7

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?

Olivier Oloa
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gymbvghjkgkjkhgfkl
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  • 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 $(5)$. – Jack D'Aurizio Jun 13 '16 at 18:42
  • And anyhow, this question is almost the same as your previous http://math.stackexchange.com/questions/1821080/showing-int-01x2n-x-over-1x-cdotdx-over-lnx-ln-left2-over – Jack D'Aurizio Jun 13 '16 at 18:45
  • @JackD'Aurizio The question really is not the same inasmuch as the integrand is substantively different from the previous one. ;-)) – Mark Viola Jun 13 '16 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. – Jack D'Aurizio Jun 13 '16 at 20:03
  • you should write your complete list of formulas ? – reuns Jun 14 '16 at 19:34

2 Answers2

10

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 $$ \int_{0}^{1}{(1-x)(x-3)\over 1+x^2}\cdot{dx\over \ln{x}}=f(1/2)+4f(1/4). $$

Olivier Oloa
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0

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

\begin{equation} \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} \end{equation}
With $\pars{1}$ and $\pars{2}$: $$ \color{#f00}{I} \equiv \int_{0}^{1}{\pars{1 - x}\pars{x - 3} \over 1 + x^{2}} \,{\dd x \over \ln\pars{x}} = \color{#f00}{\ln\pars{8 \over \Gamma^{\,4}\pars{3/4}}} $$
Felix Marin
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