Prove that $\int_0^{1/e} \frac{\mathrm{dx}}{\sqrt{(\ln x)^2-1}}=K_{0}(1)$ Prove that
$$\int_0^{1/e} \frac{\mathrm{dx}}{\sqrt{(\ln x)^2-1}}=K_{0}(1)$$
where $K_{n}(x)$ is the modified bessel function of the second kind.
Some hints/suggestions?
Thanks.
 A: Let $t = \ln(x)$. Then the integral becomes:
$$
   \int_{-\infty}^{-1} \frac{1}{\sqrt{t^2-1}} \mathrm{e}^{t} \mathrm{d} t \stackrel{u=-t}{=} \int_{1}^\infty \frac{\exp(-u)}{\sqrt{u^2-1}} \mathrm{d} u 
$$
Now compare this to eq. 10.3.28 of the DLMF handbook of special functions.
$$
  \int_{1}^\infty \frac{\exp(-u)}{\sqrt{u^2-1}} \mathrm{d} u  = \left.\frac{\Gamma\left(\nu+1/2\right)}{ \sqrt{\pi} \left(z/2\right)^{\nu}} K_\nu(z) \right|_{\nu=0, z=1} = K_0(1)
$$
A: Here is a related problem. Use the change of variable $x=e^{-(t+1)}$ and you will find that the new integrand behaves like $c\,t^{-1/2}$ as $t\to 0$.
Edit: 
Now, to evaluate the integral, we use the change of variables $x=e^{-(t+1)}$
$$ \int_0^{1/e} \frac{\mathrm{dx}}{\sqrt{(\ln x)^2-1}}=\frac{1}{e}\int _{0}^{\infty }\!{\frac {{{\rm e}^{-t}}}{\sqrt {t}\sqrt {t+2}}
}{dt} . $$
We will consider the more general integral 
$$ \int _{0}^{\infty }\!{\frac {{{\rm e}^{-st}}}{\sqrt {t}\sqrt {t+2}}
}{dt}. $$
The above integral is nothing but the Laplace transform of the function $\frac{1}{\sqrt{t^2+2t}}$ which is given by
$$e^sK_0(s).$$
So, we have
$$ \frac{1}{e}\int _{0}^{\infty }\!{\frac {{{\rm e}^{-t}}}{\sqrt {t}\sqrt {t+2}}
}{dt}=\frac{1}{e}\lim_{s\to 1} e^sK_{0}(s)=K_{0}(1). $$
