How to evaluate a certain definite integral: $\int_{0}^{\infty}\frac{\log(x)}{e^{x}+1}dx$

Question

How can I show that:

$$\int_{0}^{\infty}\frac{\log(x)}{e^{x}+1}dx=-\frac{\log^{2}(2)}{2}$$

EDIT: This is equivalent to showing that $\eta'(1)=-\ln2\gamma-\dfrac{\ln^2(2)}{2}$.

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \int_{0}^{\infty}{\ln\pars{x} \over \expo{x} + 1}\,\dd x =\lim_{\mu\to 0}\partiald{}{\mu}\int_{0}^{\infty} {x^{\mu}\expo{-x} \over 1 + \expo{-x}}\,\dd x $$

\begin{align} \int_{0}^{\infty} {x^{\mu}\expo{-x} \over 1 + \expo{-x}}\,\dd x &=\sum_{n = 0}^{\infty}\pars{-1}^{n}\int_{0}^{\infty} x^{\mu}\expo{-\pars{n + 1}x}\,\dd x =\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{n + 1}^{\mu + 1}} \int_{0}^{\infty}x^{\mu}\expo{-x}\,\dd x \\[3mm]&=\Gamma\pars{\mu + 1}\sum_{n = 1}^{\infty} {\pars{-1}^{n + 1} \over n^{\mu + 1}} =-\Gamma\pars{\mu + 1}{\rm Li}_{\mu + 1}\pars{-1} \\[3mm]&=\Gamma\pars{\mu + 1}\pars{1 - 2^{-\mu}}\zeta\pars{\mu + 1} \end{align}

\begin{align} &\color{#66f}{\large\int_{0}^{\infty}{\ln\pars{x} \over \expo{x} + 1}\,\dd x} =\lim_{\mu\ \to\ 0} \partiald{\bracks{\Gamma\pars{\mu + 1}\pars{1 - 2^{-\mu}}\zeta\pars{\mu + 1}}}{\mu} \tag{1} \\[3mm]&=\color{#66f}{\large -\,\half\,\ln^{2}\pars{2}} \approx -0.2402 \end{align}

See the Dirichlet Eta Function link and the PolyLogarithm Function Link.

ADDENDA

The limit, in expression $\pars{1}$ has to be handled carefully because $\ds{\zeta\pars{\mu + 1} \sim {1 \over \mu}}$ when $\ds{\mu \sim 0}$. It has the Laurent Expansion $$ \zeta\pars{\mu + 1}={1 \over \mu} + \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n!}\, \gamma_{n}\,\mu^{n} $$ where $\ds{\gamma_{n}}$'s are the Stieltjes Constants. Let's consider the limit evaluation. In order to evaluate the limit we just need the expansion, up to order one, of each factor in ( $\ds{\gamma}$ is the Euler-Mascheroni Constant ): $$ \Gamma\pars{\mu + 1}\,{1 - 2^{-\mu} \over \mu}\,\bracks{\mu\zeta\pars{\mu + 1}}\,, \qquad\left\vert\begin{array}{ccl} \ \Gamma\pars{\mu + 1} & \sim & 1 - \gamma\mu \\[2mm] \ {1 - 2^{-\mu} \over \mu} & \sim & \ln\pars{2} - \half\,\ln^{2}\pars{2}\mu \\[2mm] \ \mu\zeta\pars{\mu + 1} & \sim & 1 + \gamma\mu\,,\quad\gamma_{0} = \gamma \end{array}\right. $$ From these expression we see that the product $\ds{\Gamma\pars{\mu + 1}\bracks{\mu\zeta\pars{\mu + 1}} \sim 1 - \gamma^{2}\mu^{2}}$ is already of order $\ds{\mu^{2}}$. We are left with the'middle factor' $\ds{1 - 2^{-\mu} \over \mu}$ such that \begin{align}&\lim_{\mu\ \to\ 0} \partiald{\bracks{\Gamma\pars{\mu + 1}\pars{1 - 2^{-\mu}}\zeta\pars{\mu + 1}}}{\mu} =\lim_{\mu\ \to\ 0} \partiald{\bracks{\ln\pars{2} - \ln^{2}\pars{2}\mu/2}}{\mu} \\[3mm]&=\color{#66f}{\large -\,\half\,\ln^{2}\pars{2}} \end{align}

