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

To find the integral $$I=\int_{0}^{1}\ln{\left(\ln{\left(\dfrac{1}{x}+\sqrt{\dfrac{1}{x^2}-1}\right)}\right)}dx,$$ 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*} I&=\int_{0}^{\infty}e^{-x} \ln{\left(\ln{\left(e^x+\sqrt{e^{2x}-1}\right)}\right)}dx, \quad \text{ let } y=\ln{\left(e^x+\sqrt{e^{2x}-1}\right)}\\ &=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) \\ &\qquad +\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( \frac{ \left[ \left(-\frac{1}{4}+1\right)\left(-\frac{1}{4}+2\right) \cdots\left(-\frac{1}{4}+N\right) \right]^2 }{ \left[ \left(-\frac{3}{4}+1\right)\left(-\frac{3}{4}+2\right) \cdots \left(-\frac{3}{4}+N\right) \right] \left[ \left(\frac{1}{4}+1\right)\left(\frac{1}{4}+2\right) \cdots\left(\frac{1}{4}+N\right) \right] }\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.

Answer 4

To calculate $$\int_0^1\log\left(\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\right)\,{\mathrm d}x.$$

let $x=e^y$ then

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

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

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 (:

Answer 5

To calculate $$\int_0^1\log\left(\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\right)\,{\mathrm d}x.$$

let $x=e^y$ then

$$\displaystyle{\int\limits_0^\infty {{e^{ - x}}\log \left( {\log \left( {{e^x} + \sqrt {{e^{2x}} - 1} } \right)} \right)dx} = \mathop = \limits^{\log \left( {{e^x} + \sqrt {{e^{2x}} - 1} } \right) = y} = 2\int\limits_0^\infty {\frac{{{e^y}\left( {{e^{2y}} - 1} \right)}}{{{{\left( {1 + {e^{2y}}} \right)}^2}}}\log \left( y \right)dy} = }$$

$\displaystyle{ = 2\int\limits_0^\infty {\frac{{{e^{ - y}}\left( {1 - {e^{ - 2y}}} \right)}}{{{{\left( {1 + {e^{ - 2y}}} \right)}^2}}}\log \left( y \right)dy} = 2\int\limits_0^\infty {{e^{ - y}}\left( {1 - {e^{ - 2y}}} \right)\left( {\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}} \cdot n \cdot } {e^{ - 2y\left( {n - 1} \right)}}} \right)\log \left( y \right)dy} = }$

$\displaystyle{ = 2\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}} \cdot n \cdot \int\limits_0^\infty {\left( {{e^{ - y\left( {2n - 1} \right)}} - {e^{ - y\left( {2n + 1} \right)}}} \right)\log \left( y \right)dy} } = 2\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}} \cdot n \cdot \left( { - \frac{{\gamma + \log \left( {2n - 1} \right)}}{{2n - 1}} + \frac{{\gamma + \log \left( {2n + 1} \right)}}{{2n + 1}}} \right)} = }$

$\displaystyle{ = 2\gamma \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n} \cdot n \cdot \left( {\frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right)} + 2\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n} \cdot n \cdot \left( {\frac{{\log \left( {2n - 1} \right)}}{{2n - 1}} - \frac{{\log \left( {2n + 1} \right)}}{{2n + 1}}} \right)} = }$

$\displaystyle{ = \gamma \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\left( {\frac{{2n}}{{2n - 1}} - \frac{{2n}}{{2n + 1}}} \right)} + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\left( {\frac{{2n \cdot \log \left( {2n - 1} \right)}}{{2n - 1}} - \frac{{2n \cdot \log \left( {2n + 1} \right)}}{{2n + 1}}} \right)} = }$

$$\displaystyle{ = \gamma \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\left( {\frac{1}{{2n - 1}} + \frac{1}{{2n + 1}}} \right)} + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\left( {\frac{{\left( {2n - 1 + 1} \right) \cdot \log \left( {2n - 1} \right)}}{{2n - 1}} - \frac{{\left( {2n + 1 - 1} \right) \cdot \log \left( {2n + 1} \right)}}{{2n + 1}}} \right)} = }$$

$$\displaystyle{ = - \gamma + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\log \frac{{2n - 1}}{{2n + 1}}} + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\left( {\frac{{\log \left( {2n - 1} \right)}}{{2n - 1}} + \frac{{\log \left( {2n + 1} \right)}}{{2n + 1}}} \right)} = - \gamma + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\log \frac{{2n - 1}}{{2n + 1}}} = }$$

$$\displaystyle{ = - \gamma - \sum\limits_{n = 1}^\infty {\log \frac{{4n - 3}}{{4n - 1}}} + \sum\limits_{n = 1}^\infty {\log \frac{{4n - 1}}{{4n + 1}}} = - \gamma + \sum\limits_{n = 1}^\infty {\log \frac{{{{\left( {n - 1/4} \right)}^2}}}{{\left( {n - 3/4} \right)\left( {n + 1/4} \right)}}} = - \gamma + \mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = 1}^N {\log \frac{{{{\left( { - 1/4 + n} \right)}^2}}}{{\left( { - 3/4 + n} \right)\left( {1/4 + n} \right)}}} = }$$

$$\displaystyle{ = - \gamma + \mathop {\lim }\limits_{N \to \infty } \log \frac{{{{\left( {\left( { - \frac{1}{4} + 1} \right)\left( { - \frac{1}{4} + 2} \right)\left( { - \frac{1}{4} + 3} \right)..\left( { - \frac{1}{4} + N} \right)} \right)}^2}}}{{\left( {\left( { - \frac{3}{4} + 1} \right)\left( { - \frac{3}{4} + 2} \right)..\left( { - \frac{3}{4} + N} \right)} \right)\left( {\left( {\frac{1}{4} + 1} \right)\left( {\frac{1}{4} + 2} \right)..\left( {\frac{1}{4} + N} \right)} \right)}} = }$$

$$\displaystyle{ = - \gamma + \mathop {\lim }\limits_{N \to \infty } \log \left( {{{\left( {\frac{{\left( { - \frac{1}{4}} \right)\left( { - \frac{1}{4} + 1} \right)..\left( { - \frac{1}{4} + N} \right)}}{{\left( { - \frac{1}{4}} \right){{\left( {N + 1} \right)}^{ - 1/4}} \cdot N!}}} \right)}^2}\frac{{\left( { - \frac{3}{4}} \right)\frac{1}{4}{{\left( {N + 1} \right)}^{ - 3/4}}{{\left( {N + 1} \right)}^{1/4}}{{\left( {N!} \right)}^2}}}{{\left( {\left( { - \frac{3}{4}} \right)\left( { - \frac{3}{4} + 1} \right)..\left( { - \frac{3}{4} + N} \right)} \right)\left( {\frac{1}{4}\left( {\frac{1}{4} + 1} \right)..\left( {\frac{1}{4} + N} \right)} \right)}}} \right) = }$$

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

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