# Need help with $\int_0^\infty e^{-x}\ln\ln\left(e^x+\sqrt{e^{2x}-1}\right)\,dx$

I need help with this integral: $$\int_0^\infty e^{-x}\ln\ln\left(e^x+\sqrt{e^{2x}-1}\right)\,dx\approx0.20597312051214...$$ Is it possible to evaluated it in a closed form?

• What are your thoughts on the matter? Where did you encounter the ingetral, and why is it interesting? Commented Nov 25, 2013 at 4:00
• You can get it into a more manageable form by making the substitution $u = e^x$ (with $du=e^x dx$) to obtain the integral $\int_1^{\infty} \frac{1}{u^2}\log(\log(u+\sqrt{u^2+1}))du$. But from here I'm not sure where you could go. If there was only a single logarithm, you could have maybe related it back to trigonometric functions (I think) but the composed logarithms makes things trickier. Commented Nov 25, 2013 at 4:06
• $\displaystyle\int_0^\infty e^{-x}\ln\left(e^x+\sqrt{e^{2x}-1}\right)\,dx=\frac\pi2$, but here only one $\ln$ is present. Commented Nov 25, 2013 at 4:06
• @CarlMummert What a pessimistic comment... All integrals are interesting. Commented Nov 25, 2013 at 4:08
• @Vladimir Reshetnikov: that is possible, but many users on this site feel that questions should do more than merely state a problem - they should include some amount of context, as described at meta.math.stackexchange.com/questions/9959/… . Commented Nov 25, 2013 at 4:12

A useful identity here is ${\rm arccosh\,} z = \ln( z + \sqrt{z^2 -1} )$. Therefore $$\int_0^{\infty} e^{-x} \ln \ln( e^x + \sqrt{e^{2x}-1}) \;dx= \int_0^{\infty} e^{-x}\ln {\rm arccosh\,} e^x \;dx =\int_1^{\infty} \frac{\ln {\rm arccosh\,} y}{y^2} \;dy\,$$ Change integration variable $z={\rm arccosh\,}y$, $$= \int_0^{\infty} \frac{\ln z \sinh z}{ \cosh^2 z} \;dz\,.$$ Integrate by parts and use the definition of the Euler's $\gamma$ constant and $\Gamma$ function see here, $$= -\gamma + \ln\left[\frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma^2\left(\frac{3}{4}\right)} \right]\,.$$ We can further simplify this using $\Gamma(1-z)\Gamma(z)=\pi/\sin \pi z$ and $\Gamma(z)\Gamma(z+\frac{1}{2})=2^{1-2z}\sqrt{\pi}\Gamma(2z)$ for $z=3/4$ and $\sin (3\pi/4)=1/\sqrt{2}$. This gives $$=-\gamma - 3\ln 2 - 2 \ln \pi + 4 \ln \Gamma(1/4)\,.$$

• Quite fine. UpVote $0$ k. Commented Nov 25, 2013 at 5:46
• Can you be more explicit about how you integrate by parts to obtain the second-to-last line? Commented Nov 25, 2013 at 6:14
• @mjqxxxx For the integration by parts to obtain the second-to-last line see math.stackexchange.com/questions/390640/… Commented Nov 26, 2013 at 19:35

$$\int\limits_{0}^{+\infty }{e^{-x}\ln \ln \left( e^{x}+\sqrt{e^{2x}-1} \right)dx}=\int\limits_{0}^{+\infty }{e^{-x}\ln \left( \cosh ^{-1}e^{x} \right)dx}=\int\limits_{1}^{+\infty }{\frac{\ln \left( \cosh ^{-1}e^{x} \right)}{x^{2}}dx}$$

$$=\int\limits_{0}^{+\infty }{\frac{\sinh x\ln x}{\cosh ^{2}x}dx}$$

Consider $$F\left( s \right)=\int\limits_{0}^{+\infty }{\frac{x^{s}\sinh x}{\cosh ^{2}x}dx}$$ and note that $$F'\left( 0 \right)=\int\limits_{0}^{+\infty }{\frac{\sinh x\ln x}{\cosh ^{2}x}dx}=\int\limits_{0}^{+\infty }{e^{-x}\ln \ln \left( e^{x}+\sqrt{e^{2x}-1} \right)dx}$$

