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Please help me to find a closed form for the following integral: $$\int_0^1\log\left(\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\right)\,{\mathrm d}x.$$

I was told it could be calculated in a closed form.

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Do you have any idea of what could be done? Did the person that tell you there was a closed form hint you in any way? Don't you know what the result should be? –  Pedro Tamaroff May 13 '13 at 18:45
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It may or may not help to realize $\log\left(\frac{1}{x} + \sqrt{\frac{1}{x^2} - 1}\right)$ as $\operatorname{arsech} x$. This isn't a hint, just an observation. –  Stahl May 13 '13 at 18:48
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@Stahl I was thinking about that too. –  Pedro Tamaroff May 13 '13 at 18:48
    
@PeterTamaroff Unfortunately, no hints were given, except that the closed form is quite simple, although might be not elementary. –  Laila Podlesny May 13 '13 at 18:49
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Here are some equivalent forms: Since $\text{sech}^{-1}(x)=\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^{2}}-1}\right),$ we are trying to evaluate $$\int_{0}^{1}\log\left(\text{sech}^{-1}(x)\right)dx.$$ Let $u=\text{sech}^{-1}x$ so that $x=\text{sech}(u)=\frac{1}{\cosh(u)}.$ Then $dx=d(\frac{1}{\cosh(u)})$ , and we are looking at $$-\int_{0}^{\infty}\log ud\left(\frac{1}{\cosh(u)}\right)$$ which equals $$\int_{0}^{\infty}\log u\frac{\sinh(u)}{\cosh(u)^{2}}du.$$ Using integration by parts, this becomes $$\lim_{a\rightarrow0}\left(\log a+\int_{a}^{\infty}\frac{\text{sech}(x)}{u}du\right).$$ –  Eric Naslund May 13 '13 at 19:14

2 Answers 2

up vote 44 down vote accepted
+50

Answer: $$\boxed{\displaystyle\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx=-\gamma-2\ln 2-2\ln\frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}}\tag{$\heartsuit$}$$


Derivation:

After the change of variables $x=\frac{1}{\cosh u}$ the integral becomes $$\int_0^{\infty}\ln u \frac{\sinh u}{\cosh^2 u}du,$$ as was noticed above by Eric. We would like to integrate by parts to kill the logarithm but we get two divergent pieces. To go around this, let us consider another integral, $$I(s)=\int_0^{\infty}u^s \frac{\sinh u}{\cosh^2 u}du,$$ with $s>0$. The integral we actually want to compute is equal to $I'(0)$, which will be later obtained in the limit.

Indeed, integrating once by parts one finds that \begin{align} I(s)&=s\int_0^{\infty}\frac{u^{s-1}du}{\cosh u}=s\cdot 2^{1-2 s}\Gamma(s)\left[\zeta\left(s,\frac14\right)-\zeta\left(s,\frac34\right)\right]=\\ &=2^{1-2 s}\Gamma(s+1)\left[\zeta\left(s,\frac14\right)-\zeta\left(s,\frac34\right)\right], \end{align} where $\zeta(s,a)=\sum_{n=0}^{\infty}(n+a)^{-s}$ denotes Hurwitz zeta function (in the way we have used its integral representaion (5) from here).

Now to get ($\heartsuit$), it suffices to use \begin{align} &\frac{\partial}{\partial s}\left[2^{1-2 s}\Gamma(s+1)\right]_{s=0}=-2\gamma-4\ln 2,\\ &\zeta\left(0,\frac14\right)-\zeta\left(0,\frac34\right)=\frac12, \\ &\frac{\partial}{\partial s}\left[\zeta\left(s,\frac14\right)-\zeta\left(s,\frac34\right)\right]_{s=0}=-\ln\frac{\Gamma(\frac34)}{\Gamma(\frac14)}. \end{align} [See formulas (10) and (16) on the same page].

