Closed form for $\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx$

Question

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.

Answer 1

$$\boxed{\displaystyle\int_0^1\log\log\left(\frac1x+\sqrt{\frac1{x^2}-1}\right)\mathrm dx=-\gamma-2\ln\frac{2\Gamma(3/4)}{\Gamma(1/4)}}\tag{$\heartsuit$}$$

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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].

Answer 2

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*}

Answer 3

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

$$I = -\gamma - 3\ln2 - 2\ln\pi + 4\ln\Gamma\left(\frac1{4}\right)$$

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(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.