Let $\operatorname{li}x$ denote the logarithmic integral $^{[1]}$$^{[2]}$$^{[3]}$: $$\operatorname{li}x=\int_0^x\frac{dt}{\ln t}$$ and $$I=\int_0^1\frac{\ln(-\ln x)\ \operatorname{li}^2x}{x}dx\approx-4.311872263...$$ Is it possible to express the integral $I$ in a closed form (using algebraic numbers, known mathematical constants, elementary and known special functions)?


I will start from Vladimir's result and my calculation. Let

$$ F(s) = \int_{0}^{\infty} z^{s-1} \mathrm{Ei}(-z)^{2} \, dz = \frac{2\Gamma(s)}{s} \int_{0}^{\frac{1}{2}} \frac{u^{s-1}}{1 - u} \, du. $$

Then we have

\begin{align*} \int_{0}^{\infty} z^{s-1} \mathrm{Ei}(-z)^{2} \log z \, dz &= F'(s) \\ &= F(s) \left( \psi_{0}(s) - \frac{1}{s} \right) + \frac{2\Gamma(s)}{s} \int_{0}^{\frac{1}{2}} \frac{u^{s-1}\log u}{1-u} \, du. \end{align*}

Plugging $s = 1$ to both sides,

\begin{align*} I &= F(1) \left( \psi_{0}(1) - 1 \right) + 2 \int_{0}^{\frac{1}{2}} \frac{\log u}{1-u} \, du \\ &= -2(\gamma+1)\log 2 + 2 \left[ \mathrm{Li}_{2}(1-u) \right]_{0}^{\frac{1}{2}} \\ &= -\zeta(2) - \log^{2} 2 - 2(\gamma + 1) \log 2 \end{align*}

as desired. Here,

\begin{align*} \mathrm{Li}_{2}(z) = - \int_{0}^{z} \frac{\log(1 - t)}{t} \, dt \end{align*}

is the dilogarithm and we exploited the special values

\begin{align*} \mathrm{Li}_{2}(1) &= \zeta(2), \\ \mathrm{Li}_{2}\left(\tfrac{1}{2}\right) &= \frac{1}{2}\zeta(2) - \frac{1}{2}\log^{2} 2. \end{align*}


Change the integration variable $z=\ln x$, then apply the formula $(12)$. The integral becomes $$I=\int_{-\infty}^0\ln(-z)\,\operatorname{Ei}^2z\ dz,\tag1$$ where $\operatorname{Ei}z$ is the exponential integral: $$\operatorname{Ei}z=-\int_{-z}^\infty\frac{e^{-t}}t dt.\tag2$$

In this form the integral can be evaluated by Mathematica: $$I=-\frac{\pi ^2}{6}-\ln^2 2-2\,(1+\gamma)\ln2,\tag3$$ where $\gamma$ is the Euler-Mascheroni Constant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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