Integral $\int_0^1\frac{\ln x}{x-1}\ln\left(1+\frac1{\ln^2x}\right)dx$

Question

Is it possible to evaluate this integral in a closed form? $$ I \equiv \int_{0}^{1}{\ln\left(x\right) \over x - 1}\, \ln\left(1 + {1 \over \ln^{2}\left(x\right)}\right)\,{\rm d}x $$ Numerically, $$I\approx2.18083278090426462584033339029703713513\dots$$

Answer 1

Substitute $x=e^{-y}$:

$$ I=\int^\infty_0\frac{y}{e^y-1}\log(1+y^{-2})dy. $$

We Start from Binet's Second Formula: \begin{align} \log\Gamma(z) &= (z-\frac12)\log z-z+\frac12\log(2\pi)+2\int^{\infty}_0\frac{\tan^{-1}(t/z)}{e^{2\pi t}-1}dt\\ &=(z-\frac12)\log z-z+\frac12\log(2\pi)+\frac1\pi\int^{\infty}_0\frac{\tan^{-1}(\frac{y}{2\pi z})}{e^{y}-1}dy \end{align}

Integrate from $z=0$ to $\frac1{2\pi}$: \begin{align} \psi^{(-2)}\left(\frac1{2\pi}\right) &= -\frac{3}{16\pi^2}+\frac{1}{4\pi}-\frac{\log(2\pi)}{8\pi^2}+\frac{\log(2\pi)}{2\pi} +\frac1\pi\int^{\infty}_0\frac1{e^{y}-1}\left(\frac{\tan^{-1}y}{2\pi}+\frac{y\log(1+y^{-2})}{4\pi}\right)dy\\ &=\frac{I}{4\pi^2}-\frac{3}{16\pi^2}+\frac{1}{4\pi}-\frac{\log(2\pi)}{8\pi^2}+\frac{\log(2\pi)}{2\pi}+\frac1{2\pi^2}\int^{\infty}_0\left(\frac{\tan^{-1}y}{e^{y}-1}\right)dy\\ &=\frac{I}{4\pi^2}-\frac{3}{16\pi^2}+\frac{1}{4\pi}-\frac{\log(2\pi)}{8\pi^2}+\frac{\log(2\pi)}{2\pi}+\frac1{\pi}\int^{\infty}_0\left(\frac{\tan^{-1}(2\pi t)}{e^{2\pi t}-1}\right)dt. \end{align}

Therefore $$ I=4\pi^2\psi^{(-2)}\left(\frac1{2\pi}\right)+\frac{3}{4}-\pi+\frac{\log(2\pi)}{2}-2\pi\log(2\pi)-4\pi\int^{\infty}_0\left(\frac{\tan^{-1}(2\pi t)}{e^{2\pi t}-1}\right)dt.$$

We use Binet's second formula again: $$ \log\Gamma(\frac{1}{2\pi})=(\frac{1}{2\pi}-\frac12)\log \frac{1}{2\pi}-\frac{1}{2\pi}+\frac12\log(2\pi)+2\int^{\infty}_0\frac{\tan^{-1}(2\pi t)}{e^{2\pi t}-1}dt\\ $$

Therefore $4\pi\int^{\infty}_0\frac{\tan^{-1}(2\pi t)}{e^{2\pi t}-1}dt=2\pi\log\Gamma(\frac{1}{2\pi})+1-(2\pi-1)\log(2\pi)$, and $$ I=4\pi^2\psi^{(-2)}\left(\frac1{2\pi}\right)-\frac{1}{4}-\pi-\frac{\log(2\pi)}{2}-2\pi\log\Gamma(\frac{1}{2\pi}).$$

Also, we have $$ \psi^{(-2)}(z)=\frac{z}{2}\log(2\pi)+(z-1)\log\Gamma(z)-z(z-1)/2-\log G(z), $$

Therefore $$ 4\pi^2\psi^{(-2)}(\frac1{2\pi})=\pi\log(2\pi)-(4\pi^2-2\pi)\log\Gamma(\frac{1}{2\pi})-\frac{1}{2}+\pi-4\pi^2\log G(\frac{1}{2\pi}) $$

And we have the final answer $$ I=\left(\pi-\frac12\right)\log(2\pi)-\frac{3}{4}-4\pi^2\left(\log\Gamma\left(\frac{1}{2\pi}\right)+\log G\left(\frac{1}{2\pi}\right)\right).$$

Answer 2

$$I=\left(\pi-\frac12\right)\ln(2\pi)-\frac34-4\pi^2\left(\ln\Gamma\left(\frac1{2\pi}\right)+\ln G\left(\frac1{2\pi}\right)\right),$$ where $G(z)$ denotes the Barnes G-function.