How does one show that these integrals are equal?
$$ \int_0^1 e^{A(x)}~dx $$
$$ \frac{1}{2}\int_{-\infty}^\infty e^{B(x)}~dx $$
$A(x)=\frac{1}{\log(x)}$
$B(x)=-\sqrt{4+x^2}$
I tried to represent the first integral in a different way and go from there:
$$I:=\int_0^1\exp\frac{1}{\log x}dx=\int_0^\infty\frac{1}{y^2}\exp -(y+y^{-1})dy=\frac12\int_0^\infty(1+1/y^2)\exp -(y+y^{-1})dy$$
where I used the substitution $y=-1/\log(x)$ and then averaged with $1/y.$
Then this is equivalent to the well known Bessel representation by making the substitution $y=e^{-t}$:
$$\int_{-\infty}^\infty\exp[-2\cosh t]\cosh tdt$$
From here I am stuck...I don't see how to proceed to obtain: $$ \frac{1}{2}\int_{-\infty}^\infty e^{B(x)}~dx $$
You've managed to show that $$ \int_0^1 e^{\frac{1}{\ln(x)}}\, \mathrm{d}x = \frac12\int_0^{\infty} e^{-\left(\frac{1}{y} +y \right)}\, \left( \frac{1}{y^2} +1\right)\mathrm{d}y $$ From here take the substitution $ u = \frac{1}{y} -y\implies \, -\mathrm{d}u= \left( \frac{1}{y^2} +1\right)\mathrm{d}y$ and noticing that $$ \left(\frac{1}{y}+y \right)^2 = \frac{1}{y^2} \color{blue}{+ 2} + y^2 = \frac{1}{y^2} \color{blue}{- 2} + y^2 \color{blue}{+4} =\left(\frac{1}{y}-y \right)^2 +4 $$ we conclude $$ \int_0^1 e^{\frac{1}{\ln(x)}}\, \mathrm{d}x = \frac12\int_0^{\infty} e^{-\left(\frac{1}{y} +y \right)}\, \left( \frac{1}{y^2} +1\right)\mathrm{d}y \overset{u = 1/y -y}{=} \frac{1}{2} \int_{-\infty}^{\infty} e^{-\sqrt{u^2 +4}}\, \mathrm{d}u $$ as desired.