Do you really need "contour" here? (2)

To save some effort, we can generalise it first.

Consider this integral: $I=\int_{-\infty }^{\infty }\frac{\cos{bx}}{a^{2}+x^{2}}\: dx.$ $\text{Let }u=\frac{1}{a^{2}+x^{2}}\, ,\: \: \frac{dv}{dx}=\cos{bx},$ $\Rightarrow \frac{du}{dx}=-\frac{2x}{(a^{2}+x^{2})^{2}}\, ,\: \: v=\frac{\sin{bx}}{b}\, ,$ $\text{Using Integration by Parts, giving}$ $I=\left [ \frac{\sin{bx}}{a^{2}+x^{2}} \right ]_{-\infty }^{\infty }+\frac{2}{b}\int_{-\infty }^{\infty }\frac{x\sin{bx}}{(a^{2}+x^{2})^{2}}\: dx.$ $\text{Let }f(x)=\frac{\sin{bx}}{a^{2}+x^{2}}\, ,\: \: \text{then we have }f(x)=f(-x),$ $\text{Thus }f(x)\: \: \text{is an even function and }\left [ \frac{\sin{bx}}{a^{2}+x^{2}} \right ]_{-\infty }^{\infty }=0.$ $\therefore I=\frac{2}{b}\int_{-\infty }^{\infty }\frac{x\sin{bx}}{(a^{2}+x^{2})^{2}}\: dx\, ,$ $\Rightarrow bI=2\int_{-\infty }^{\infty }\frac{x\sin{bx}}{(a^{2}+x^{2})^{2}}\: dx.\; \; \; \; \; \; [1]$ $\text{Differentiate both sides with respect to }b\text{ gives:}$ $b\, \frac{dI}{db}+I=2\int_{-\infty }^{\infty }\frac{x^{2}\cos{bx}}{(a^{2}+x^{2})^{2}}\, dx=2\left ( \int_{-\infty }^{\infty }\frac{\cos{bx}}{a^{2}+x^{2}}\: dx-\int_{-\infty }^{\infty }\frac{\cos{bx}}{(a^{2}+x^{2})^{2}} \right ),$ $\Rightarrow b\, \frac{dI}{db}+I=2I-2\int_{-\infty }^{\infty }\frac{\cos{bx}}{(a^{2}+x^{2})^{2}},$ $\Rightarrow b\, \frac{dI}{db}-I=2\int_{-\infty }^{\infty }\frac{\cos{bx}}{(a^{2}+x^{2})^{2}}.$ $\text{Differentiate both sides again with respect to }b\text{ gives:}$ $b\, \frac{d^{2}I}{db^{2}}=2\int_{-\infty }^{\infty }\frac{x\sin{bx}}{(a^{2}+x^{2})^{2}}.\; \; \; \; \; \; [2]$ $[1] - [2]\text{ gives }I-\frac{d^{2}I}{db^{2}}=0\:\: \Rightarrow I=A e^{ab}+Be^{-ab}.$ $\text{Since }\lim_{b\rightarrow \infty }I=\lim_{b\rightarrow \infty }\frac{2}{b}\int_{-\infty }^{\infty }\frac{x\sin{bx}}{(a^{2}+x^{2})^{2}}\: dx=0,\text{ we have }A=0.$ $\text{Also since setting }b=0\text{ gives }A+B=\int_{-\infty}^{\infty }\frac{1}{a^{2}+x^{2}}\, dx=\frac{\pi}{a}\: \: \Rightarrow B=\frac{\pi}{a}.$ $\text{Thus, } I=\frac{\pi}{a}\, e^{-ab}=\boxed{\frac{\pi}{ae^{ab}}}\: .$

Now, setting $a=1$ and $b=1$ gives us the answer to this particular problem, which is $\frac{\pi}{e}=\boxed{1.156}\: ,$ to 3 significant figures.

Since, assuming that $a > 0$ $\int_{-\infty}^\infty e^{-a|t|} e^{-ixt}\,dt \; = \; \int_0^\infty \left(e^{-(a+ix)t} + e^{-(a-ix)t}\right)\,dt \;=\; \frac{1}{a+ix} +\frac{1}{a-ix} \; = \; \frac{2a}{a^2 + x^2}$ for any $x \in \mathbb{R}$ , the theory of Fourier transforms tells us that $\frac{1}{2\pi} \int_{-\infty}^\infty \frac{2a}{a^2 + x^2}e^{ixt}\,dx \; = \; e^{-a|t|} \hspace{2cm} t \in \mathbb{R}$ (the function $e^{-a|t|}$ is sufficiently well-behaved for this to be true pointwise, and not just in a Lebesgue-integral sense). Putting $t=1$ , we see that $\int_{-\infty}^\infty \frac{\cos x}{a^2 + x^2}\,dx \; = \; \int_{-\infty}^\infty \frac{e^{i x}}{a^2 + x^2}\,dx \; = \; \tfrac{\pi}{a}e^{-a}$ Thus the answer to the question is $\tfrac{\pi}{e} = \boxed{1.156}$ .