Closed form for $\int_0^1 e^{\frac{1}{\ln(x)}}dx$?

Question

I want to evaluate and find a closed form for this definite integral:$$\int_0^1 e^{\frac{1}{\ln(x)}}dx.$$

I don't know where to start. I've tried taking the natural logarithm of the integral, substitution and expressing the integrand in another way but they haven't led anywhere. The approximation should be between $.2$ and $.3,$ and is probably a transcendental number.

Thanks.

Answer 1

Let's verify Robert Israel's find. Observe that

$$I:=\int_0^1\exp\frac{1}{\ln x}dx=\int_0^\infty\frac{1}{y^2}\exp -(y+y^{-1})dy=\frac12\int_0^\infty(1+1/y^2)\exp -(y+y^{-1})dy,$$where we have substituted $y=-1/\ln x$ and then averaged with $y\mapsto 1/y$.

In view of the integral $K_\alpha(x)=\int_0^\infty\exp[-x\cosh t]\cosh(\alpha t)dt$ and the substitution $y=\exp -t$, $$2K_1(2)=\int_{-\infty}^\infty\exp[-2\cosh t]\cosh tdt=\frac12\int_0^\infty\exp [-y-1/y](1+1/y^2)dy.$$