Question:- Prove that $$\int_0^1 \frac {x-1}{\log(x) (1+x^3)} \, dx = \frac {\log(3)}{2}$$
I saw this problem as an comment on a youtube video few hours ago but I don't know how to prove this one as integration by parts doesn't works here. Also I wasn't able to figure out any proper subsitution that would simplify the integral.
Can someone suggests me some hints?
Note that $\int_0^1 x^s\,ds=\frac{x-1}{\log(x)}$. Then, we have
$$\int_0^1\frac{x-1}{\log(x)(x^3+1)}\,dx=\int_0^1\int_0^1 \frac{x^s}{(x^3+1)}\,ds\,dx$$
Now we apply Fubini's Theorem to interchange the order of integration to reveal
$$\int_0^1\frac{x-1}{\log(x)(x^3+1)}\,dx=\int_0^1\int_0^1 \frac{x^s}{(x^3+1)}\,dx\,ds$$
Next, we expand the denominator in a geometric series to find that
$$\begin{align} \int_0^1\frac{x-1}{\log(x)(x^3+1)}\,dx&=\sum_{n=0}^\infty (-1)^n\int_0^1\int_0^1 x^{s+3n}\,dx\,ds\\\\ &=\sum_{n=0}^\infty (-1)^n \log\left(\frac{3n+2}{3n+1}\right) \end{align}$$
Can you finish now?
BONUS:
To evaluate the final series we appeal to the digamma function, its relationship with the Gamma function, and Euler's reflection formula. Proceeding, we write
$$\begin{align} \sum_{n=0}^\infty (-1)^n\log\left(\frac{3n+2}{3n+1}\right)&=\int_0^1 \sum_{n=0}^\infty (-1)^n \frac1{s+3n+1}\,ds\\\\ &=\int_0^1 \sum_{n=0}^\infty\left(\frac1{6n+s+1}-\frac1{6n+s+4}\right)\,ds\\\\ &=\frac16\int_0^1\left(\psi((s+4)/6)-\psi((s+1)/6)\right)\,ds\\\\ &=\log\left(\frac{\Gamma(5/6)\Gamma(1/6)}{\Gamma(2/3)\Gamma(1/3)}\right)\\\\ &=\log\left(\frac{\sin(2\pi/3)}{\sin(5\pi/6)}\right)\\\\ &=\log(\sqrt 3) \end{align}$$
as expected!
Let $J(a) = \int_{0}^{\infty} \frac {x^{a-1}-1}{\ln x (1+x^3)}dx$. Then $J’(a) = \int_{0}^{\infty} \frac {x^{a-1}}{1+x^3}dx=\frac\pi3\csc\frac{\pi a}3 $
\begin{align} &\int_{0}^{1} \frac {x-1}{\ln x (1+x^3)}dx \overset{x\to\frac1x}= \frac12\int_{0}^{\infty} \frac {x-1}{\ln x (1+x^3)}dx\\ =&\ \frac12 J(2) =\frac12\int_1^2J’(a)da=\frac\pi6\int_1^2\csc\frac{\pi a}3da=\frac12\ln3 \end{align}
Alternatively to Mark Viola's approach, use the geometric series to see $$\small\int_0^1\frac{x-1}{x^3+1}\frac{{\rm d}x}{\log x}=\sum_{n\ge0}(-1)^n\int_0^1\frac{x^{3n+1}-x^{3n}}{\log x}\,{\rm d}x=\sum_{n\ge0}(-1)^{n+1}\int_0^\infty\frac{e^{-(3n+2)x}-e^{-(3n+1)x}}x\,{\rm d}x$$ The latter is a Frullani integral and evaluates as $$\int_0^\infty\frac{e^{-(3n+2)x}-e^{-(3n+1)x}}x\,{\rm d}x=-\log\left(\frac{3n+2}{3n+1}\right)$$ and thus arriving at $$\int_0^1\frac{x-1}{x^3+1}\frac{{\rm d}x}{\log x}=\sum_{n\ge0}(-1)^n\log\left(\frac{3n+2}{3n+1}\right)$$ aswell.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1}{x - 1 \over \ln\pars{x}\pars{1 + x^{3}}} \,\dd x} = \int_{0}^{1}{1 \over 1 + x^{3}}\ \overbrace{\int_{0}^{1}x^{t}\,\dd t}^{\ds{x - 1 \over \ln\pars{x}}}\ \dd x \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1}{x^{t} - x^{t + 3} \over 1 - x^{6}}\,\dd x\,\dd t = {1 \over 6} \int_{0}^{1}\int_{0}^{1}{x^{t/6 - 5/6} - x^{t/6 - 1/3} \over 1 - x} \,\dd x\,\dd t \\[5mm] = &\ {1 \over 6}\int_{0}^{1}\pars{\int_{0}^{1}{1 - x^{t/6 - 1/3} \over 1 - x} \,\dd x - \int_{0}^{1}{1 - x^{t/6 - 5/6} \over 1 - x} \,\dd x}\,\dd t \\[5mm] = &\ {1 \over 6}\int_{0}^{1}\bracks{\Psi\pars{{t \over 6} + {2 \over 3}} - \Psi\pars{{t \over 6} + {1 \over 6}}}\,\dd t = \left. \ln\pars{\Gamma\pars{t/6 + 2/3} \over \Gamma\pars{t/6 + 1/6}}\right\vert_{\ 0}^{\ 1}\label{1}\tag{1} \\[5mm] = &\ \ln\pars{{\Gamma\pars{5/6} \over \Gamma\pars{1/3}}\,{\Gamma\pars{1/6} \over \Gamma\pars{2/3}}} = \ln\pars{\sin\pars{\pi/3} \over \sin\pars{\pi/6}} = \ln\pars{\root{3}/2 \over 1/2}\label{2}\tag{2} \\[5mm] = & \bbx{\large {\ln\pars{3} \over 2}} \approx 0.5493 \\ & \end{align}
(\ref{1}): See Digamma $\ds{\Psi}$ Identity $\ds{\bf\color{black}{6.3.22}}$.
