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I'm struggling with this integral $$I=\int_0^1\frac{1-x^2+\left(1+x^2\right)\ln x}{\left(x+x^2\right)\ln^3x}dx.\tag1$$ Mathematica could not evaluate it in a closed form. Its numeric value is approximately $I\approx0.7804287418294087023386965471512328112...$$^\text{[more]}$, but I could not find a plausible closed form for this number using inverse symbolic calculators available online.


Update: Based on Raymond Manzoni's comment below, there is actually a conjectural closed form, numerically matching up to at least $10^3$ decimal digits: $$I\stackrel?=6\ln A-\frac{\ln4}3-\frac14,\tag2$$ where $A$ is the Glaisher-Kinkelin constant: $$A=\exp\left(\frac1{12}-\zeta'(-1)\right).\tag3$$ Could you suggest a proof of the conjecture $(2)$?

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    $\begingroup$ I'll conjecture that the answer is $$\displaystyle \frac 14-\frac 23\log(2)-6\zeta'(-1)$$ (correct to the $1000$ digits provided). $\endgroup$ Mar 9, 2014 at 23:31
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    $\begingroup$ @RaymondManzoni How did you discover this closed form? BTW, it can be expressed using Glaisher-Kinkelin constant as $6\ln A-\frac23\ln2-\frac14$. $\endgroup$ Mar 9, 2014 at 23:46
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    $\begingroup$ @VladimirReshetnikov: I searched linear dependencies using PSLQ and my own tables. I noticed too the relation with G-K constant but without direct integral transformation I could see yet. Cheers, $\endgroup$ Mar 9, 2014 at 23:49

2 Answers 2

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We start from the first Binet's formula: $$\ln\Gamma(z)=\left(z-\tfrac12\right)\ln z-z+\frac{\ln(2\pi)}2+\int_0^\infty\left(\frac12-\frac1t+\frac1{e^t-1}\right)\frac{e^{-t\,z}}t dt.\tag1$$ Change variable $t=-2\ln x$: $$\ln\Gamma(z)=\left(z-\tfrac12\right)\ln z-z+\frac{\ln(2\pi)}2+\frac12\int_0^1x^{2z-1}\frac{1-x^2+(1+x^2)\ln x}{(x^2-1)\ln^2x}dx.\tag2$$ Integrate on interval $0<z<\frac12$: $$\psi^{(-2)}\left(\tfrac12\right)=\frac1{16}+\frac{\ln8}8+\frac{\ln\pi}4+\frac14\int_0^1\frac{1-x^2+(1+x^2)\ln x}{(x+x^2)\ln^3x}dx.\tag3$$ Use formula $(5)$ from here connecting Barnes G-Function and negapolygamma, and let $z=\frac12$: $$\ln G\left(\tfrac12\right)=\frac{\ln(2\pi)}4+\frac18-\frac{\ln\pi}4-\psi^{(-2)}\left(\tfrac12\right).\tag4$$ Use formula $(19)$ from the same page to get a closed form for $\ln G\left(\tfrac12\right)$ in terms of the Glaisher-Kinkelin constant $A$: $$\ln G\left(\tfrac12\right)=-\frac{3\ln A}2-\frac{\ln\pi}4+\frac18+\frac{\ln2}{24}.\tag5$$ Comparing $(4)$ and $(5)$ we can get: $$\psi^{(-2)}\left(\tfrac12\right)=\frac{3\ln A}2+\frac{\ln\pi}4+\frac{5\ln2}{24}\tag6.$$ Finally, from $(3)$ and $(6)$ it follows: $$\int_0^1\frac{1-x^2+(1+x^2)\ln x}{(x+x^2)\ln^3x}dx=6\ln A-\frac{\ln4}3-\frac14,\tag7$$ that proves the conjecture.

