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 $\int^{\infty}_0 \frac{e^{-x}}{x^s+1}\,dx$.

$$I=\int_{0}^{\infty} \frac{\ln x}{1+x^s} dx.$$
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.
