Does anyone know how to calculate the following integral?

Following the analysis here:

$$\begin{align} F(x) &= \int_0^{\infty} \frac{dy}{y} e^{-y-\frac{x}{y}}\end{align}$$

Sub $u=y+\frac{x}{y}$, then

$$y = \frac12 \left (u \pm \sqrt{u^2-4 x}\right )$$ $$dy = \frac12 \left ( 1 \pm \frac{u}{\sqrt{u^2-4 x}} \right ) du$$

Then

$$\begin{align}F(x) &= \frac1{4 x} \int_{\infty}^{2 \sqrt{x}} du \left ( 1 - \frac{u}{\sqrt{u^2-4 x}} \right )\left (u + \sqrt{u^2-4 x}\right ) e^{-u} \\ &+ \frac1{4 x} \int_{2 \sqrt{x}}^{\infty} du \left ( 1 + \frac{u}{\sqrt{u^2-4 x}} \right )\left (u - \sqrt{u^2-4 x}\right )e^{-u}\\ &= 2 \int_{2 \sqrt{x}}^{\infty} du \frac{e^{-u}}{\sqrt{u^2-4 x}}\\ &= 2 \int_0^{\infty} dv \, e^{-2 \sqrt{x} \cosh{v}}\\ &= 2 K_0(2 \sqrt{x})\end{align}$$

By using recurrence relations for the $K_n$, we see that the above simple expression is equivalent to the one derived using differentiation under the integral sign.