Evaluate by Residue theory $$\int_0^\infty \frac{x^{a-1}\cos(\log_e x)}{e^x+1}dx , \quad 0<a<1$$
My try -
$$I= \int_0^\infty \frac{x^{a-1}\cos(\log_e x)}{e^x+1}dx ,\quad 0<a<1$$ $$I= \int_0^\infty \frac{x^a\cos(\log_e x)}{x(e^x+1)}dx$$ Take $0<\epsilon<R$. Consider the integration path $C$ formed by the interval $[\epsilon,R]$, the circle $C_R$ of radius $R$ counterclockwise, the interval $[R,\epsilon]$ and the circle $c_\epsilon$ of radius $\epsilon$ clockwise. The function $$ f(z)=\frac{z^a\cos(\log_e z)}{z(e^z+1)} $$ is holomorphic in the interior of $C$ (we consider the branch of $z^{a}$ defined on $\mathbb{C}\setminus[0,\infty)$) except for poles at $i\pi (2n+1),n\in \mathbb{Z}$ and branch points of $\log_e z$. Then $$ \int_C f(z)\,dz=\int_0^Rf(z)\,dz+\int_{C_R} f(z)\,dz+\int_R^0 f(z)\,dz+\int_{c_\epsilon} f(z)\,dz. $$
Sketch of the residue appproach
Your integral is $\Re w,\,w:=\int_0^\infty g(z)dz$ with $g(z):=\frac{z^{a-1+i}}{e^z+1}$. Replacing $f$ with $g$ in your reasoning, use$$\lim_{\epsilon\to0^+}\int_0^{2\pi}g(\epsilon e^{i\phi})i\epsilon e^{i\phi}d\phi=\lim_{\epsilon\to0^+}\int_0^{2\pi}\frac{\epsilon^{a+i}e^{(-1+ai)\phi}dx}{e^{\epsilon e^{i\phi}}+1}id\phi=0$$to neglect the $c_\epsilon$ contribution; handle $C_R$ similarly. That just leaves$$\int_0^\infty(g(x)-g(xe^{2\pi i}))dx=(1-e^{2\pi (1+(a-1)i)})I.$$Hence$$I=\frac{2\pi i}{1-e^{2\pi (1+(a-1)i)}}\sum_{p\in S}\operatorname{Res}_{z=p}g(z),$$with $S$ the set of poles of $g$, since they each have winding number $1$. I leave you try this using $S=\{\pi i(2n+1)|n\in\Bbb Z\}$.
Alternative approach
@metamorphy suggested using$$\begin{align}\int_0^\infty\frac{z^{s-1}dz}{e^z+1}&=\Gamma(s)\eta(s)\\\implies\int_0^\infty\frac{z^{a-1}\cos\ln zdz}{e^z+1}&=\frac{\Gamma(a-1+i)\eta(a-1+i)+\Gamma(a-1-i)\eta(a-1-i)}{2}.\end{align}$$