Evaluate this integrate $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\ln{(\ln{\tan{x}})}dx} $$
My friend tian_275461 proposed this integrate,but I have no idea about it.
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cant we use $\int logx = xlog(x) -x + constant$ identity? – Learner Mar 08 '13 at 12:12
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@Learner Not with this badboy – L. F. Mar 08 '13 at 12:13
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do we know it is converging ? – Dominic Michaelis Mar 08 '13 at 12:15
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For the record, this is known as Vardi's integral. There are other evaluations of it on this site. – L. F. Mar 13 '13 at 20:11
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Let $u = \ln \tan x$, so that $\frac{\pi}{4} < x< \frac{\pi}{2}$ is mapped to $u>0$, $x=\arctan(\exp(u))$ and $\mathrm{d}x = \frac{\mathrm{d}u}{2 \cosh(u)}$. Then $$ \int_{\pi/4}^{\pi/2} \ln( \ln(\tan x))\, \mathrm{d}x = \frac{1}{2} \int_0^\infty \frac{\ln (u)}{\cosh(u)} \mathrm{d}u = \frac{1}{2} \lim_{s \to 0^+}\frac{\mathrm{d}}{\mathrm{d} s} \int_0^\infty \frac{u^s}{\cosh(u)} \mathrm{d}u $$ The latter parametric integral is evaluated by expanding $\cosh(u)$ into exponential and using Euler's gamma-integral: $$\begin{eqnarray} \int_0^\infty \frac{u^s}{2\cosh(u)} \mathrm{d}u &=& \int_0^\infty u^s \frac{\exp(-u)}{1+\exp(-2u)}\mathrm{d}u = \sum_{n=0}^\infty (-1)^n \int_0^\infty u^{s} \exp(-(2n+1)u) \,\mathrm{d}u \\ &=& \Gamma(s+1) \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^{s+1}} = \Gamma(s+1) 2^{-2s-2} \left( \zeta\left(s+1,\frac{1}{4}\right) - \zeta\left(s+1,\frac{3}{4}\right) \right) \end{eqnarray} $$ Near $s=0$: $$ \zeta(1+s,a) = \frac{1}{s} - \psi(a) - \gamma_1(a) s + \mathcal{o}(s) $$ where $\psi(a)$ is the digamma function, and $\gamma_1(a)$ is the first generalized Stieltjes constant. Differentiating and taking the limit we have $$ \begin{eqnarray} \int_{\pi/4}^{\pi/2} \ln( \ln(\tan x))\, \mathrm{d}x &=& \frac{1}{4} \left( \left(\log(4) + \gamma\right)\left(\psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4}\right)\right) - \left(\psi_1\left(\frac{1}{4}\right)-\psi_1\left(\frac{3}{4}\right)\right)\right) \\ &=& \frac{\pi}{4} \log \left( \frac{4 \pi^3}{\Gamma\left(\frac{1}{4}\right)^4} \right) \approx -0.260443 \end{eqnarray} $$ where the latter equality is given my Mathematica.
Sasha
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In what situations (or rather, for what other functions) can we use the same trick of substituting $$\ln(u)=\lim_{s \to 0^+} u^s$$ – Vincent Tjeng Mar 08 '13 at 13:30
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Thanks, @GEdgar. What I meant was the trick $$\ln(u)= \frac{d}{ds} \lim_{s \to 0^+} u^s$$ – Vincent Tjeng Mar 08 '13 at 15:40
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Or find it in Gradshteyn-Rizhyk 4.371 $$ \int_0^\infty \frac{\log x}{\cosh x}\;dx = \pi\log\left[ \frac{\sqrt{2\pi}\;\Gamma(3/4)}{\Gamma(1/4)}\right] $$ half of that is your answer, agreeing with Sasha.
GEdgar
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