Answer 2

Since for any $\alpha$ such that $\Re(\alpha)>-1$ we have: $$f(\alpha)=\int_{0}^{+\infty}\frac{x^\alpha}{e^x+1}\,dx = \left(1-\frac{1}{2^{\alpha}}\right)\Gamma(1+\alpha)\zeta(1+\alpha)\tag{1}$$ by expanding the integrand function as a geometric series, we just need to find the limit of the derivative of the RHS of $(1)$ when $\alpha\to 0^+$. It is worth to consider logarithmic derivatives and exploit the identity $g'(z)=g(z)\cdot\frac{d}{dz}\log g(z)$, since the RHS of $(1)$ is a product and $f(0)=\log 2$. Now we have, in a neighbourhood of zero: $$ 2^z-1 = z \log2 + \frac{z^2}{2}\log^2 2 + o(z^3),$$ $$ \Gamma(z+1) = 1-\gamma z + O(z^2),$$ $$ \zeta(z+1) = \frac{1}{z}+\gamma+O(z),$$ hence their product is $$ \log 2 + \frac{z}{2}\log^2 2 + O(z^2)$$ and the value of the logarithmic derivative in zero of $\left(1-\frac{1}{2^{z}}\right)\Gamma(1+z)\zeta(1+z)$ is just $-\frac{\log 2}{2}$.

This gives:

$$\int_{0}^{+\infty}\frac{\log x}{e^x+1}\,dx = -\frac{\log^2 2}{2}$$

as wanted.

Answer 3

\begin{align} \int^\infty_0\frac{\color\red{\log{x}}}{e^x+1}dx &=\color\red{\lim_{a \to 1}\frac{d}{da}}\int^\infty_0\frac{\color\red{x^{a-1}}e^{-x}}{1+e^{-x}}dx\tag1\\ &=\lim_{a \to 1}\frac{d}{da}\sum_{n \ge 0}(-1)^n\int^\infty_0x^{a-1}e^{-(n+1)x}dx\tag2\\ &=\lim_{a \to 1}\sum_{n \ge 0}(-1)^n\frac{d}{da}\frac{\Gamma(a)}{(n+1)^a}\tag3\\ &=\lim_{a \to 1}\sum_{n \ge 0}(-1)^n\frac{\Gamma(a)\psi(a)-\Gamma(a)\log(n+1)}{(n+1)^a}\tag4\\ &=\psi(1)\sum_{n \ge 0}\frac{(-1)^n}{n+1}-\sum_{n \ge 0}\frac{(-1)^n\log(n+1)}{n+1}\tag5\\ &=-\gamma\log{2}-\left(\frac{1}{2}\log^22-\gamma\log{2}\right)\tag6\\ &=-\frac{1}{2}\log^22\\ \end{align} Explanation
$(1)$: Divide numerator and denominator by $e^x$
$(2)$: Expand the integrand as a geometric series, swap the order of integration and summation
$(3)$: Recognise the gamma function
$(4)$: Quotient rule
$(5)$: Apply the limit and split the sum into two.
$(6)$: For the first sum, $\psi(1)=-\gamma$, and $\displaystyle\ln(1+1)=\sum_{n \ge 0}\frac{(-1)^n}{n+1}1^{n+1}$. I will now show how to evaluate the second sum. \begin{align} \sum_{n \ge 0}\frac{(-1)^n\log(n+1)}{n+1} &=\sum_{n \ge 1}\frac{(-1)^{n-1}\log{n}}{n}\tag7\\ &=\lim_{s \to 1}\frac{d}{ds}\sum_{n \ge 1}\frac{(-1)^{n}}{n^s}\\ &=-\eta'(1)\tag8\\ &=\frac{1}{2}\log^22-\gamma\log{2}\tag9\\ \end{align} $(7)$: Index shift
$(8),(9)$: See Dirichlet eta function, Stieltjes Constants

Answer 4

So we would like to find $\eta'(1)$ which in essence mean computing $\sum_{k=1}^{\infty} \frac{(-1)^{k-1}\log k}{k}$.

