How does one prove $\int_0^\infty \frac{\log(x)}{1 + e^{ax}} \, dx = -\frac{\log(2)(2\log(a) + \log(2))}{2a}$ for $a > 0$?

Question

Link to WolframAlpha's assertion. Here's my attempt. Using the substitution $t = ax$, we can show the integral is equal to $$ \frac{1}{a} \int_0^\infty \frac{\log(t)}{1 + e^t}\, dt -\frac{\log(a)}{a} \int_0^\infty \frac{dt}{1 + e^{t} } .$$

The second integral is equal to $\log(2)$ using another substitution $v = e^t$ and partial fractions.

So I'm left with the first integral. I'll switch to complex variables for notation. I make two observations:

(I) The denominator has simple poles when $z = t = (2k-1)\cdot i \pi, \, k \in \mathbb{N}.$

(II) The numerator has a branch point $z = 0$.

Answer 1

Let $I$ be the integral given by

$$\bbox[5px,border:2px solid #C0A000]{I=\int_0^\infty\frac{\log(x)}{1+e^x}\,dx} \tag 1$$

Expanding the denominator of $(1)$ in a series of $e^{-nx}$ and interchanging the order of summation and integration reveals

$$\begin{align} I&=\int_0^\infty\frac{e^{-x}\log(x)}{1+e^{-x}}\,dx\\\\ &=\int_0^\infty\log(x)\sum_{n=0}^\infty (-1)^ne^{-(n+1)x}\,dx\\\\ &=\sum_{n=0}^\infty (-1)^n \int_0^\infty\log(x)e^{-(n+1)x}\,dx\\\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{n+1}\int_0^\infty e^{-x}(\log(x)-\log(n+1))\,dx \\\\ &=\sum_0^\infty \frac{(-1)^{n+1}(\gamma+\log(n+1))}{n+1}\\\\ &=-\gamma\log(2)+\color{blue}{\sum_{n=1}^\infty \frac{(-1)^{n}\log(n)}{n}}\\\\ &=-\gamma\log(2)+\color{blue}{\eta'(1)}\\\\ &=-\gamma\log(2)+\gamma\log(2)-\frac12\log^2(2)\\\\ &=-\frac12\log^2(2) \end{align}$$

where $\eta'(s)$ is the derivative of the Dirichlet Eta Function with $\eta'(1)=\gamma\log(2)-\frac12\log^2(2)$ (SEE THIS ANSWER).

Therefore, we have the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty\frac{\log(x)}{1+e^x}\,dx=-\frac12\log^2(2)}$$

which agrees with that obtained using Wolfram Alpha!

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NOTE $1$: DIRECT EVALUATION OF THE SUM $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}\log(n)}{n}$

Here, we provide a direct evaluation of the sum $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}\log(n)}{n}$ facilitated by use of the Euler-Maclaurin Summation Formula (EMSF). To that end we proceed.

First, we note that we can write for any $N\ge 1$

$$\begin{align} \sum_{n=1}^{2N}\frac{(-1)^{n}\log(n)}{n}&=2\sum_{n=1}^{N}\frac{\log(2n)}{2n}-\sum_{n=1}^{2N}\frac{\log(n)}{n}\\\\ &=\log(2)\sum_{n=1}^N\frac1n -\sum_{n=N+1}^{2N}\frac{\log(n)}{n} \tag 2 \end{align}$$

Using the EMSF, we expand the second sum on the right-hand side of $(2)$ to obtain

$$\begin{align} \sum_{n=N+1}^{2N}\frac{\log(n)}{n}&=\int_N^{2N}\frac{\log(x)}{x}\,dx+O\left(\frac{\log(N)}{N}\right)\\\\ &=\frac12\log^2(2N)-\frac12\log^2(N)+O\left(\frac{\log(N)}{N}\right)\\\\ &=\log(2)\log(N)+\frac12\log^2(2)+O\left(\frac{\log(N)}{N}\right) \tag 3 \end{align}$$

Substituting $(3)$ in $(2)$ reveals

$$\begin{align} \sum_{n=1}^{\infty}\frac{(-1)^{n}\log(n)}{n}&=\lim_{N\to \infty}\sum_{n=1}^{2N}\frac{(-1)^{n}\log(n)}{n}\\\\ &=\lim_{N\to \infty}\left(\log(2)\sum_{n=1}^n\frac1n -\log(2)\log(N)-\frac12\log^2(2)+O\left(\frac{\log(N)}{N}\right)\right)\\\\ &=\log(2)\lim_{N\to \infty}\left(\sum_{n=1}^N \frac1n -\log(N)\right)-\frac12\log^2(2)\\\\ &=\gamma \log(2)-\frac12\log^2(2) \end{align}$$

Therefore, we find that

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^{\infty}\frac{(-1)^{n}\log(n)}{n}=\gamma \log(2)-\frac12\log^2(2)}$$

Answer 2

Here is another solution, although it is not as efficient as Dr. MV's solution.

