# Prove that $\int_{0}^{1}{(1-x)(x-3)\over 1+x^2}\cdot{dx\over \ln{x}}={\ln{8\over \Gamma^4(3/4)}}$

Prove

$$I=\int_{0}^{1}{(1-x)(x-3)\over 1+x^2}\cdot{dx\over \ln{x}}=\color{blue}{\ln{8\over \Gamma^4(3/4)}}\tag1$$

$(1-x)(x-3)=-x^2+4x-3$

$${1\over 1+x^2}=\sum_{n=0}^{\infty}(-1)^nx^{2n}\tag2$$

$$I=-\sum_{n=0}^{\infty}(-1)^n\int_{0}^{\infty}{x^{2n+2}-4x^{2n+1}+3x^{2n}\over \ln{x}}dx\tag3$$

Rewrite (3) to apply Frullani's theorem

$$I=-\sum_{n=0}^{\infty}(-1)^n\int_{0}^{\infty}{x^{2n+2}-x^{2n+1}-3x^{2n+1}+3x^{2n}\over \ln{x}}dx\tag4$$

$$I=\sum_{n=0}^{\infty}(-1)^{n-1}\ln\left({2n+3\over (2n+2)^4}\cdot{(2n+1)^3}\right)\tag5$$

This method is a bit boring method! I converted (1) into series and using Frullani's theorem and again have to solve the series is another step before we can reached our answer.

How can I solve (1) without using series?

• I think you cannot. Here $\Gamma(3/4)$ comes from an infinite product, hence $\log\Gamma(3/4)$ is a series of logarithms related to $(5)$. Jun 13, 2016 at 18:42
• And anyhow, this question is almost the same as your previous math.stackexchange.com/questions/1821080/… Jun 13, 2016 at 18:45
• @JackD'Aurizio The question really is not the same inasmuch as the integrand is substantively different from the previous one. ;-)) Jun 13, 2016 at 20:01
• @Dr.MV: the meaning of the previous "almost the same" is "a question that can be solved like the other one, with really minor changes". Abstract duplicate, for short. Jun 13, 2016 at 20:03
• you should write your complete list of formulas ? Jun 14, 2016 at 19:34

Hint. One may set $$f(s):=\int_0^1 \frac{x^{4s}-1}{(1+x^2)\ln x}dx, \quad s>0. \tag1$$ In order to get rid of the factor $\ln x$ in the denominator, we may differentiate under the integral sign getting $$f'(s)=4\int_0^1 \frac{x^{4s}}{1+x^2}dx=4\int_0^1 \frac{x^{4s}(1-x^2)}{1-x^4}dx, \quad s>0, \tag2$$ giving \begin{align} f'(s)&=\psi\left(s+\frac34\right)-\psi\left(s+\frac14\right),\tag3 \end{align} where we have used the standard integral representation of the digamma function $$\int_{0}^{1}{1 - t^{s - 1} \over 1 - t}\,dt \, = \psi (s)+ \gamma, \quad s>0,$$ $\gamma$ being the Euler-Mascheroni constant.

Then integrating $(3)$, observing that as $s \to 0^+$, $f(s) \to 0$, one gets

$$f(s)=\int_0^1 \frac{x^{4s}-1}{(1+x^2)\ln x}dx=\log\left(\frac{\Gamma\left(\frac14\right)\Gamma\left(s+\frac34\right)}{\Gamma\left(\frac34\right)\Gamma\left(s+\frac14\right)}\right), \quad s>0, \tag4$$

from which you deduce the value of your initial integral by writing $$\int_{0}^{1}{(1-x)(x-3)\over 1+x^2}\cdot{dx\over \ln{x}}=f(1/2)+4f(1/4).$$

• How do you cancel the logarithm after differentiating f(s)? Jun 13, 2016 at 19:14
• Thank you @Jack D'Aurizio. Jun 13, 2016 at 19:14
• Oh, of course, thanks Jun 13, 2016 at 19:20
• Very well done! +1 -Mark Jun 13, 2016 at 19:54
• @China cat You are welcome. Jun 13, 2016 at 22:29