We will present a solution which is in some sense reminscent of one of the proofs of the fact $\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k}=\log 2$. Namely rewriting $\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k}=\sum_{k=1}^{2n} \frac{1}{k}-2\sum_{k=1}^{n} \frac{1}{2k}=H_{2n}-H_{n}=(H_{2n}-\log 2n)-(H_n-\log n)+\log 2$ ($H_n$ represents the $n$-th harmonic number) and using the fact $\lim_{n\to \infty}H_n-\log n$ exists (and is in fact equal to Euler-Mascheroni $\gamma$ constant).

First of all we prove that the following sequence is convergent - $b_n=\sum_{k=1}^{n}\frac{\log k}{k}-\frac{\log(n)^2}{2}$ (analog of $H_n-\log n$, see here). Rewrite $b_n=\frac{\log 2}{2}-\frac{\log^2 3}{2}+\sum_{k=3}^{n}\frac{\log k}{k}-\int_{3}^{n}\frac{\log x}{x}\mathrm{d}x$ and denote the last two summands as $c_n$ so it is equivalent to prove $c_n$ is convergent. Observe that $\frac{\log x}{x}$ is decreasing for $x\ge 3$. Now $c_{n}-c_{n-1}=\frac{\log n}{n}-\int_{n-1}^n \frac{\log x}{x}\mathrm{d}x\le \frac{\log n}{n}-\frac{\log n}{n}=0$, so the sequence is decreasing. Moreover $c_n=\sum_{k=3}^{n}\frac{\log k}{k}-\int_{3}^{n}\frac{\log x}{x}\mathrm{d}x=\sum_{k=3}^{n}\frac{\log k}{k}-\sum_{k=3}^{n-1}\int_{k}^{k+1}\frac{\log x}{x}\mathrm{d}x\ge \sum_{k=3}^{n}\frac{\log k}{k}-\sum_{k=3}^{n-1}\frac{\log k}{k}=\frac{\log n}{n}\ge 0$,

so the sequence is bounded from below, and it is convergent. (the proof is the same as that of the general fact here.)

Now $$a_{2n}=\sum_{k=1}^{2n}\frac{(-1)^{k-1}\log k}{k}=\sum_{k=1}^{2n}\frac{\log k}{k}-2\sum_{k=1}^{n}\frac{\log 2k}{2k}=\sum_{k=1}^{2n}\frac{\log k}{k}-\sum_{k=1}^{n}\frac{\log k}{k}-H_n\log 2=\left(\sum_{k=1}^{2n}\frac{\log k}{k}-\frac{\log^2 2n}{2}\right)-\left(\sum_{k=1}^{n}\frac{\log k}{k}-\frac{\log^2 n}{2}\right)-(H_n-\log n)\log 2+\left(-\log n\log2+\frac{\log^2 2n}{2}-\frac{\log^2 n}{2}\right)\to -\gamma \log 2 + \frac{\log^2 2}{2}$$ as $n$ goes to $\infty$.

In a similar manner the series $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\log^d n}{n}$$ for integer $d$ (i.e. the $d$-th derivative of eta at $1$) could be evaluated, however the Stieltjes constants (which we referred to earlier) will be present in the answer.

