Evaluating the integral $\int_0^\infty \frac{e^{-(t+\frac{1}{t})} dt}{t}$ How can the closed form of the following integral be evaluated?
$$\int_0^\infty \frac{e^{-(t+\frac{1}{t})} dt}{t}$$ 
I could not find any substitution nor factorisation that can be applied here. 
I am not getting any idea regarding the evaluation of this integral. I referred a similiar question posted here, yet I could not relate it with this definite integral. 
What should be done to find the above definite integral?

To elaborate on the answer in the comments, substitute $t = e^u$, giving
$$\int_0^\infty \frac{e^{-(t+\frac 1t)}}{t} dt = \int_{-\infty}^\infty  e^{-(e^u + e^{-u})} du \\ = 2\int_0^\infty e^{-2\cosh u} du \\ = 2K_0(2)$$