Let $$I = \int\limits_{0}^{\infty} \frac{\ln(x)}{1+e^{ax}} \mathrm{d}x$$ and \begin{align} I_{1} &= \int\limits_{0}^{\infty} \frac{x^{b}}{1+e^{ax}} \mathrm{d}x = \int\limits_{0}^{\infty} \frac{x^{b}e^{-ax}}{1+e^{-ax}} \mathrm{d}x \\ &= \sum\limits_{n=0}^{\infty} (-1)^{n} \int\limits_{0}^{\infty} x^{b} e^{-(a+na)x} \mathrm{d}x \end{align}

We designate the last integral on the right as $I_{2}$ and make the substitution $y=(a+na)x$ \begin{align} I_{2} &= \int\limits_{0}^{\infty} x^{b} e^{-(a+na)x} \mathrm{d}x \\ &= \frac{1}{(a+na)^{b+1}} \int\limits_{0}^{\infty} y^{b} e^{-y} \mathrm{d}y \\ &= \frac{\Gamma(b+1)}{(a+na)^{b+1}} \\ &= \frac{\Gamma(b+1)}{a^{b+1}} \frac{1}{(n+1)^{b+1}} \end{align}

Now $I_{1}$ becomes \begin{equation} I_{1} = \frac{\Gamma(b+1)}{a^{b+1}} \sum\limits_{n=0}^{\infty} (-1)^{n} \frac{1}{(n+1)^{b+1}} = \frac{\Gamma(b+1)}{a^{b+1}} \eta(b+1) \end{equation} and we have \begin{align} I &= \lim_{b \to 0} \frac{\partial I_{1}}{\partial b} \\ &= \lim_{b \to 0} \int\limits_{0}^{\infty} \frac{x^{b} \ln(x)}{1+e^{ax}} \mathrm{d}x \\ &= \lim_{b \to 0}\frac{\partial}{\partial b}\frac{\Gamma(b+1)\eta(b+1)}{a^{b+1}} \\ &= \lim_{b \to 0} \frac{a^{b+1} \Big[ \Gamma(b+1)\eta^{\prime}(b+1) + \Gamma^{\prime}(b+1)\eta(b+1) \Big] - \Gamma(b+1)\eta(b+1)a^{b+1}\ln(a)}{\left(a^{b+1}\right)^{2}} \\ &= \lim_{b \to 0} \frac{\Gamma(b+1)\eta^{\prime}(b+1) + \Gamma^{\prime}(b+1)\eta(b+1) - \Gamma(b+1)\eta(b+1)\ln(a)}{a^{b+1}} \\ \tag{a} &= \frac{1}{a} \left(\Big[\gamma \ln(2) - \frac{1}{2} \ln^{2}(2)\Big] -\gamma \ln(2) - \ln(2)\ln(a) \right) \\ &= -\frac{\ln(2)}{a} \left(\frac{1}{2} \ln(2) + \ln(a) \right) \end{align}

In step (a) we have \begin{align} \lim_{b \to 0} \Gamma^{\prime}(b+1)\eta(b+1) &= \lim_{b \to 0} \Gamma^(b+1)\psi(b+1)\eta(b+1) \\ &= \Gamma(1)\psi(1)\eta(1) \\ &= -\gamma \ln(2) \end{align} and \begin{align} \lim_{b \to 0} \Gamma(b+1)\eta^{\prime}(b+1) &= \lim_{s \to 1} \eta^{\prime}(s) \\ &= \lim_{s \to 1} \sum\limits_{n=0}^{\infty} (-1)^{n} \frac{\ln(n)}{n^{s}} \\ &= \gamma \ln(2) - \frac{1}{2} \ln^{2}(2) \end{align} See Dr. MV's solution for a proof of this result.

Notes:

1. $\Gamma(z)$ is the Gamma function.
2. $\eta(s)$ is the Dirichlet eta function.
3. $\zeta(s)$ is the Riemann zeta function.
4. $\psi(z)$ is the digamma function.
5. $\gamma$ is the Euler-Mascheroni constant.