(\ref{2}): Euler Reflection Formula $\ds{\bf\color{black} {6.1.17}}$.
Note the Digamma $\ds{\Psi}$ Function Definition in terms of the Gamma Function $\ds{\Gamma}$: $$ \Psi\pars{z} = \totald{\ln\pars{\Gamma\pars{z}}}{z} $$ which was used in (\ref{1}).
Feynman’s Trick
Let‘s consider the parameterised integral $$ I(a)=\int_0^1 \frac{x^a-1}{\ln x\left(1+x^3\right)} d x, $$ then differentiating $I(a)$ w.r.t. $a$ brings us $$ \begin{aligned} I^{\prime}(a)&= \int_0^1 \frac{x^2}{1+x^3} d x\\ & =\sum_{n=0}^{\infty} \frac{(-1)^n}{a+3 n+1} \\ & =\sum_{n=0}^{\infty}\left(\frac{1}{6 n+a+1}-\frac{1}{6 n+a+4}\right) \\ & =\frac{1}{6} \sum_{n=0}^{\infty}\left(\frac{1}{n+\frac{a+1}{6}}-\frac{1}{n+\frac{a+4}{6}}\right) \\ & =\frac{1}{6}\left[\psi\left(\frac{a+4}{6}\right)-\psi\left(\frac{a+1}{6}\right)\right] \end{aligned} $$ Integrating back from $a=0$ to $1$ yields $$ \begin{aligned} \int_0^1 \frac{x^a-1}{\ln x\left(1+x^3\right)} d x& =I(1) -I(0) \\ & =\frac{1}{6} \int_0^1\left[\psi\left(\frac{a+4}{6}\right)-\psi\left(\frac{a+1}{6}\right)\right] d a \\ & =\left[\ln \Gamma\left(\frac{a+4}{6}\right)-\ln \Gamma\left(\frac{a+1}{6}\right)\right]_0^1 (\textrm{ By } \psi(x)=\frac{d}{d x} \ln \Gamma(x) )\\ & =\ln \left[\frac{\Gamma\left(\frac{5}{6}\right) \Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{1}{3}\right) \Gamma\left(\frac{2}{3}\right)}\right]\\&= \ln \left(\frac{\pi \csc \frac{\pi}{6}}{\pi \csc \frac{\pi}{3}}\right) \textrm{ (By reflection property of Gamma function, $\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$)} \\&=\frac{\ln 3}{2} \end{aligned} $$
$\newcommand{\d}{\,\mathrm{d}}\newcommand{\res}{\operatorname{Res}}\newcommand{\artanh}{\operatorname{artanh}}$Transform variables twice; take $x=e^{-u},u=e^t$ to obtain: $$I=\int_{\Bbb R}\frac{e^{2e^t}-e^{e^t}}{e^{3e^t}+1}\d t$$Observe that if $t$ jumps to $t+\pi i$ we get $-I$; thus by taking a contour integral along the rectangle between $\Bbb R$ and $\Bbb R+\pi i$ in the complex plane we obtain: $$I=\pi i\cdot\sum\res$$Where, strictly speaking, "$\sum\res$" is a symmetric limit over the sum of residues in the finite strip $|\Re z|\le R,0\le\Im z\le\pi$ as $R\to\infty$. This is correct since the error terms, the integrals $\pm R\to\pm R+\pi i$, obviously vanish as $R\to\infty$.
Note that if $e^{3e^z}=-1,3e^z=\pi i+2\pi Ni$ for some integer $N$ and $z=\frac{\pi i}{2}+2\pi iM+\ln|\pi(1+2N)/3|$ for some integers $M,N$. Evidently such a $z$ lands in the given strip only for $M=0$, but for arbitrary $N$. Well, we only get distinct $z$, note, if $N$ ranges in $\Bbb N_0$.
Ok, and at such a $z=z_N$ what shall the residue be? The pole is simple and differentiating the denominator and evaluating at $z_N$ gives: $$\res(z_N)=\frac{e^{2\pi i(1+2N)/3}-e^{\pi i(1+2N)/3}}{-\pi i(1+2N)}$$Thus: $$\begin{align}I&=\sum_{n=0}^\infty\frac{e^{\pi i(1+2n)/3}-e^{2\pi i(1+2n)/3}}{1+2n}\\&=2\sum_{n=0}^\infty\frac{\cos(\pi(1+2n)/3)}{1+2n}\\&=1-\frac{2}{3}+\frac{1}{5}+\frac{1}{7}-\frac{2}{9}+\frac{1}{11}+\cdots\\&\overset{?}{=}\frac{1}{2}\ln3\end{align}$$I admit the final equality is a bit mysterious. An alternative approach is to recognise: $$I=\lim_{\delta\to0^+}\artanh(e^{\pi i/3-\delta})-\artanh(e^{2\pi i /3-\delta})\\=\lim_{\delta\to0^+}\artanh\frac{e^{\pi i/3-\delta}-e^{2\pi i/3-\delta}}{1-e^{\pi i-2\delta}}\\=\artanh(1/2)\\=\frac{1}{2}\ln3$$
