TITLE: Closed form for $\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx$
QUESTION [68 upvotes]: 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.

REPLY [2 votes]: Following the comments above, there is another path with digamma function. Recalling this identity of Euler-Mascheroni constant
$$-\int_{0}^{\infty} {e^{-u}\ln u \>\mathrm{d}u} = \gamma$$
and
$$\frac{\mathrm{d}(e^{-u}\tanh u)}{\mathrm{d}u} = e^{-u} - \frac{\sinh u}{\cosh^{2}\!u}$$
We deduce
$$\begin{aligned}
I + \gamma 
& = \int_{0}^{\infty} {\left( \frac{\sinh u}{\cosh^{2}\!u}-e^{-u} \right)\ln u \>\mathrm{d}u}\\
& = -(e^{-u}\tanh u\ln u) \bigr|_{u=0}^{\infty} + \int_{0}^{\infty} {\frac{e^{-u}\tanh u}{u} \mathrm{d}u}\\
& = \int_{0}^{\infty} {\frac{e^{-u}\tanh u}{u} \mathrm{d}u}
\end{aligned}$$
Introduce a parameterized integral
$$J(a) = \int_{0}^{\infty} {\frac{e^{-au}\tanh u}{u} \mathrm{d}u}$$
Take derivative of $J(a)$
$$\begin{aligned}
\frac{\mathrm{d}J(a)}{\mathrm{d}a}
& = -\int_{0}^{\infty} {e^{-au}\tanh u \>\mathrm{d}u}\\
& = \int_{0}^{\infty} {e^{-au} \>\mathrm{d}u} - \int_{0}^{\infty} {e^{-au}(1+\tanh u) \>\mathrm{d}u}
\end{aligned}$$
where we have
$$\begin{aligned}
\int_{0}^{\infty} {e^{-au}(1+\tanh u) \>\mathrm{d}u}
& = 2\int_{0}^{\infty} {\frac{e^{-au}}{1+e^{-2u}} \mathrm{d}u}\\
& = 2\int_{0}^{\infty} {\frac{e^{-au}(1-e^{-2u})}{1-e^{-4u}} \mathrm{d}u}\\
& = \frac1{2} \int_{0}^{\infty} {\frac{e^{-\tfrac{a}{4}u}}{1-e^{-u}} \mathrm{d}u} - \frac1{2} \int_{0}^{\infty} {\frac{e^{-\tfrac{a+2}{4}u}}{1-e^{-u}} \mathrm{d}u}
\end{aligned}$$
Using the integral representation of digamma function
$$\psi(z) = \int_{0}^{\infty} {\left( \frac{e^{-u}}{u} - \frac{e^{-zu}}{1-e^{-u}} \right) \mathrm{d}u}$$
we have
$$\frac{\mathrm{d}J(a)}{\mathrm{d}a} = \frac1{a} + \frac1{2}\psi\bigg(\frac{a}{4}\bigg) - \frac1{2}\psi\left(\frac{a+2}{4}\right)$$
with $\lim_{a\to\infty}J(a)=0$ and $I+\gamma=J(1)$
$$\begin{aligned}
J(1) & = -\int_{1}^{\infty} {J'(a) \>\mathrm{d}a}\\
& = -\left( \ln a + 2\ln\Gamma\bigg(\frac{a}{4}\bigg) - 2\ln\Gamma\left(\frac{a+2}{4}\right) \right) \biggr|_{a=1}^{\infty}
\end{aligned}$$
Notice the asymptotic series of $\ln\Gamma(z) = (z-\tfrac1{2})\ln z - z + \ln2\pi + o(z^{-1})$ which indicates
$$\lim_{a\to\infty} {\left( \ln a + 2\ln\Gamma\bigg(\frac{a}{4}\bigg) - 2\ln\Gamma\left(\frac{a+2}{4}\right) \right)} = 2\ln2$$
thus
$$I + \gamma = -2\ln2 + 2\ln\frac{\Gamma(1/4)}{\Gamma(3/4)}$$
Recalling reflection formula, we can finally deduce
$$I = -\gamma - 3\ln2 - 2\ln\pi + 4\ln\Gamma\left(\frac1{4}\right)$$

(Edit for another path)
Occasionally, I find a direct solution which is almost equivalent to the method I used above, where let $\frac1{t} = \frac1{x} + \sqrt{\frac1{x^{2}}-1}$, which gives
$$
 x = \frac{2t}{1+t^{2}}, \quad \mathrm{d}x = -\frac{2(t^{2}-1)}{(t^{2}+1)^{2}} \mathrm{d}t
 $$
hence
$$
 \int_{0}^{1} {\ln \left(\ln \left(\frac1{x} + \sqrt{\frac1{x^{2}}-1} \right) \right) \mathrm{d}x} = -2\int_{0}^{1} {\frac{t^{2}-1}{(t^{2}+1)^{2}} \ln\left(\ln\left(\frac1{t}\right)\right) \mathrm{d}t}
 $$
on the other hand, with integration by parts
$$
 \begin{aligned}
 \int_{0}^{1} {\frac{t^{2}-1}{t^{2}+1} \frac{\mathrm{d}t}{\ln t}}
 & = \frac{t(t^{2}-1)}{t^{2}+1}\ln\left(\ln\left(\frac1{t}\right)\right)\biggr|_{t=0}^{1} - \int_{0}^{1} {\frac{t^{4}+4t^{2}-1}{(t^{2}+1)^{2}} \ln\left(\ln\left(\frac1{t}\right)\right) \mathrm{d}t}\\
 & = -\int_{0}^{1} {\ln\left(\ln\left(\frac1{t}\right)\right) \mathrm{d}t} - 2\int_{0}^{1} {\frac{t^{2}-1}{(t^{2}+1)^{2}} \ln\left(\ln\left(\frac1{t}\right)\right) \mathrm{d}t}
 \end{aligned}
 $$
thus
$$
 \int_{0}^{1} {\ln \left(\ln \left(\frac1{x} + \sqrt{\frac1{x^{2}}-1} \right) \right) \mathrm{d}x} = \int_{0}^{1} {\ln\left(\ln\left(\frac1{t}\right)\right) \mathrm{d}t} + \int_{0}^{1} {\frac{t^{2}-1}{t^{2}+1} \frac{\mathrm{d}t}{\ln t}}
 $$
the first item is literally $-\gamma$, the second can be find in this post, where, actually, the integral is cracked by similar fashion used in this post above.