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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.

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    @LevLivnev: I thought of $x=e^u$. I'll go that way right now. Thanks.2012-12-26

2 Answers 2

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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) $

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    just brilliant!2012-12-26
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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). $

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    @Chris'ssister: I added more materials.2012-12-27