# Vardi's Integral

• Dec 30th 2012, 11:23 PM
Ozymandias
Vardi's Integral
Prove...

$\int_0^\infty \frac{1}{1+x^2}\ln \ln \frac{1}{x} dx= \frac{\pi}{2}\ln \left\{ \sqrt{2\pi}\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac {1}{4})}\right\}$

• Dec 31st 2012, 12:12 AM
sbhatnagar
Re: Vardi's Integral
The evaluation of such integrals involves a lot of number theory. We will need the Dirichlet L Function

$L(s)=1-\frac{1}{3^s}+\frac{1}{5^s}-\cdots = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$

The gamma function,

$\frac{\Gamma(s) (-1)^n}{(2n+1)^s}=\int_0^\infty t^{s-1}(-1)^n e^{-(2n+1)t}dt$

$\Gamma(s) L(s)=\int_0^\infty t^{s-1} \sum_{n=0}^\infty (-1)^n e^{-(2n+1)t} \ dt$

Now let $z=e^{-t}$

$\Gamma(s) L(s)=\int_01 \left( \ln\frac{1}{z}\right)^{s-1} \sum_{n=0}^\infty (-1)^{n} z^{-(2n+1)} \ \frac{dz}{z}$

$\Gamma(s) L(s)=\int_0^1\left( \ln\frac{1}{z}\right)^{s-1} \left( \frac{1}{1+z^2}\right) \ dz$

Differentiate with respect to the parameter $s$.

$\frac{d}{ds}\Gamma(s) L(s)=\int_0^1\left( \ln\frac{1}{z}\right)^{s-1} \left( \frac{1}{1+z^2}\right) \ln \ln \frac{1}{z}\ dz$

Hence we conclude

$\int_0^1 \frac{1}{1+x^2}\ln \ln \frac{1}{x} dx= \frac{d}{ds}\Gamma(s)L(s) \Bigg|_{s=1}$

Evaluating $d/ds (\Gamma(s) L(s))$ is it not a "walk in the park" and requires the help of number theory.