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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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I'll conjecture that the answer is $$\displaystyle \frac 14-\frac 23\log(2)-6\zeta'(-1)$$ (correct to the $1000$ digits provided). –  Raymond Manzoni Mar 9 at 23:31
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@RaymondManzoni How did you discover this closed form? BTW, it can be expressed using Glaisher-Kinkelin constant as $6\ln A-\frac23\ln2-\frac14$. –  Vladimir Reshetnikov Mar 9 at 23:46
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@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, –  Raymond Manzoni Mar 9 at 23:49

2 Answers 2

up vote 27 down vote accepted
+350

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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Nice work, but I respectfully disagree that evaluating the integral in Mathematica proves anything. –  Ron Gordon Mar 10 at 18:28
    
@RonGordon: Could you do it without contour integration, the residue theorem and other complex analysis tools? –  Lucian Mar 10 at 18:38
    
@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. –  Ron Gordon Mar 10 at 18:41
    
Chen, the trouble with Mathematica is that unfortunately it does not reveal to us the methods by which it arrives at the result... and we're really curious as to precisely that particular piece of information. :-) –  Lucian Mar 10 at 18:42
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Here we can relate $\psi^{(-2)}(x)$ to the Barnes G-function, and use the values of the G-function on the page. –  Chen Wang Mar 10 at 22:19

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