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$$\begin{align*} I&=\int^1_0\frac{1-x^2+(1+x^2)\log x}{x+1}\frac{dx}{x\log^3x}\\ &=\left.\frac{-1}{2\log^2x}\frac{1-x^2+(1+x^2)\log x}{x+1}\right|^1_0-\int^1_0\frac{-1}{2\log^2x}\frac{\partial}{\partial x}\left(\frac{1-x^2+(1+x^2)\log x}{x+1}\right)dx\\ &=\int^1_0\frac{1}{2\log^2x}\frac{\partial}{\partial x}\left(1-x+\frac{(1+x^2)\log x}{x+1}\right)dx\\ &=\frac12\int^1_0\frac{1}{\log^2x}\left(-1+\frac{(1+x^2)}{x(x+1)}+\frac{(x^2+2x-1)\log x}{(x+1)^2}\right)dx\\ &=\frac12\int^1_0\left(\frac{1-x}{x(x+1)\log^2x}+\frac{1}{\log x}-\frac{2}{(x+1)^2\log x}\right)dx\\ &=\frac12\int^1_0\left(\frac{1-x}{x(x+1)\log^2x}+\frac{2}{(x+1)^2\log x}\right)dx+\frac12\int^1_0\left(\frac{1}{\log x}-\frac{4}{(x+1)^2\log x}\right)dx\\ &=\frac12(I_1+I_2). \end{align*}$$

$$\begin{align*} I_1&=\int^1_0\left(\frac{1-x}{x(x+1)\log^2x}+\frac{2}{(x+1)^2\log x}\right)dx\\ &=\int^1_0\left(\frac{1-x^2+2x\log x}{(x+1)^2}\right)\frac{dx}{x\log^2x}\\ &=\left.\frac{-1}{\log x}\frac{1-x^2+2x\log x}{(x+1)^2}\right|^1_0-\int^1_0\frac{-1}{\log x}\frac{\partial}{\partial x}\left(\frac{1-x^2+2x\log x}{(x+1)^2}\right)dx\\ &=\int^1_0\frac{-1}{\log x}\frac{-2(1-x)\log x}{(x+1)^3}dx\\ &=2\int^1_0\frac{1-x}{(x+1)^3}dx=\frac12. \end{align*}$$

Mathematica is able to evaluate $I_2=12\log A-1-\frac43\log2$, so the conjectured closed form is proved.

Edit: to evaluate $I_2$, we write $F(a)=\int^1_0\frac{x^a-1}{(x+1)^2\log x}dx$, so that $F(0)=0$, and $F'(a)=\int^1_0\frac{x^a}{(x+1)^2}dx$. We have $$\begin{align*} I_2&=\int^1_0\left(\frac{1}{\log x}-\frac{4}{(x+1)^2\log x}\right)dx\\ &=\int^1_0\frac{x^2+2x-3}{(x+1)^2\log x}dx=F(2)+2F(1). \end{align*}$$

