How can I prove $\int_{0}^{1} \frac {x-1}{\log(x) (1+x^3)}dx=\frac {\log3}{2}$

Solution 1:

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!

Solution 2:

Note

$$I=\int_{0}^{1} \frac {x-1}{\ln x (1+x^3)}dx \overset{x\to\frac1x}= \frac12\int_{0}^{\infty} \frac {x-1}{\ln x (1+x^3)}dx$$

Let $J(a) = \int_{0}^{\infty} \frac {x^a-1}{\ln x (1+x^3)}dx$. Then $J’(a) = \int_{0}^{\infty} \frac {x^a}{1+x^3}dx=\frac\pi3\csc\frac{\pi(a+1)}3 $. Thus,

$$I=\frac12 J(1) =\frac12\int_0^1J’(a)da=\frac\pi6\int_0^1\csc\frac{\pi(a+1)}3da=\frac{\ln3}2 $$

Solution 3:

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.

Solution 4:

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}).