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{I} & \equiv \int_{0}^{1}{\pars{1 - x}\pars{x - 3} \over 1 + x^{2}} \,{\dd x \over \ln\pars{x}} = \int_{0}^{1}{3 - x \over 1 + x^{2}}\,\ \overbrace{{x - 1 \over \ln\pars{x}}}^{\ds{4\int_{0}^{1/4}x^{4y}\,\dd y}}\ \,\dd x \\[3mm] & = 4\int_{0}^{1/4}\int_{0}^{1}{3x^{4y} - x^{4y + 1} \over 1 + x^{2}}\,\dd x\,\dd y = 4\int_{0}^{1/4}\int_{0}^{1} {\pars{3x^{4y} - x^{4y + 1}}\pars{1 - x^{2}} \over 1 - x^{4}}\,\dd x\,\dd y \\[3mm] & = 4\int_{0}^{1/4}\int_{0}^{1} {3x^{4y} - 3x^{4y + 2} - x^{4y + 1} + x^{4y + 3}\over 1 - x^{4}}\,\dd x\,\dd y \\[3mm] & \stackrel{x^{4}\ \mapsto\ x}{=}\ \int_{0}^{1/4}\int_{0}^{1} {3x^{y - 3/4} - 3x^{y - 1/4} - x^{y - 1/2} + x^{y}\over 1 - x}\,\dd x\,\dd y \\[8mm] & = \int_{0}^{1/4}\left(% 3\int_{0}^{1}{1 - x^{y - 1/4}\over 1 - x}\,\dd x + \int_{0}^{1}{1 - x^{y - 1/2}\over 1 - x}\,\dd x\right. \\[3mm] & \left.\phantom{\int_{0}^{1/4}\pars{}}- 3\int_{0}^{1}{1 - x^{y - 3/4}\over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{y}\over 1 - x}\,\dd x\right)\,\dd y \\[8mm] & = \int_{0}^{1/4}\bracks{3\,\Psi\pars{y + {3 \over 4}} + \Psi\pars{y + \half} - 3\,\Psi\pars{y + {1 \over 4}} -\Psi\pars{y + 1}}\,\dd y \\[3mm] & = \left.\ln\pars{\Gamma^{\,3}\pars{y + 3/4}\Gamma\pars{y + 1/2} \over \Gamma^{\,3}\pars{y + 1/4}\Gamma\pars{y + 1}}\right\vert_{\ 0}^{\ 1/4} = \ln\pars{{\Gamma^{\,3}\pars{1}\Gamma\pars{3/4} \over \Gamma^{\,3}\pars{1/2}\Gamma\pars{5/4}} {\Gamma^{\,3}\pars{1/4}\Gamma\pars{1} \over \Gamma^{\,3}\pars{3/4}\Gamma\pars{1/2}}} \\[3mm] & = \ln\pars{{1 \over \Gamma^{\,4}\pars{1/2}}\ {\Gamma^{\,3}\pars{1/4} \over \Gamma\pars{5/4}\Gamma^{\,2}\pars{3/4}}}\tag{1} \end{align}

$$\left\lbrace\begin{array}{rcl} \ds{1 \over \Gamma^{\,4}\pars{1/2}} & \ds{=} & \ds{1 \over \pi^{2}}\quad \mbox{because}\quad \ds{\Gamma\pars{\half} = \root{\pi}} \\[5mm] \ds{\Gamma^{\,3}\pars{1/4} \over \Gamma\pars{5/4}\Gamma^{\,2}\pars{3/4}} & \ds{=} & \ds{{\Gamma^{\,3}\pars{1/4} \over \pars{1/4}\Gamma\pars{1/4}\Gamma^{\,2}\pars{3/4}} = 4\bracks{\Gamma\pars{1/4} \over \Gamma\pars{3/4}}^{2}} \\[1mm] & \ds{=} & \ds{4\bracks{{1 \over \Gamma\pars{3/4}}\,{\pi \over \Gamma\pars{3/4}\sin\pars{\pi/4}}}^{2} = {8\pi^{2} \over \Gamma^{\,4}\pars{3/4}}} \end{array}\right.\tag{2}$$
With $\pars{1}$ and $\pars{2}$: $$\color{#f00}{I} \equiv \int_{0}^{1}{\pars{1 - x}\pars{x - 3} \over 1 + x^{2}} \,{\dd x \over \ln\pars{x}} = \color{#f00}{\ln\pars{8 \over \Gamma^{\,4}\pars{3/4}}}$$