Confirm result of $2^n\int_{0}^{1}{x^{2^n-1} \over 1+x^{2^n}}\ln{(-\ln{x})}\mathrm dx=-{2n+1\over 2}\cdot\ln^2{2}$

Question

Consider the integral

$$2^n\int_{0}^{1}{x^{2^n-1} \over 1+x^{2^n}}\ln{(-\ln{x})}\mathrm dx=-{2n+1\over 2}\cdot\ln^2{2}\tag1$$ $n\ge0$

An attempt:

$u=x^{2^n}$ $\implies$ $du=2^n\cdot x^{2^n-1}dx$

$(1)$ becomes

$$\int_{0}^{1}{\mathrm du\over 1+u}\cdot\ln{\left(-{1\over 2^n}\ln{u}\right)}\tag2$$

Another sub:

$v=-{1\over 2^n}\ln{u}$ $\implies$ $-2^nudv=du$

$(2)$ becomes

$$2^n\int_{0}^{\infty}{\mathrm dv\over 1+e^{2^nv}}\cdot\ln{v}\tag3$$

Maybe I could apply geometric series and integrate term by term?

How else can we confirm the result of $(1)?$

Answer 1

Starting from your $(3)$ and using the substitution $v=\frac{u}{2^n}$ it is enough to compute the integrals $$ I=\int_{0}^{+\infty}\frac{du}{1+e^u}=\log(2),\qquad J=\int_{0}^{+\infty}\frac{\log u}{1+e^u}\,du.\tag{4} $$ Since $$ g(\alpha)=\int_{0}^{+\infty}\frac{u^\alpha}{1+e^u}\,du = (1-2^{-\alpha})\,\Gamma(1+\alpha)\,\zeta(1+\alpha) \tag{5}$$ for any $\alpha>-1$, we have: $$ J=g'(0) = g(0)\cdot\left.\frac{d}{d\alpha}\log g(\alpha)\right|_{\alpha=0}\tag{6}$$ and the claim follows from known values for $\Gamma,\zeta,\Gamma'$ and $\zeta'$ or simply $\eta$ and $\eta'$. Alternative methods for the evaluation of $\int_{0}^{+\infty}\frac{\log x}{e^x+1}\,dx$ are given here, here and here - thanks to Hazem Orabi.