Answer 5

\begin{align*} &\int_0^\infty \ln(x)e^{-x}dx = -\gamma\\ &\mathop{\lim}_{s\to 1}\left(\zeta(s) - \frac{1}{s-1}\right) = \gamma\\ &\Rightarrow\int_0^\infty \ln(x)e^{-sx}dx = \mathop{\lim}_{s\to 1}\left(\frac{1}{s}\int_0^\infty \ln\left(\frac{y}{s}\right)e^{-y}dy\right) \\ &\qquad= -\frac{\gamma+\ln(s)}{s}\\ &\Rightarrow\int_0^\infty \frac{\ln(x)}{e^x+1}dx = \int_0^\infty \ln(x)e^{-x}\left(\sum_{n=0}^\infty (-1)^ne^{-nx}\right)dx = \int_0^\infty \ln(x)\left(\sum_{n=1}^\infty (-1)^{n-1}e^{-nx}\right)dx \\ &\qquad=\sum_{n=1}^\infty (-1)^{n-1}\int_0^\infty \ln(x)e^{-nx}dx \\ &\qquad=\sum_{n=1}^\infty (-1)^n\left(\frac{\gamma+\ln(n)}{n}\right) = \gamma\sum_{n=1}^\infty \frac{(-1)^n}{n} + \sum_{n=1}^\infty \frac{(-1)^n\ln(n)}{n} \\ &\Rightarrow\Rightarrow \boxed{\int_0^\infty \frac{\ln(x)}{e^x+1}dx = -\gamma \ln(2) + \sum_{n=1}^\infty \frac{(-1)^n\ln(n)}{n}} \\ &\sum_{n=1}^\infty (-1)^n\frac{1}{n^s} = \left(\frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{6^s} + \ldots\right) - \left(1 + \frac{1}{3^s} + \frac{1}{5^s} + \frac{1}{7^s} + \ldots\right) \\ &\qquad= 2\left(\frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{6^s} + \ldots\right) - \left(1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \ldots\right) \\ &\qquad= 2\frac{1}{2^s}\zeta(s) - \zeta(s) = (2^{1-s}-1)\zeta(s) \\ &\Rightarrow\frac{d}{ds}\left(\sum_{n=1}^\infty (-1)^n\frac{1}{n^s}\right) = -\ln(2) \cdot 2^{1-s}\zeta(s) + (2^{1-s}-1)\zeta(s) \\ &\Rightarrow \boxed{\sum_{n=1}^\infty (-1)^n\frac{\ln(n)}{n^s} = \ln(2)\cdot 2^{1-s}\zeta(s) + (1-2^{1-s})\zeta'(s)} \\ &\zeta(s) - \frac{1}{s-1} = \gamma \Rightarrow \frac{(s-1)\zeta(s)-1}{s-1} = \gamma \\ &\Rightarrow \boxed{\zeta(s)+(s-1)\zeta'(s) = \gamma}\text{ and }\boxed{\zeta(s)(s-1)=1} \\ &\sum_{n=1}^\infty (-1)^n\frac{\ln(n)}{n^s} = \ln(2) \cdot 2^{1-s}\zeta(s) + (1-2^{1-s})\zeta'(s)\\ &\Rightarrow \sum_{n=1}^\infty (-1)^n\frac{\ln(n)}{n} = \mathop{\lim}_{s\to 1}\left(\ln(2)\cdot 2^{1-s}\zeta(s) + (1-2^{1-s})\zeta'(s)\right) \\ &= \mathop{\lim}_{s\to 1}\left(\ln(2) \cdot 2^{1-s}\zeta(s) + \frac{1-2^{1-s}}{s-1} \qquad-\frac{1-2^{1-s}}{s-1}\zeta(s) \right) \\ &= \gamma \ln(2) + \mathop{\lim}_{s\to 1}\left(\frac{(s-1)\ln(2) \cdot 2^{1-s} - (1-2^{1-s})}{(s-1)^2}\right)\zeta(s)\\ &= \gamma \ln(2) + \mathop{\lim}_{s\to 1}\left(\frac{(s-1)\ln(2) \cdot 2^{1-s} - (1-2^{1-s})}{(s-1)^2}\right) \\ &= \gamma \ln(2) - \frac{1}{2}{\ln^2}(2) \\ &\Rightarrow \boxed{\sum_{n=1}^\infty (-1)^n\frac{\ln(n)}{n} = \gamma \ln(2) - \frac{1}{2}{\ln^2}(2)} \\ &\text{finally}\int_0^\infty \frac{\ln(x)}{e^x+1}dx = -\gamma \ln(2) + \sum_{n=1}^\infty (-1)^n\frac{\ln(n)}{n} \\ &= -\gamma \ln(2) + \gamma \ln(2) - \frac{1}{2}{\ln^2}(2)\\ &\Rightarrow \boxed{\int_0^\infty \frac{\ln(x)}{e^x+1}dx = -\frac{1}{2}{\ln^2}(2)} \end{align*}