Is it possible to obtain a closed form of the following integral? $$\int_0^1\log (- \log x)\log \left(\frac{1+x}{1-x}\right)\,dx$$ I've made the change of variable $$t=\frac{1+x}{1-x} $$ but I feel like I'm turning in circles...
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We have the following closed form.
Proposition. $$ \int_0^1\log (- \log x)\log \left(\frac{1+x}{1-x}\right)\,dx=\gamma_1-2\ln^2 2-2\gamma \ln 2 -\gamma_1\Big({1,\small\frac12}\Big)\tag{$\star$} $$
where $\gamma_1$ is the Stieltjes constant, $$\gamma_1 = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log n}n-\int_1^N\frac{\log t}t\:dt\right)$$ and where $\gamma_1(a,b)$ is the poly-Stieltjes constant, $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\int_1^N\frac{\log t}t\:dt\right)\!.$$
Proof. One may recall the classic integral representation of the Euler gamma function $$ \frac{\Gamma(s)}{(a+1)^s}=\int_0^\infty t^{s-1} e^{-(a+1)t}\:dt, \qquad s>0,\, a>-1. \tag1 $$ By differentiating $(1)$ with respect to $s$, putting $s=1$ and making the change of variable $x=e^{-t}$, we get $$ \int_0^1x^a\log\left(-\log x\right)\:dx=-\frac{\gamma+\log(a+1)}{a+1},\qquad a>-1, \tag2 $$
where $\displaystyle \gamma=\lim_{N\to+\infty}\left(\sum_{n=1}^N \frac1n-\int_1^N\frac{dt}t\right)$ is the Euler-Mascheroni constant.
From the standard Taylor series expansion, $$ -\log (1-x)= \sum_{n=1}^{\infty} \frac{x^n}n, \qquad |x|<1,\tag3 $$ one gets $$ \log (1+x)-\log (1-x)=2 \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}, \qquad |x|<1.\tag4 $$ One may write the given integral as $$ \int_0^1\log (- \log x)\log \left(\frac{1+x}{1-x}\right)\,dx =\int_0^1\log (- \log x) \left(\log (1+x)-\log (1-x)\right)dx $$ then, inserting $(4)$ into the latter integrand and using $(2)$, we obtain $$ \begin{align} \int_0^1\log (- \log x)\log \left(\frac{1+x}{1-x}\right)\,dx&=2\int_0^1\log (- \log x) \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}\:dx\\ &=2\sum_{n=0}^{\infty} \frac1{2n+1}\int_0^1 x^{2n+1}\log (- \log x)\:dx\\ &=-2\sum_{n=0}^{\infty} \frac{\gamma+\log(2n+2)}{(2n+1)(2n+2)}\\ &=-\sum_{n=0}^{\infty} \frac{2\left(\gamma+\ln 2 \right)}{(2n+1)(2n+2)}-\sum_{n=1}^{\infty} \frac{2\log (n+1)}{(2n+1)(2n+2)}.\tag5 \end{align} $$ On the one hand, using Abel's theorem and using $(3)$, one has $$ \begin{align} \sum_{n=0}^{\infty} \frac2{(2n+1)(2n+2)}&=\lim_{x \to 1^-}\sum_{n=0}^{\infty} \frac{2x^{2n+2}}{(2n+1)(2n+2)}\\ &=\lim_{x \to 1^-}\left( (1+x)\log(1+x)+(1-x)\log(1-x)\right)\\ &=2\ln2.\tag6 \end{align} $$ On the other hand, using Theorem 2 here one has $$ \begin{align} \sum_{n=1}^{\infty} \frac{2\log (n+1)}{(2n+1)(2n+2)}=-\gamma_1+\gamma_1\Big({1,\small\frac12}\Big),\tag7 \end{align} $$ since $\gamma_1(1,1)=\gamma_1$.
Finally, bringing all the steps together gives $(\star)$.
Olivier Oloa
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Putting the two and two together, it would follow that it is possible to generalize $S(k)=\displaystyle\sum_{n=2}^{k+1}\frac{\zeta(n)}{2^{n-1}}$ to non-natural arguments using the aforementioned $\gamma$ constants, would it not ? – Lucian Dec 25 '15 at 21:03
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@Lucian It is not difficult to generalize $S(k)=\displaystyle\sum_{n=2}^{k+1}\frac{\zeta(n)}{2^{n-1}}$ to non-natural arguments: switch the two summations. Then differentiating your $I(k)$ with respect to $k$ and putting $k=0$, you arrive at $(5)$ above (with no proof). But, in my humble opinion, to get an analytic closed-form here, the step of poly-Stieltjes constants is decisive. – Olivier Oloa Dec 26 '15 at 10:57
Another interesting fact is $$I=2\ln2\ln\pi-\pi\int^1_0\ln\Gamma(x)\left(\frac\pi 2\cos(\pi x)+\ln\cot\left(\frac{\pi x}{2}\right)\sin(\pi x)\right)\ dx$$
– M.N.C.E. Dec 25 '15 at 07:19