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

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.

• 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
• 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
• @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
• 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

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

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].

• Beautiful answer ! +1 indeed. Cheers :) – Claude Leibovici Jul 17 '14 at 9:16
• Maybe this formula is better and shorter? $$-\gamma - \ln {\frac {8 \pi ^ 2} {\Gamma (\frac {1} {4}) ^ 4 }}$$ – Ivan Kochurkin Feb 16 '16 at 23:28
• @KvanTTT I guess it depends. In my research context, the ratio $\Gamma(z) / \Gamma(1-z)$ appears very often - there is even a special notation $\gamma(z)$ for it. – Start wearing purple Feb 17 '16 at 7:28

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*}

• Very nice! I upvoted. – user98186 Nov 29 '15 at 0:21

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.