Then we need to give closed form of $F(a)$. First we have $$\begin{align*} \int^1_0\frac{x^adx}{(1-zx)^2}&=\int^1_0\sum_{n=0}^{\infty}(n+1)z^nx^{n+a}dx\\ &=\sum_{n=0}^{\infty}\frac{(n+1)z^n}{n+a+1}\\ &=\sum_{n=0}^{\infty}z^n-a\sum_{n=0}^{\infty}\frac{z^n}{n+a+1}\\ &=\frac{1}{1-z}-\frac{a}{a+1}~_2F_1(1,a+1;a+2\mid z) \end{align*}$$ for all $|z|<1$. Taking limits, we have $$\begin{align*} F'(a)&=\frac12-\frac{a}{a+1}~_2F_1(1,a+1;a+2\mid-1)\\ &=\frac12-\frac{a}{2}\left(\psi\left(\frac{a+2}{2}\right)-\psi\left(\frac{a+1}{2}\right)\right) \end{align*}$$ and therefore $$\begin{align*} F(a)&=\int^a_0F'(b)~db\\ &=\frac{a}{2}-\int^a_0\frac{b}{2}\left(\psi\left(\frac{b+2}{2}\right)-\psi\left(\frac{b+1}{2}\right)\right)db\\ &=\frac{a}{2}-2\int^{a/2}_{0}c\left(\psi(c+1)-\psi(c+\frac12)\right)dc\\ &=\frac{a}{2}+\int^{a/2}_{0}(2\psi(c+1)-\psi(c+\frac12))dc-2\int^{a/2}_{0}\left((c+1)\psi(c+1)-(c+\frac12)\psi(c+\frac12)\right)dc\\ &=\frac{a}{2}+(2\log\Gamma(a/2+1)-2\log\Gamma(1)-\log\Gamma(a/2+1/2)+\log\Gamma(1/2))\\ &-2\left(\int^{a/2+1}_{1}-\int^{a/2+1/2}_{1/2}\right)c\psi(c)dc\\ &=\frac{a}{2}+\log\Gamma(1/2)+2\log\Gamma(a/2+1)-\log\Gamma(a/2+1/2)-2\left.(c\log\Gamma(c)-\psi^{(-2)}(c))\right|^{a/2+1,1/2}_{a/2+1/2,1}\\ &=\frac{a}{2}+\log\Gamma(1/2)+2\log\Gamma(a/2+1)-\log\Gamma(a/2+1/2)-(a+2)\log\Gamma(a/2+1)\\&+(a+1)\log\Gamma(a/2+1/2)-\log\Gamma(1/2)+2\left.(\psi^{(-2)}(c))\right|^{a/2+1,1/2}_{a/2+1/2,1}\\ &=\frac{a}{2}-a\log\Gamma(a/2+1)+a\log\Gamma(a/2+1/2)+2\left.(\psi^{(-2)}(c))\right|^{a/2+1,1/2}_{a/2+1/2,1}\\ &=\frac{a}{2}-a\log\Gamma(a/2+1)+a\log\Gamma(a/2+1/2)\\ &+2\left(\psi^{(-2)}(a/2+1)-\psi^{(-2)}(a/2+1/2)+\psi^{(-2)}(1/2)-\psi^{(-2)}(1)\right).\\ \end{align*}$$ Here $\psi^{(-2)}(a)=\int^a_0\log\Gamma(x)dx$. We can now give a closed form for $I_2$: $$\begin{align*} I_2&=F(2)+2F(1)\\ &=(1-2\log\Gamma(2)+2\log\Gamma(3/2))+2\left(1/2-\log\Gamma(3/2)+\log\Gamma(1)\right)\\ &+2(\psi^{(-2)}(2)-\psi^{(-2)}(3/2)+\psi^{(-2)}(1/2)-\psi^{(-2)}(1))\\ &+4(\psi^{(-2)}(3/2)-\psi^{(-2)}(1)+\psi^{(-2)}(1/2)-\psi^{(-2)}(1))\\ &=2+2\psi^{(-2)}(2)+2\psi^{(-2)}(3/2)+6\psi^{(-2)}(1/2)-10\psi^{(-2)}(1). \end{align*}$$ Using the known values $\psi^{(-2)}(2)=\log2+\log\pi-1$, $\psi^{(-2)}(1)=\frac12\log2+\frac12\log\pi$, $\psi^{(-2)}(3/2)=\frac34\log\pi+\frac{5}{24}\log2+\frac32\log A-\frac12$ and $\psi^{(-2)}(1/2)=\frac14\log\pi+\frac{5}{24}\log 2+\frac32\log A$, we finally conclude that $I_2=12\log A-1-\frac43\log2$, and therefore $$I=\frac12(I_1+I_2)=6\log A-\frac23\log2-\frac14.$$

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    $\begingroup$ Nice work, but I respectfully disagree that evaluating the integral in Mathematica proves anything. $\endgroup$
    – Ron Gordon
    Mar 10, 2014 at 18:28
  • $\begingroup$ @RonGordon: Could you do it without contour integration, the residue theorem and other complex analysis tools? $\endgroup$
    – Lucian
    Mar 10, 2014 at 18:38
  • $\begingroup$ @Lucian: In fact, I have so far worked it out to a certain point as a real integral, but am currently stuck and unfortunately, work calls. To be devil's advocate, there is likely a way that would use contour integration, but that way is not obvious to me yet. $\endgroup$
    – Ron Gordon
    Mar 10, 2014 at 18:41
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    $\begingroup$ The second part is done using polygamma functions. $\endgroup$
    – Chen Wang
    Mar 10, 2014 at 22:09
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    $\begingroup$ Here we can relate $\psi^{(-2)}(x)$ to the Barnes G-function, and use the values of the G-function on the page. $\endgroup$
    – Chen Wang
    Mar 10, 2014 at 22:19

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