It is known that
$$\int\_ 0^{\infty}\frac {e^{-x - \frac {1} {x}}} {x} dx=2 K\_0(2),$$ but now I want to get the closed form approximate result of 
$$\int\_ 0^a\frac {e^{-x - \frac {1} {x}}} {x} dx.$$
I have searched the classic Table of Integrals, Series, and Products, but there is no pattern match this situation.
Is there some approaches to the problem?



Is there a way to get the closed form approximate result of $\int_ 0^a\frac {e^{-x - \frac {1} {x}}} {x}\, dx$
![](https://ilorentz.org/beenakker/MO/BesselK.png)


The blue curve is the desired integral $\int\_ 0^{a}\frac {e^{-x - \frac {1} {x}}} {x} dx$, the orange curve is the approximate answer $2K\_0(2)a^3(1+a^3)^{-1}$. 


You could use this for a global approximation. In the small-$a$ region the asymptotics is $e^{-1/a}(a-2a^2)$, which is quite accurate up to about $a=0.2$, see plot below. Incidentally, the small-$a$ and large-$a$ asymptotics (see Brendan McKay's comment) are the same upon replacement of $a$ by $1/a$.


![](https://ilorentz.org/beenakker/MO/BesselK_2.png)


