac{1}{2\pi} \left[ \frac{1}{1 - i\omega} e^{(1 - i\omega)x} \right]_{-\infty}^{0} \\ &= \frac{1}{2\pi} \left(0 + \frac{1}{1 + i\omega}\right) - \frac{1}{2\pi} \left(\frac{1}{1 - i\omega} - 0\right) \\ &= \frac{1}{2\pi} \left( \frac{1}{1 + i\omega} - \frac{1}{1 - i\omega} \right) \\ &= \frac{1}{2\pi} \cdot \frac{(1 - i\omega) - (1 + i\omega)}{(1 - i\omega)(1 + i\omega)} \\ &= \frac{-i\omega}{\pi(1 + \omega^2)}. \end{aligned}

Plancherels formel säger att

\frac{1}{2\pi} \int_{-\infty}^{\infty} |f(x)|^2 \, dx = \int_{-\infty}^{\infty} |F(\omega)|^2 \, d\omega.

I detta fall har vi

\begin{aligned} \frac{1}{2\pi} \int_{-\infty}^{\infty} |f(x)|^2 \, dx &= \frac{1}{2\pi} \int_{0}^{\infty} e^{-2x} \, dx + \frac{1}{2\pi} \int_{-\infty}^{0} e^{2x} \, dx = \frac{1}{\pi} \int_{0}^{\infty} e^{-2x} \, dx \\ &= \frac{1}{\pi} \left[ -\frac{1}{2} e^{-2x} \right]_{0}^{\infty} = \frac{1}{\pi} \left(0 + \frac{1}{2} \right) = \frac{1}{2\pi} \end{aligned}

samt

\int_{-\infty}^{\infty} |F(\omega)|^2 \, d\omega = \int_{-\infty}^{\infty} \frac{\omega^2}{\pi^2 (1 + \omega^2)^2} \, d\omega

så

\frac{1}{2\pi} = \frac{1}{\pi^2} \int_{-\infty}^{\infty} \frac{\omega^2}{\pi^2 (1 + \omega^2)^2} \, d\omega.

Den sökta integralen har alltså värdet

\int_{-\infty}^{\infty} \frac{x^2}{(1 + x^2)^2} \, dx = \frac{\pi}{2}.