We are asked to evaluate [ I(x) = \int_0^\infty \frac{\cos(\alpha x) + \alpha \sin(\alpha x)}{1 + \alpha^2} , d\alpha ] using Fourier integral representation. Let's proceed step by step. --- Consider the Fourier integral representation for a function (f(t)): [ f(x) = \frac{1}{\pi} \int_0^\infty F_c(\alpha) \cos(\alpha x), d\alpha + \frac{1}{\pi} \int_0^\infty F_s(\alpha) \sin(\alpha x), d\alpha ] where (F_c(\alpha) = \int_{-\infty}^{\infty} f(t) \cos(\alpha t), dt) and (F_s(\alpha) = \int_{-\infty}^{\infty} f(t) \sin(\alpha t), dt). Notice that the given integral has the form [ I(x) = \int_0^\infty \frac{\cos(\alpha x)}{1+\alpha^2} , d\alpha + \int_0^\infty \frac{\alpha \sin(\alpha x)}{1+\alpha^2} , d\alpha. ] --- Step 1: Extend to the whole real line. Let’s define the function [ F(\alpha) = \frac{1}{1+\alpha^2}, \quad \alpha \in \mathbb{R}. ] Then we can write [ \frac{\cos(\alpha x) + \alpha \sin(\alpha x)}{1+\alpha^2} = \operatorname{Re}\left{\frac{e^{i \alpha x}(1 - i \alpha)}{1+\alpha^2}\right}. ] Check: [ e^{i \alpha x}(1 - i \alpha) =