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I need to evaluate the following integral with a high precision: $$ I=\int_{0}^{\infty}\left[% {\pi^{2} \over 6} - {\rm Li}_2\left({\rm e}^{-x}\right) -{\rm Li}_2\left({\rm e}^{-1/x}\right)\right]\,{{\rm d}x \over x}, $$ where ${\rm Li}_{2}$ denotes the dilogarithm $\displaystyle{% \left(~\mbox{note that}\ {\rm Li}_{2}\left(1\right) = {\pi^{2} \over 6}~\right)}$.

Unfortunately, a numerical integration in my CAS is only able to produce $3$ stable digits $I \approx 3.77\ldots$ that I do not even sure to be provably correct.

So, if only I were so lucky that a closed form existed for this integral, then, hopefully, it could be used to easily evaluate $I$ with a much higher precision. Could you suggest how to find a closed form ( if one exists )?

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2 Answers 2

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$$ I=\int^{\infty}_0 x^{-1}\sum^{\infty}_{k=1}\frac{1-e^{-kx}-e^{-kx^{-1}}}{k^2}dx\\ =\sum^{\infty}_{k=1}k^{-2}\int^{\infty}_0 \frac{1-e^{-kx}-e^{-kx^{-1}}}{x}dx\\ =2\sum^{\infty}_{k=1}k^{-2}(\gamma+\log k)\\ =2\gamma\zeta(2)-2\zeta'(2)\\ =2\zeta(2)(12\log A-\log2\pi). $$

Here $A$ is the Glaisher-Kinkelin constant.

Edit: $$\int^{\infty}_0 (1-e^{-kx}-e^{-kx^{-1}})\frac{dx}{x}\\ =\int^{1}_0 \frac{1-e^{-kx}}{x}dx-\int^{1}_0 \frac{e^{-kx^{-1}}}{x}dx+\int^{\infty}_1 \frac{1-e^{-kx^{-1}}}{x}dx-\int^{\infty}_1 \frac{e^{-kx}}{x}dx\\ =\int^{1}_0 \frac{1-e^{-kx}}{x}dx-\int^{\infty}_1 \frac{e^{-ky}}{y}dy+\int^{1}_0 \frac{1-e^{-ky}}{y}dy-\int^{\infty}_1 \frac{e^{-kx}}{x}dx\\ =2\int^{1}_0 \frac{1-e^{-kx}}{x}dx-2\int^{\infty}_1 \frac{e^{-kx}}{x}dx\\ =2\int^{k}_0 \frac{1-e^{-x}}{x}dx-2\int^{\infty}_k \frac{e^{-x}}{x}dx\\ =2(\operatorname{Ein}(k)-E_1(k))=2(\gamma+\log k). $$ Here $E_1$ and $\operatorname{Ein}$ are exponential integrals, see §6.2 of DLMF.

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    $\begingroup$ If you could perhaps expand on how to solve that integral on the second row. $\endgroup$
    – Lucian
    Feb 27, 2014 at 9:53
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\bbox[10px,#ffd]{I \equiv \int_{0}^{\infty}{\pi^{2}/6 - {\rm Li}_{2}\pars{\expo{-x}} -{\rm Li}_{2}\pars{\expo{-1/x}} \over x}\,\dd x}:\ {\large ?}}$


\begin{align} \color{#c00000}{I} & = -\int_{0}^{\infty}\ln\pars{x}\left[% -{\rm Li}_{2}'\pars{\expo{-x}}\pars{-\expo{-x}} -\right. \\ & \phantom{\color{#c00000}{I} = -\int_{0}^{\infty}\ln\left(x\right)\left[\right.} \left.{\rm Li}_{2}'\pars{\expo{-1/x}}\pars{\expo{-1/x} \over x^{2}}\right]\!\dd x \\[5mm] & = -\int_{0}^{\infty}\ln\pars{x}{\rm Li}_{2}'\pars{\expo{-x}}\expo{-x}\,\dd x \\[2mm] &\ +\int_{\infty}^{0}\ln\pars{1/x}{\rm Li}_{2}'\pars{\expo{-x}}x^{2}\expo{-x} \,\pars{-\,{\dd x \over x^{2}}} \\[5mm]&=-2\int_{0}^{\infty}\ln\pars{x}{\rm Li}_{2}'\pars{\expo{-x}}\expo{-x}\,\dd x \\[5mm] & = -2\int_{0}^{\infty}\ln\pars{x}\, {{\rm Li}_{1}\pars{\expo{-x}} \over \expo{-x}}\expo{-x}\,\dd x \\[5mm] & = \color{#c00000}{2\int_{0}^{\infty}\ln\pars{x} \ln\pars{1 - \expo{-x}}\,\dd x} \end{align} where we used $\ds{{\rm Li}_{s + 1}'\pars{z} = {{\rm Li}_{s}\pars{z} \over z}}$ and $\ds{{\rm Li}_{1}\pars{z} = -\log\pars{1 - z}}$. See [this web page](http://en.wikipedia.org/wiki/Polylogarithm).

\begin{align} \color{#00f}{\large I}&=2\int_{0}^{\infty}\ln\pars{x} \ln\pars{1 - \expo{-x}}\,\dd x = 2\sum_{\ell = 1}^{\infty} \pars{-\,{1 \over \ell}}\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{\infty}x^{\mu}\expo{-\ell x}\,\dd x \\[3mm]&=2\sum_{\ell = 1}^{\infty} \pars{-\,{1 \over \ell}}\lim_{\mu \to 0}\partiald{}{\mu} \bracks{\Gamma\pars{\mu + 1} \over \ell^{\mu + 1}} \\[3mm]&=2\sum_{\ell = 1}^{\infty} \pars{-\,{1 \over \ell}}\lim_{\mu \to 0} \bracks{{\Gamma\pars{\mu + 1}\Psi\pars{\mu + 1} \over \ell^{\mu + 1}} -{\Gamma\pars{\mu + 1}\ln\pars{\ell} \over \ell^{\mu + 1}}} \\[3mm]&=2\sum_{\ell = 1}^{\infty}\pars{-\,{1 \over \ell}}\bracks{% {-\gamma \over \ell} - {\ln\pars{\ell} \over \ell}} =2\gamma\sum_{\ell = 1}^{\infty}{1 \over \ell^{2}} +2\sum_{\ell = 1}^{\infty}{\ln\pars{\ell} \over \ell^{2}} \\[3mm]&=\color{#00f}{\large 2\bracks{\gamma\zeta\pars{2} - \zeta\,'\pars{2}}} \approx {\tt 3.7741} \end{align}

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  • $\begingroup$ this answer should have more upvotes (+1) $\endgroup$
    – tired
    Dec 16, 2016 at 11:05
  • $\begingroup$ @tired Thanks. I agree. $\endgroup$ Dec 17, 2016 at 22:25

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