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, $$I\approx2.18083278090426462584033339029703713513\dots$$

  • 1
    Have you tried Taylor expansion? – user122283 Mar 13 '14 at 0:22
  • Is $\ln^2(x) = (\ln(x))^2$ or $\ln(\ln(x))$? – Yiyuan Lee Mar 13 '14 at 0:46
  • @YiyuanLee The numerical value agrees with $(\ln x)^2$. – Vladimir Reshetnikov Mar 13 '14 at 1:51
  • @YiyuanLee $\ln^2x=(\ln x)^2$ – Nik Z. Mar 13 '14 at 3:00
up vote 20 down vote accepted

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

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

  • I upvoted the other answer, because it shows how the result is gotten. This is magic. – marty cohen Mar 26 '15 at 6:34

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.