That looks ugly and beautiful simultaneously

We will solve via a contour integration, using a semi-circular anti-clockwise contour called $C$ , with the straight part on the real axis, and the curved part called $\gamma_R$ . We will have radius $R>1$ as there will be poles at $\pm i$ .

Let $f(z) = \dfrac {e^{iz}}{z^2+1}$ . We have

$\displaystyle \oint_{C} f(z) dz = \int_{-R}^R f(x) dx + \int_{\gamma_R} f(z) dz$

We will now try to determine the poles of $f(z)$ . Since $e^{iz}$ is always differentiable in the complex plane, it won't contain any poles. However, when $z^2+1=0$ , $z=\pm i$ , so $f$ has poles at $\pm i$ . Notice that only one of these poles lies inside $C$ , and the other one doesn't, so we only need to worry about one of them (WLOG $i$ is inside $C$ ).

We will now try to solve the LHS by using the Residue Theorem. First, we will find the residue of $f(z)$ at $z=i$ .

$\begin{aligned} \displaystyle \lim_{z\to i} (z-i) \dfrac {e^{iz}}{z^2+1} &= \lim_{z\to i} \dfrac {e^{iz}}{z+i}\\ &=\dfrac{1}{2ei}\\ \end{aligned}$

Now, we can apply the Residue Theorem to get

$\begin{aligned} \displaystyle \oint_C f(z) dz &= 2\pi i \left ( \dfrac{1}{2ei} \right )\\ &= \dfrac {\pi}{e}\\ \end{aligned}$

We will now evaluate the integral furthest to the right using the Estimation Lemma. By the Estimation Lemma, we have

$\left | \displaystyle \int_{\gamma_R} f(z) dz \right |\leq \text{length }(\gamma_R) \cdot \max_{z \in \gamma_R} |f(z)|$

Now, the length of $\gamma_R$ is just $\pi R$ , because it is the perimeter of half a circle (a circle's circumference is $2\pi R$ ). Now we just need to determine $\displaystyle \max_{z \in \gamma_R} |f(z)|$ .

Firstly, by the reverse Triangle Inequality $|z^2 + 1| \geq |z|^2 - 1$ . Now, $|z|=R$ whenever $z$ is on $\gamma_R$ because it is on the curved side of a semicircle, so $|z^2 + 1| \geq R^2 - 1$ . Thus, $\dfrac {1}{|z^2 + 1|} \leq \dfrac {1}{R^2 - 1}$ .

Now, we will try to bound the numerator $e^{iz}$

$\begin{aligned} |e^{iz}| &= |e^{i(\Re(z) + \Im(z)i)}|\\ & = |e^{\Re(z)i}| |e^{- \Im(z)}| \\ &= |e^{- \Im(z)}|\\ &= e^{-\Im(z)}\\ &\leq 1\\ \end{aligned}$

The third line comes from the fact that since $\Re(z)$ is a real number, it follows that $e^{i\Re(z)} = \cos(\Re(z)) + i \sin(\Re(z))$ , so $|e^{i\Re(z)}| = 1$ . The fourth line is because the exponential term is always non-negative, so we don't need an absolute value sign. In the fifth line, we used the fact that $\gamma_{R}$ is defined to have $\Im(z) \geq 0$ .

Thus,

$\dfrac {|e^{iz}|}{|z^2+1|} \leq \dfrac {1}{R^2-1}$

Therefore, we have

$\left | \displaystyle \int_{\gamma_R} f(z) dz \right | \leq \dfrac {\pi R}{R^2 - 1}$

As $R$ tends to infinity, we find that the RHS tends to 0, so the LHS is just 0. We are now left with

$\begin{aligned} \displaystyle \int_{-R}^R f(x) dx + \int_{\gamma_R} f(z) dz &= \oint_{C} f(z) dz\\ \int_{-\infty}^{\infty} f(x) dx &= \dfrac {\pi}{e}\\ \int_{-\infty}^{\infty} \dfrac {e^{ix}}{x^2+1} dx &= \dfrac {\pi}{e}\\ \int_{-\infty}^{\infty} \dfrac{\cos x + i \sin x}{x^2 + 1} dx &= \dfrac {\pi}{e}\\ \int_{-\infty}^{\infty} \dfrac {\cos x}{x^2 +1} dx + i \int_{-\infty}^{\infty} \dfrac {\sin(x)}{x^2 + 1} dx &= \dfrac {\pi}{e}\\ \end{aligned}$

Note that since $\sin(x)$ is an odd function, the integral of it must be 0. Thus, we are left with

$\int_{-\infty}^{\infty} \dfrac {\cos x}{x^2 + 1} dx = \dfrac {\pi}{e}$

For a Fourier transform approach and a Laplace transform approach, see Mark Henning's solution here and my solution (in the same link), respectively.

There exist at least another two solutions, by Mark Hennings and me (and a generalisation of $\, \int_{-\infty }^{\infty }\frac{\cos{bx}}{a^{2}+x^{2}}\: dx$ is also given), in here .