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+1, Very nice solution. –  Eric Naslund May 13 '13 at 22:27
    
Beautiful answer ! +1 indeed. Cheers :) –  Claude Leibovici Jul 17 at 9:16

Find the integral $$I=\int_{0}^{1}\ln{\left(\ln{\left(\dfrac{1}{x}+\sqrt{\dfrac{1}{x^2}-1}\right)}\right)}dx$$ solution:since let $$x=e^{-y}$$ then $$I=\int_{0}^{\infty}e^{-y}\ln{\ln{\left(e^y+\sqrt{e^{2y}-1}\right)}}dy$$ so \begin{align*} &\int_{0}^{\infty}e^{-x}\ln{(\ln{(e^x+\sqrt{e^{2x}-1})})}dx=_{y=\ln{(e^x+\sqrt{e^{2x}-1})}}=2\int_{0}^{\infty}\dfrac{e^y(e^{2y}-1)}{(1+e^{2y})^2}\ln{y}dy\\ &=2\int_{0}^{\infty}\dfrac{e^{-y}(1-e^{-2y})}{(1+e^{-2y})^2}\ln{y}dy= 2\int_{0}^{\infty}e^{-y}(1-e^{-2y})\left(\sum_{n=1}^{\infty}(-1)^{n-1}ne^{-2y(n-1)}\right)\ln{y}dy\\ &=2\sum_{n=1}^{\infty}(-1)^{n-1}n\int_{0}^{\infty}\left(e^{-y(2n-1)}-e^{-y(2n+1)}\right)\ln{y}dy =2\sum_{n=1}^{\infty}(-1)^{n-1}\cdot n\left(-\dfrac{\gamma+\ln{(2n-1)}}{2n-1}+\dfrac{\gamma+\ln{(2n+1)}}{2n+1}\right)\\ &=2\gamma\sum_{n=1}^{\infty}(-1)^n\cdot n\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)+2\sum_{n=1}^{\infty}(-1)^n \cdot n\left(\dfrac{\ln{(2n-1)}}{2n-1}-\dfrac{\ln{(2n+1)}}{2n+1}\right)\\ &=\gamma\sum_{n=1}^{\infty}(-1)^n\left(\dfrac{2n}{2n-1}-\dfrac{2n}{2n+1}\right)+\sum_{n=1}^{\infty}(-1)^n \left(\dfrac{2n\ln{(2n-1)}}{2n-1}-\dfrac{2n\cdot\ln{(2n+1)}}{2n+1}\right)\\ &=\gamma\sum_{n=1}^{\infty}(-1)^n\left(\dfrac{1}{2n-1}+\dfrac{1}{2n+1}\right)+\sum_{n=1}^{\infty}(-1)^n\left( \dfrac{(2n-1+1)\ln{(2n-1)}}{2n-1}-\dfrac{(2n+1-1)\ln{(2n+1)}}{2n+1}\right)\\ &=-\gamma+\sum_{n=1}^{\infty}(-1)^n\ln{\dfrac{2n-1}{2n+1}}+\sum_{n=1}^{\infty}(-1)^n\left(\dfrac{\ln{(2n-1)}}{2n-1} +\dfrac{\ln{(2n+1)}}{2n+1}\right)=-\gamma+\sum_{n=1}^{\infty}(-1)^n\ln{\dfrac{2n-1}{2n+1}}\\ &=-\gamma-\sum_{n=1}^{\infty}\ln{\dfrac{4n-3}{4n-1}}+\sum_{n=1}^{\infty}\ln{\dfrac{4n-1}{4n+1}}=-\gamma+ \sum_{n=1}^{\infty}\ln{\dfrac{(n-1/4)^2}{(n-3/4)(n+1/4)}}\\ &=-\gamma+\lim_{N\to\infty}\ln{\left( \dfrac{((-\frac{1}{4}+1)(-\dfrac{1}{4}+2)\cdots(-\dfrac{1}{4}+N))^2}{((-\dfrac{3}{4}+1)(-\dfrac{3}{4}+2)\cdots (-\dfrac{3}{4}+N))((\dfrac{1}{4}+1)(\dfrac{1}{4}+2)\cdots(\dfrac{1}{4}+N))}\right)}\\ &=-\gamma+\ln{\dfrac{-3\Gamma{(-3/4)}\Gamma{(1/4)}}{\Gamma^2{(-1/4)}}} =-\gamma+\ln{\dfrac{4\Gamma^2{(1/4)}}{\Gamma^2(-1/4)}}=-\gamma+4\ln{\Gamma{\left(\dfrac{1}{4}\right)}} -3\ln{2}-2\ln{\pi} \end{align*}

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