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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?

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Have you tried changing variables to $u=\ln x$, and then going by parts? –  user54147 Dec 26 '12 at 19:08
@LevLivnev: I thought of $x=e^u$. I'll go that way right now. Thanks. –  Chris's sis the artist Dec 26 '12 at 19:09

2 Answers 2

up vote 3 down vote accepted

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! –  Chris's sis the artist Dec 26 '12 at 19:21

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


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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I'm not sure I understand your point. Does it prove that $\int_0^{1/e} \frac{\mathrm{dx}}{\sqrt{(\ln x)^2-1}}=K_{0}(1)$? –  Chris's sis the artist Dec 26 '12 at 19:18
This answer does not seem to address the OP's question at all. –  Sasha Dec 26 '12 at 19:22
@Chris'ssister: I added more materials. –  Mhenni Benghorbal Dec 27 '12 at 1:55

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