Evaluating the integral $\int_0^\infty \frac{\ln(x)}{e^x+1}$

Question

I'm interested in evaluating the integral of

$$ \int_0^\infty \frac{\ln(x)}{e^x+1}\ $$

TLDR: How do you evaluate $ \eta'(1) $?

I found and adapted this integral from this video, unfortunately, they didn't fully evaluate the integral and I'm interested in the result.

This question is similar to this, but my question goes more in-depth on the way that I tried to evaluate this integral.

This was solved by @J.G. in the answers, and by @Zacky in the comments.

I'll run through what I did and what the video did to evaluate the integral

$$ I =\int_0^\infty \frac{\ln(x)}{e^x+1}$$

Multiplying by $ \frac{e^{-x}}{e^{-x}}\ $ results in

$$ = \int_0^\infty\frac{e^{-x}\ln(x)}{e^{-x}+1} $$

Expanding the denominator into a geometric sum results in

$$ = \int_0^\infty\ e^{-x}\ln(x)\sum_{k\ge0}{-e^{-x^k}} $$

Interchanging our signs of integration of summation and simplifying results in

$$ \sum_{k \ge 0}(-1)^k \int_0^\infty\ e^{-x(k+1)}\ln(x) $$

Substituting $ x(k+1) = t $ and $ dx = \frac{dt}{k+1} $ results in

$$ \sum_{k \ge 0} (-1)^k \int_0^\infty e^{-t}\ln(\frac t{k+1})\frac{dt}{k+1} $$

Expanded and simplified you get

$$ \sum_{k \ge 0} \frac{(-1)^k}{k+1} (\int_0^\infty \ln(t)e^{-t}dt - \ln(k+1)\int_0^\infty e^{-t}dt) $$

Once again simplified we can see that our first integral is the $ \mathcal L\{\ln(t)\} $ at $s = 1$ and our second integral is $\Gamma(1)$ in total this results in.

Where $ \mathcal L $ is the Laplace transform and $\Gamma$ is the Gamma function.

$$ \sum_{k \ge 0} \frac{(-1)^k}{k+1}(\mathcal L\{\ln(t)\}-\ln(k+1)\Gamma(1)) $$

This then simplifies to

$$ \sum_{k \ge 0} \frac{(-1)^k}{k+1}(-\gamma-\ln(k+1)) $$

With $ \gamma $ being the euler-mascheroni constant

Increasing our lower bound on our summation we get

$$ \sum_{k \ge 1} \frac{(-1)^{k-1}}k(-\gamma-\ln(k)) $$

Realizing that the Dirichlet Eta function is defined as

$$ \eta(s) = \sum_{k \ge 1} \frac{(-1)^{k-1}}{k^s} $$

We can then see that our equation is

$$ \eta(1)(-\gamma-\ln(k)) $$

Knowing that $ \eta(1) = \ln(2) $ we can then distribute our eta to get

$$ -\gamma\ln(2)-\ln(k)\eta(1) $$

Which is

$$ -\gamma\ln(2) - \sum_{k \ge 1} \frac{(-1)^{k-1}}k\ln(k) $$

Taking the derivative of the eta function we can see that this is

$$ -\gamma\ln(2)+\eta'(1) $$

According to Wikipedia $ \eta'(1) $ is equal to $ \gamma\ln(2)-\frac{\ln(2)^2}2$

Meaning that our integral $I$ will evaluate to

$$ I = -\frac{\ln(2)^2}2$$

But how can we evaluate $ \eta'(1) $ without encountering the pole at $s = 1$ of the Zeta function?

Are there any other interesting ways to evaluate this integral not using the eta function?

Thank you for your time and patience!

Answer 1

For $s\approx1$, $\zeta(s)\approx\frac{1}{s-1}+\gamma$ and $\zeta^\prime(s)\approx\frac{-1}{(s-1)^2}$, so$$\begin{align}\eta^\prime(s)&=2^{1-s}\ln2\cdot\zeta(s)+(1-2^{1-s})\zeta^\prime(s)\\&\approx(1+(1-s)\ln 2)\ln2\left(\frac{1}{s-1}+\gamma\right)-\frac{\ln 2}{s-1}+\frac12\ln^22\\&=\gamma\ln 2-\frac12\ln^22.\end{align}$$Note we only need to expand $2^{1-s}$ to $O(1-s)^2$ viz.$$2^{1-s}=1+(1-s)\ln 2+\frac12(1-s)^2\ln^22+o((1-s)^2)$$in the $\zeta^\prime$ term.