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Find $\displaystyle A =\int \frac{x \, dx}{e^x +1 }$

Using Wolfram Alpha this yields $-\operatorname{Li}_2(-e^x) + \frac{x^2}{2}-x \ln(e^x +1)$, where $\operatorname{Li}$ is the polylogaritm function. I am pretty really new to polylogarithms but I wanted to know how WA arrived at this result and if possible if there are any books that could help me grasp the use of these special functions in integration.

Martin Sleziak
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user362617
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3 Answers3

5

Note that $$\begin{align} \int\,\frac{x\,\exp(x)}{\exp(x)+1}\,\text{d}x&=\int\,x\,\text{d}\ln\big(\exp(x)+1\big)\\&=x\,\ln\big(\exp(x)+1\big)-\int\,\ln\big(\exp(x)+1\big)\,\text{d}x\,. \end{align}$$ Then, we have that $$\int\,\ln\big(\exp(x)+1\big)\,\text{d}x=\int\,\frac{\ln\Big(1-\big(-\exp(x)\big)\Big)}{\big(-\exp(x)\big)}\,\text{d}\big(-\exp(x)\big)=-\text{Li}_2\big(-\exp(x)\big)+C\,,$$ where $C$ is a constant (as $\text{Li}_2(z)=-\int_0^z\,\frac{\ln(1-t)}{t}\,\text{d}t$, according to this). Thus, $$\begin{align}\int\,\frac{x}{\exp(x)+1}\,\text{d}x &=\int\,\left(x-\frac{x\,\exp(x)}{\exp(x)+1}\right)\,\text{d}x \\&=\frac{1}{2}x^2-x\,\ln\big(\exp(x)+1\big)-\text{Li}_{2}\big(-\exp(x)\big)+C\,.\end{align}$$

You can prove $\text{Li}_2(z)=-\int_0^z\,\frac{\ln(1-t)}{t}\,\text{d}t$ directly via $$\begin{align}\text{Li}_2(z)&=\sum_{k=1}^\infty\,\frac{z^k}{k^2}=\sum_{k=1}^\infty\,\int_0^z\frac{t^{k-1}}{k}\,\text{d}t=\int_0^z\,\frac{1}{t}\,\sum_{k=1}^\infty\,\frac{t^k}{k}\,\text{d}t \\ &=\int_0^z\,\frac{-\ln(1-t)}{t}\,\text{d}t\,, \end{align}$$ for all $z\in\mathbb{C}$ with $|z|<1$. Then, analytic continuation handles the rest.

Batominovski
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Since $\int xe^{-nx}\,\mathrm{d}x=-\frac1{n^2}(nx+1)e^{-nx}+C$, $$ \begin{align} \int \frac{x\,\mathrm{d}x}{e^x+1} &=\int x\left(e^{-x}-e^{-2x}+e^{-3x}+\dots\right)\mathrm{d}x\tag{1}\\ &=C-(x+1)e^{-x}+\tfrac14(2x+1)e^{-2x}-\tfrac19(3x+1)e^{-3x}+\dots\tag{2}\\ &=C-x\left(e^{-x}-\tfrac12e^{-2x}+\tfrac13e^{-3x}-\dots\right)+\left(-e^{-x}+\tfrac14e^{-2x}-\tfrac19e^{-3x}+\dots\right)\tag{3}\\ &=C-x\log\left(1+e^{-x}\right)+\mathrm{Li}_2\left(-e^{-x}\right)\tag{4}\\ &=C+x^2-x\log\left(e^x+1\right)-\tfrac{\pi^2}6-\mathrm{Li}_2\!\left(-e^x\right)-\tfrac12x^2\tag{5}\\ &=C_2+\tfrac12x^2-x\log\left(e^x+1\right)-\mathrm{Li}_2\!\left(-e^x\right)\tag{6} \end{align} $$ Explanation:
$(1)$: write $\frac1{e^x+1}$ as a geometric series
$(2)$: use the formula from above
$(3)$: separate terms
$(4)$: use the series for $\log(1+x)$ and $\mathrm{Li}_2(x)$
$(5)$: use $(5)$ from this answer
$(6)$: collect terms

Community
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robjohn
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\begin{align} I&=\int\frac{x\ dx}{e^x+1}=\int\frac{xe^{-x}}{1+e^{-x}}\ dx\overset{e^{-x}=y}{=}\int\frac{\ln y}{1+y}\ dy\\ &\overset{IBP}{=}\ln(1+y)\ln y-\int\frac{\ln(1+y)}{y}\ dy\\ &=\ln(1+y)\ln y-(-\operatorname{Li}_2(-y))\\ &=\ln(1+e^{-x})\ln(e^{-x})+\operatorname{Li}_2(-e^{-x})\\ &\{\color{red}{\text{use }\operatorname{Li}_2(-1/x)=-\operatorname{Li}_2(-x)-\frac12\ln^2x-\zeta(2)}\}\\ &=-x\ln\left(\frac{e^x+1}{e^x}\right)-\operatorname{Li}_2(-e^{x})-\frac12\ln^2(e^x)-\zeta(2)\\ &=-x\ln(1+e^x)+x^2-\operatorname{Li}_2(-e^{x})-\frac12x^2-\zeta(2)\\ &=\frac12x^2-x\ln(1+e^x)-\operatorname{Li}_2(-e^{x})-\zeta(2)+C \end{align}

Note: The identity in red is proved here but just replace $x$ with $-x$.

Ali Shadhar
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