Compute integral $\int_{-\infty}^{\infty}\frac{e^{-ixt}}{\sqrt{1+x^2}} dx$

Yes, this integral may be done using complex analysis. Sort of. What we end up with are two different representations of a modified Bessel function. That said, the complex analysis route results in a transformation of the oscillatory Fourier integral - which is harder to compute - into a very quickly convergent integral representation.

Consider the contour intgeral

$$\oint_C dz \frac{e^{i t z}}{\sqrt{1+z^2}} $$

where $C$ is the following contour:

Note that the radius of the large circular arcs is $R$ and the small circle about the branch point at $z=i$ is $\epsilon$. Note that we avoid crossing the branch cut along the imaginary axis. By Cauchy's theorem, this contour integral is zero. (Why?) On the other hand, the contour integral is equal to, after parametrizing the various pieces:

$$\int_{-R}^R dx \frac{e^{i t x}}{\sqrt{1+x^2}} + i R \int_0^{\pi/2} d\theta \, e^{i \theta} \frac{e^{i t R e^{i \theta}}}{\sqrt{1+R^2 e^{i 2 \theta}}} \\ + i \int_R^{1+\epsilon} dy \, \frac{e^{-t y}}{i \sqrt{y^2-1}} + i \epsilon \int_{\pi/2}^{-3 \pi/2} d\phi \, e^{i \phi} \frac{e^{i t (i+\epsilon e^{i \phi})}}{\sqrt{1+(i+\epsilon e^{i \phi})^2}} \\ + i \int_{1+\epsilon}^R dy \, \frac{e^{-t y}}{-i \sqrt{y^2-1}}+ i R \int_{\pi/2}^{\pi} d\theta \, e^{i \theta} \frac{e^{i t R e^{i \theta}}}{\sqrt{1+R^2 e^{i 2 \theta}}}$$

We now consider the limits as $R \to \infty$ and $\epsilon \to 0$. We first note that we may combine the second and sixth integrals and show that they vanish in this limit when $t \gt 0$. Consider the absolute value of the combined integral and it is bounded from above as follows:

$$\frac{2 R}{\sqrt{R^2-1}} \int_0^{\pi/2} d\theta \, e^{-t R \sin{\theta}} \le \frac{2 R}{\sqrt{R^2-1}} \int_0^{\pi/2} d\theta \, e^{-2 t R \theta/\pi} \le \frac{\pi}{ t \sqrt{R^2-1}}$$

which clearly vanishes as $R \to \infty$. It is straightforward to show that the fourth integral vanishes as $\epsilon \to 0$. This, because the contour integral is zero, we are left with the following expression for the original Fourier integral:

$$\int_{-\infty}^{\infty} dx \, \frac{e^{i x t}}{\sqrt{1+x^2}} = 2 \int_1^{\infty} dy \, \frac{e^{-t y}}{\sqrt{y^2-1}} = 2 \int_0^{\infty} du \, e^{-t \cosh{u}}$$

The latter integral is $2 K_0(t)$. That said, note that the integral we obtained is a far easier one to handle numerically if it came to it. Accordingly, complex analysis is very useful in general for converting oscillatory Fourier integrals that are hard to evaluate numerically into exponential integrals that are very easy.

Note also that the above required $t \gt 0$. When $t \lt 0$, we must use a different contour below the real axis, but the result is the same. Therefore,

$$\int_{-\infty}^{\infty} dx \, \frac{e^{i x t}}{\sqrt{1+x^2}} = 2 K_0(|t|)$$

$\frac{e^{-ixt}}{\sqrt{1+x^2}}$ does not belong to $L^1(\mathbb{R})$, but $$ \lim_{M\to +\infty}\int_{-M}^{M}\frac{e^{-ixt}}{\sqrt{1+x^2}}\,dx =2\, K_0(|t|)$$ where $K_0$ is a modified Bessel function of the second kind (see eq 11). You can prove such identity by noticing that for $t>0$ both sides are solutions of the same differential equation.