$\int_{0}^{1}{2n-x-x^3-x^5-\cdots-x^{4n-1}\over 1+x^2}\cdot{\mathrm dx\over \ln{x}}$

Question

Considere

$$\int_{0}^{1}{2n-x-x^3-x^5-\cdots-x^{4n-1}\over 1+x^2}\cdot{\mathrm dx\over \ln{x}}=I\tag1$$ $n\ge1$

¿Cómo se puede demostrar que $I=2n\ln{\Gamma(3/4)\over \Gamma(5/4)}-\ln{[8^n(2n-1)!!]}?$

Un intento:

$J=x+x^3+x^5+x^7+\cdots+$

$J=x(1+x^2+x^4+x^6+\cdots+)$

Series geométricas

$1+x+x^2+x^3+\cdots x^{n-1}={x(1-x^n)\over 1-x}$

$1+x^2+x^4+x^6+\cdots+x^{2n-2}={x^2(1-x^{2n})\over 1-x^2}$

$I$ se convierte en

$$2n\int_{0}^{1}{1\over 1+x^2}\cdot{\mathrm dx\over \ln{x}}-\int_{0}^{1}{1-x^{2n}\over 1+x^2}\cdot{x^3\over \ln{x}}\mathrm dx=I\tag2$$

$x=\tan{y}$ entonces $dx=\sec^2{y}dy$

$$\int_{0}^{\pi/4}{1\over \ln{\tan{y}}}\mathrm dy-\int_{0}^{\pi/4}{1-\tan^{2n}{y}\over \ln{\tan{y}}}\cdot\tan^3{y}\mathrm dy\tag3$$

$$\int_{0}^{\pi/4}{1-\tan^3{y}\over \ln{\tan{y}}}\mathrm dy-\int_{0}^{\pi/4}{\tan^{2n}{y}\over \ln{\tan{y}}}\cdot\tan^3{y}\mathrm dy\tag4$$

$1-x^3=(1-x)(1+x+x^2)$

No estoy seguro de qué hacer a continuación

Answer 1

El primer término en el LHS de su $(2)$ es una integral divergente: mejor evitarlas.
En cambio, podemos considerar que:

$$ I(k)=\int_{0}^{1}\frac{1-x^{2k-1}}{1+x^2}\cdot\frac{dx}{\log(x)}=-\int_{0}^{+\infty}\frac{e^{-x}-e^{-2kx}}{1+e^{-2x}}\cdot\frac{dx}{x} $$ puede tratarse fácilmente mediante una expansión en serie geométrica y Teorema de Frullani lo que lleva a..: $$ I(k) = \log\frac{2k}{1}-\log\frac{2k+2}{3}+\log\frac{2k+4}{5}-\log\frac{2k+6}{7}+\ldots $$ entonces a: $$ I(k) = -\log\left(\frac{2k}{1}\cdot\frac{3}{2k+2}\cdot\frac{2k+4}{5}\cdot\frac{7}{2k+6}\cdots\right) $$ y finalmente a: $$ I(k) = (-1)^{k+1}\log\sqrt{\pi} +\log\left(\frac{(k-2)!!\,\Gamma\left(\frac{3}{4}\right)}{2(k-1)!!\,\Gamma\left(\frac{5}{4}\right)}\right)$$ que es simple de sumar sobre $k$ .

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Básicamente, todo el problema se reduce a evaluar la siguiente integral:

\begin{align} &\int_{0}^{1}{1 - x^{\mu} \over 1 + x^{2}}\,{\dd x \over \ln\pars{x}} = \int_{0}^{1}{1 - x^{2} - x^{\mu} + x^{\mu + 2}\over 1 - x^{4}} \,{\dd x \over \ln\pars{x}} \\[5mm] \stackrel{x^{4}\ \mapsto\ x}{=} & \int_{0}^{1}{x^{-3/4} - x^{-1/4} - x^{\mu/4 - 3/4} + x^{\mu/4 - 1/4} \over 1 - x}\,{\dd x \over \ln\pars{x}} \\[5mm] = &\ \int_{0}^{1}{x^{-3/4} - x^{-1/4} - x^{\mu/4 - 3/4} + x^{\mu/4 - 1/4} \over 1 - x}\pars{-\int_{0}^{\infty}x^{t}\,\dd t}\,\dd x \\[5mm] = &\ \int_{0}^{\infty}\int_{0}^{1} {-x^{t - 3/4} + x^{t - 1/4} + x^{t + \mu/4 - 3/4} - x^{t + \mu/4 - 1/4} \over 1 - x}\,\dd x\,\dd t \\[5mm] = &\ \int_{0}^{\infty}\bracks{\Psi\pars{t + {1 \over 4}} - \Psi\pars{t + {3 \over 4}} - \Psi\pars{t + {\mu + 1\over 4}} + \Psi\pars{t + {\mu + 3 \over 4}}}\dd t \\[5mm] = &\ \left. \ln\pars{\Gamma\pars{t + 1/4}\Gamma\pars{t + \bracks{\mu + 3}/4} \over \Gamma\pars{t + 3/4}\Gamma\pars{t + \bracks{\mu + 1}/4}} \right\vert_{\ t\ =\ 0}^{\ t\ \to\ \infty} = \bbx{\ds{-\ln\pars{{\root{2} \over 2\pi}\,\Gamma^{2}\pars{1 \over 4}\, {\Gamma\pars{\bracks{\mu + 3}/4} \over \Gamma\pars{\bracks{\mu + 1}/4}}}}} \end{align}