then $$F\left( s \right)=\int\limits_{0}^{+\infty }{\frac{x^{s}\sinh x}{\cosh ^{2}x}dx}=s\int\limits_{0}^{+\infty }{\frac{x^{s-1}}{\cosh x}dx}=2s\int\limits_{0}^{+\infty }{\frac{e^{-x}x^{s-1}}{1+e^{-2x}}dx}$$

$$=2s\int\limits_{0}^{+\infty }{e^{-x}x^{s-1}\sum\limits_{n=0}^{+\infty }{\left( -e^{-2x} \right)^{n}}dx=}2s\sum\limits_{n=0}^{+\infty }{\left( -1 \right)^{n}\int\limits_{0}^{+\infty }{e^{-x-2nx}x^{s-1}dx}}$$

$$=2s\Gamma \left( s \right)\sum\limits_{n=0}^{+\infty }{\frac{\left( -1 \right)^{n}}{\left( 1+2n \right)^{s}}}=2^{1-s}s\Gamma \left( s \right)\sum\limits_{n=0}^{+\infty }{\frac{\left( -1 \right)^{n}}{\left( \frac{1}{2}+n \right)^{s}}}$$

$$S=\sum\limits_{n=0}^{+\infty }{\frac{\left( -1 \right)^{n}}{\left( \frac{1}{2}+n \right)^{s}}}=\frac{\left( -1 \right)^{0}}{\left( \frac{1}{2}+0 \right)^{s}}+\frac{\left( -1 \right)^{1}}{\left( \frac{1}{2}+1 \right)^{s}}+\frac{\left( -1 \right)^{2}}{\left( \frac{1}{2}+2 \right)^{s}}+\frac{\left( -1 \right)^{3}}{\left( \frac{1}{2}+3 \right)^{s}}+\frac{\left( -1 \right)^{4}}{\left( \frac{1}{2}+4 \right)^{s}}+\frac{\left( -1 \right)^{5}}{\left( \frac{1}{2}+5 \right)^{s}}+...$$

$$=\left( \frac{1}{\left( \frac{1}{2} \right)^{s}}+\frac{1}{\left( \frac{1}{2}+2 \right)^{s}}+\frac{1}{\left( \frac{1}{2}+4 \right)^{s}}+... \right)-\left( \frac{1}{\left( \frac{1}{2}+1 \right)^{s}}+\frac{1}{\left( \frac{1}{2}+3 \right)^{s}}+\frac{1}{\left( \frac{1}{2}+5 \right)^{s}}+... \right)$$

$$=\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( \frac{4n+1}{2} \right)^{s}}}-\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( \frac{4n+3}{2} \right)^{s}}}=\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( 2n+\frac{1}{2} \right)^{s}}}-\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( 2n+\frac{3}{2} \right)^{s}}}$$

$$=\frac{1}{2^{s}}\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( n+\frac{1}{4} \right)^{s}}}-\frac{1}{2^{s}}\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( n+\frac{3}{4} \right)^{s}}}=\frac{1}{2^{s}}\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)$$

$$F\left( s \right)=2^{1-s}s\Gamma \left( s \right)\sum\limits_{n=0}^{+\infty }{\frac{\left( -1 \right)^{n}}{\left( \frac{1}{2}+n \right)^{s}}}=2^{1-2s}\Gamma \left( s+1 \right)\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)$$

$$=2^{1-2s}\Gamma \left( s+1 \right)\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)$$

$$F'\left( s \right)=\frac{d}{ds}2^{1-2s}\Gamma \left( s+1 \right)\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)$$

$$=\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)\frac{d}{ds}\left( 2^{1-2s}\Gamma \left( s+1 \right) \right)+2^{1-2s}\Gamma \left( s+1 \right)\frac{d}{ds}\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)$$

$$=2^{1-2s}\Gamma \left( s+1 \right)\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)\left( \psi ^{\left( 0 \right)}\left( s+1 \right)-2\log 2 \right)+2^{1-2s}\Gamma \left( s+1 \right)\frac{d}{ds}\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)$$

