The special function $P(s)=\int^\infty_0 \frac{\ln(x)dx}{1+x^s}$ 
Evaluate the integral
$$P(s)=\int^\infty_0 \frac{\ln(x) \,dx}{1+x^s}$$

I tried integration by parts, using $u=\ln(x),du=\frac{x}{dx},v=\frac{1}{1+x^s}$, but I did not manage to get anything.
I thought about using a power series, but I did not find one that converges on the entire domain.
Using the substitution $u=\ln(x),du=\frac{dx}{x}$ we get $$\int^\infty_{-\infty} \frac{ue^{-u}}{1+e^{su}} ,$$
which doesn't seem useful either.
The previous similar question Integral of $\int^{\infty}_0 \frac{x^n}{x^s+1}dx$ gives
$$\int^\infty_0 \frac{x^n \,dx}{1+x^s}=\frac{\Gamma(\frac{1}{s})\Gamma\left(1-\frac{1}{s}\right)}{s(n+1)} .$$
Another possibly relevant question is Integral of $\int^{\infty}_0 \frac{e^{-x}}{x^s+1}\,dx$.
 A: Differentiating$$\int_0^\infty\frac{x^{t-1}dx}{1+x^s}=\frac1s\int_0^\infty\frac{y^{t/s-1}dy}{1+y}=\frac{\pi}{s}\csc\frac{\pi t}{s}$$with respect to $t$ gives$$\int_0^\infty\frac{x^{t-1}\ln x\,dx}{1+x^s}=-\frac{\pi^2}{s^2}\csc\frac{\pi t}{s}\cot\frac{\pi t}{s}.$$Set $t=1$ to get$$\int_0^\infty\frac{\ln x\,dx}{1+x^s}=-\frac{\pi^2}{s^2}\csc\frac{\pi}{s}\cot\frac{\pi}{s}.$$The case $s=2$ is a famous sanity check, for which the integral is $0$.
A: To see a possible pattern, I think that we need to explore for larger values of $s$.
For example
$$P(7)=-\frac{4 \pi ^2 \left(1-3 \sin \left(\frac{\pi }{14}\right)+3 \sin \left(\frac{3 \pi
   }{14}\right)\right)}{49 \left(3+6 \sin \left(\frac{\pi }{14}\right)-4 \sin
   \left(\frac{3 \pi }{14}\right)\right)}$$ which can be simplified nicely.
In fact, a CAS gives the beautiful
$$\color{blue}{P(s)=\int^\infty_0 \frac{\log(x)}{1+x^s}dx=-\pi ^2\frac{ \cot \left(\frac{\pi }{s}\right) \csc \left(\frac{\pi
   }{s}\right)}{s^2}}$$
A: An algorithm to get a solution.
Step 1: As it was mentioned in math.stackexchange.com/questions/3709298  by Calvin Khor for natural $n$, it's easy to make a substituion $y = x^{n+1}$ in the integral $\int_{0}^{\infty }\frac{x^n}{x^s + 1}$ and get the intergal of such type: $\int_{0}^{\infty }\frac{1}{y^s + 1}dy$, which is known (case $n=0$). But this idea works not only for natural $n$. Hence, we may find
$$I(a) = \int_{0}^{\infty }\frac{x^a}{x^s + 1}dx$$
for real $a$.
Step 2. We have
$$I'(a) = \int_{0}^{\infty }\frac{x^a \ln x}{x^s + 1}dx.$$
So it's sufficient to put $a=1$.
A: The $p$-test implies that this integral diverges for $s \leq 1$, so we assume that $s > 1$.
Hint This integral is a standard application of the Residue Theorem. In this case, we can take the contours $\Gamma_R$ to be the boundaries of the sectors, centered at the origin, of radius $R$ and central angle $\frac{2 \pi}{s}$. (A convenient choice is to take one boundary line segment along the positive real axis and the other along the ray through $e^{2 \pi i / s}$.) Then, the contour contains a single pole, at $e^{\pi i / s}$. Proceeding as usual by rewriting the contour integral as a sum of three integrals, taking the limit as $R \to \infty$ (which eliminates one of the integrals), rearranging, and taking real and imaginary parts gives values both of the given integral,
$$\int_0^\infty \frac{\log x \,dx}{1 + x^s} ,$$
and, as a welcome bonus, the related integral,
$$\int_0^\infty \frac{\,dx}{1 + x^s} .$$

Carrying out the above procedure gives that the relevant residue is $$\operatorname{Res}\left(\frac{\log z}{1 + z^s}, z = e^{\pi i / s}\right) = -\frac{\pi}{s^2} \exp \left(\frac{s + 2}{2 s} \pi i\right)$$ and then that the integral has value $$\int_0^\infty \frac{\log x \,dx}{1 + x^s} = -\frac{\pi^2}{s^2} \cot \frac{\pi}{s} \csc \frac{\pi}{s} .$$

The above technique is essentially robjohn's approach in his answer to this question, which treats the special case $s = 3$. Ron Gordon's approach there, that is, using instead a keyhole contour, applies at least in the special case that $s$ is an integer (necessarily $\geq 2$). Marko Riedel's approach there is similar in spirit to J.G.'s answer to this question.
Remark This integral takes on special values where $\frac{\pi}{s}$ does, including at various rational numbers with small numerator and denominator. In particular for $s = 2$ the integral vanishes, which can be shown using a slick but easier argument.
A: $$I=\int_{0}^{\infty} \frac{\ln x}{1+x^s} dx.$$
Let $x=e^t$, then
$$I=\int_{-\infty}^{\infty} \frac{t e^t}{1+e^{st}}=\int_{-\infty}^{0} \frac{t e^t}{1+e^{st}} dt+\int_{0}^{\infty} \frac{t e^t}{1+e^{st}} dt$$
In the first one let $t=-z$, then
$$I=-\sum_{k=0}^{\infty}\int_{0}^{\infty} ze^{-(1+ks)z} dz+\sum_{0}^{\infty}
\int_{0}^{\infty} te^{-(s+ks-1)t} dt$$
$$I=-\sum_{k=0}^{\infty} \frac{1}{(1+ks)^2}+ \sum_{k=0}^{\infty} \frac{1}{[s(1+k)-1]^2}=\frac{\psi^{(1)}(1-1/s)-\psi^{(1)}(1/s)}{s^2}$$
Using the property of poly-Gamma function:
https://en.wikipedia.org/wiki/Polygamma_function
Hoping to come back.
