It is well-known that $$\int^\infty_{-\infty} \frac{\cos x}{x^2 + 1}\, dx = \frac{\pi}{e}$$ More generally, $$\int^\infty_{-\infty} \frac{\cos ax}{x^2 + b^2}\, dx = \frac{\pi}{b} e^{-ab} \;\forall\: a,b>0$$ There is a common solution using complex analysis that simply involves parameterising a semicircular contour $C$ and integrating over it using the Residue Theorem.
$$\mathbf{The\;Complex\;Solution}$$
Define $f(z) = \frac{e^{iaz}}{z^2 + b^2}$. Notice that since $e^{iaz}$ is holomorphic over $\mathbb{C}$, $f(z)$ has simple poles $\pm ib$. Consider the residue at $z = ib$. $$\mathop{\mathrm{Res}}_{z = ib}\, f(z) = \lim_{z\rightarrow ib} \,(z-ib)\frac{e^{iaz}}{z^2 + b^2} = \lim_{z\rightarrow ib}\, \frac{e^{iaz}}{z+ib} = \frac{e^{-ab}}{2ib}$$ Let $R$ be an arbitrary real number greater than $b$. Define an arc $\gamma_R = Re^{i\theta}$ where $\theta\in [0, \pi]$. Then define a closed curve $C_R = \gamma_R \,\cup [-R,R]$. This gives us a semicircular closed curve with radius $R$ centred about the origin. This curve encloses the point $ib$. By the Residue Theorem, $$\int_{C_R} f(z)\, dz = 2\pi i \cdot \mathop{\mathrm{Res}}_{z = ib}\, f(z) = \frac{\pi}{b} e^{-ab}$$ However, we also see that $$\int_{C_R} f(z)\, dz = \int_{-R}^R f(z)\, dz + \int_{\gamma_R} f(z)\, dz$$ Hence, $$\int_{-R}^R f(z)\, dz = \frac{\pi}{b} e^{-ab} - \int_{\gamma_R} f(z)\, dz$$ By Jordan's Lemma, we have $$\lim_{R\rightarrow \infty} M_R = \lim_{R\rightarrow \infty} \mathop{\mathrm{max}}_{\theta\in [0, \pi]} \left|\frac{1}{(Re^{i\theta})^2 + b^2}\right| = \lim_{R\rightarrow \infty} \mathop{\mathrm{max}}_{\theta\in [0, \pi]} \frac{1}{\left|(Re^{i\theta})^2 + b^2\right|} = \lim_{R\rightarrow \infty} \frac{1}{R^2 + b^2} = 0$$ $$\implies \lim_{R\rightarrow\infty} \left|\int_{\gamma_R} f(z)\, dz \right| \leq \lim_{R\rightarrow\infty}\frac{\pi}{a}M_R = 0$$ $$\int_{-\infty}^\infty f(z)\, dz = \lim_{R\rightarrow \infty}\int_{-R}^R f(z)\, dz = \frac{\pi}{b} e^{-ab} - \lim_{R\rightarrow \infty}\int_{\gamma_R} f(z)\, dz$$ We know that $\int_{\gamma_R} f(z)\, dz \in \mathbb{R}$ as $\int_{C_R} f(z)\, dz$ is real. Thus, $$\int_{-\infty}^\infty f(z)\, dz = \frac{\pi}{b} e^{-ab} - 0 = \frac{\pi}{b} e^{-ab}$$ $$\int^\infty_{-\infty} \frac{\cos(ax)}{x^2 + b^2} \,dx = \mathfrak{R}\left(\int_{-\infty}^\infty f(z)\, dz\right) = \frac{\pi}{b} e^{-ab}$$
$$\mathbf{An\;Attempt\;At\;The\;Real\;Solution}$$
Let $$I(b) = \int^\infty_{-\infty} \frac{\cos ax}{x^2 + b^2}\, dx$$ $$I'(b) = \frac{\partial}{\partial b}\int^\infty_{-\infty} \frac{\cos ax}{x^2 + b^2}\, dx = \int^\infty_{-\infty} \frac{\partial}{\partial b} \frac{\cos ax}{x^2 + b^2}\, dx = \int^\infty_{-\infty} -\frac{2b\cos ax}{(x^2 + b^2)^2}\, dx = -\int^\infty_{-\infty} \frac{\cos ax}{x^2 + b^2}\frac{2b}{x^2 + b^2}\, dx$$
Let $u = \frac{2b}{x^2 + b^2}$ and $dv = \frac{\cos ax}{x^2 + b^2} dx$. Then $du = -\frac{4bx}{(x^2 + b^2)^2} dx$ and $v = \int \frac{\cos ax}{x^2 + b^2}\, dx$.
Integrating by parts, we get $$I'(b) = -\int^\infty_{-\infty} \frac{\cos ax}{x^2 + b^2}\frac{2b}{x^2 + b^2}\, dx = -\left(\frac{2b}{x^2 + b^2}\int \frac{\cos ax}{x^2 + b^2}\, dx - \int \left(\int \frac{\cos ax}{x^2 + b^2}\, dx\right)\left(-\frac{4bx}{(x^2 + b^2)^2}\right) dx\right) \Bigg|^\infty_{-\infty} = -\frac{2b}{x^2 + b^2}I(b)\Bigg|^\infty_{-\infty} - \int^\infty_{-\infty} \left(\int \frac{\cos ax}{x^2 + b^2}\, dx\right)\left(\frac{4bx}{(x^2 + b^2)^2}\right) dx = - \int^\infty_{-\infty} \left(\int \frac{\cos ax}{x^2 + b^2}\, dx\right)\left(\frac{4bx}{(x^2 + b^2)^2}\right) dx$$
I'm stuck here. We could try substituting $\int \frac{\cos ax}{x^2 + b^2}\, dx = \int^x_{-\infty} \frac{\cos at}{t^2 + b^2}\, dt$, but I don't see how that would help.