where $$\zeta \left( s,q \right)=\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( q+n \right)^{s}}}$$ $$\frac{d}{ds}\zeta \left( s,\frac{1}{4} \right)=\frac{d}{ds}\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( n+\frac{1}{4} \right)^{s}}}=-\left( \frac{\log \left( \frac{1}{4} \right)}{\left( \frac{1}{4} \right)^{s}}+\frac{\log \left( 1+\frac{1}{4} \right)}{\left( 1+\frac{1}{4} \right)^{s}}+\frac{\log \left( 2+\frac{1}{4} \right)}{\left( 2+\frac{1}{4} \right)^{s}}+\frac{\log \left( 3+\frac{1}{4} \right)}{\left( 3+\frac{1}{4} \right)^{s}}+... \right)$$

$$\frac{d}{ds}\zeta \left( s,\frac{3}{4} \right)=\frac{d}{ds}\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( n+\frac{3}{4} \right)^{s}}}=-\left( \frac{\log \left( \frac{3}{4} \right)}{\left( \frac{3}{4} \right)^{s}}+\frac{\log \left( 1+\frac{3}{4} \right)}{\left( 1+\frac{3}{4} \right)^{s}}+\frac{\log \left( 2+\frac{3}{4} \right)}{\left( 2+\frac{3}{4} \right)^{s}}+\frac{\log \left( 3+\frac{3}{4} \right)}{\left( 3+\frac{3}{4} \right)^{s}}+... \right)$$

$$F'\left( 0 \right)=2\Gamma \left( 1 \right)\left( \zeta \left( 0,\frac{1}{4} \right)-\zeta \left( 0,\frac{3}{4} \right) \right)\left( \psi ^{\left( 0 \right)}\left( 1 \right)-2\log 2 \right)+2\Gamma \left( 1 \right)\frac{d}{ds}\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)\left| _{s=0} \right.$$

$$F'\left( 0 \right)=-\left( \gamma +2\log 2 \right)+2\frac{d}{ds}\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)\left| _{s=0} \right.$$

$$=-\left( \gamma +2\log 2 \right)-2\log \left( \frac{\Gamma \left( \frac{3}{4} \right)}{\Gamma \left( \frac{1}{4} \right)} \right)=-\left( \gamma +2\log 2 \right)-2\log \left( \frac{\sqrt{2}\pi }{\Gamma ^{2}\left( \frac{1}{4} \right)} \right)$$

$$=-\gamma -3\log 2-2\log \pi +4\log \Gamma \left( \frac{1}{4} \right)$$

Done (:

This is a partial answer only.

Let $x=\log u$; then $dx=du/u$, and $$I=\int_{1}^{\infty} \frac{1}{u^2}\log\log(u+\sqrt{u^2-1})du.$$ Next let $u=\cosh v$; then $du=\sinh v dv$, and $u+\sqrt{u^2-1}=\cosh v + \sinh v=e^v$, so $$I=\int_{0}^{\infty}\frac{\sinh v \log v}{\cosh^2 v} dv=\int_{0}^{a}\log{v}\cdot d\left(1-\frac{1}{\cosh v}\right) - \int_{a}^{\infty}\log v \cdot d\left(\frac{1}{\cosh v}\right),$$ where we can choose $a\in(0,\infty)$. Now integrate by parts to get $$I=\log{a}\left(1-\frac{1}{\cosh a}\right)-\int_{0}^{a}\left(\frac{1}{v}-\frac{1}{v\cosh v}\right)dv +\frac{\log a}{\cosh a} +\int_{a}^{\infty}\frac{dv}{v\cosh v}= \\\log a - \int_{0}^{a}\left(\frac{1}{v}-\frac{1}{v\cosh v}\right)dv+\int_{a}^{\infty}\frac{dv}{v \cosh v}.$$ If we let $a\rightarrow 0$, this eliminates the middle term: $$I=\lim_{a\rightarrow 0+}\left[\log{a} + \int_{a}^{\infty}\frac{dv}{v\cosh v}\right].$$ Perhaps someone can take it from here?