# Evaluate the Integral $$\int_{0}^{\infty} \frac{\ln(x)}{1+e^{ax}} \mathrm{d}x$$

How to evaluate

I = \int\limits_{0}^{\infty} \frac{\ln(x)}{1+e^{ax}} \mathrm{d}x
\tag{1}
\label{eq:161008-1}

was a question posed at Mathematics Stack Exchange. As of the time of writing this post, there was a good solution posted there. Here is my solution.

Let
\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

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)

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 here 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.