= Solution
Use the <rectangular contour for an exponential beta integral>. For $F(z)=e^{az}/(1+e^z)$, integrate counterclockwise around the rectangle with vertices $-R,R,R+2\pi i,-R+2\pi i$. Its only pole is at $z=\pi i$, with
$$
\operatorname{Res}_{z=\pi i}F(z)=-e^{\pi ia}.
$$
The right vertical side is bounded in modulus by $2\pi e^{aR}/(e^R-1)$, and the left by $2\pi e^{-aR}/(1-e^{-R})$. Both tend to zero because $0<a<1$. On the top edge $F(x+2\pi i)=e^{2\pi ia}F(x)$, and that edge is traversed right to left. The <residue theorem> therefore gives
$$
(1-e^{2\pi ia})\int_{-\infty}^{\infty}\frac{e^{ax}}{1+e^x}\,dx
=-2\pi i e^{\pi ia}.
$$
Since $1-e^{2\pi ia}=-2i e^{\pi ia}\sin(\pi a)$,
$$
\boxed{\int_{-\infty}^{\infty}\frac{e^{ax}}{1+e^x}\,dx=\frac{\pi}{\sin(\pi a)}.}
$$
The same inequalities establish convergence at both real infinities. With $t=e^x$, this integral becomes $\int_0^\infty t^{a-1}/(1+t)\,dt$. Taking $a=1/6$ gives
$$
\boxed{\int_0^\infty\frac{dt}{t^{5/6}(1+t)}=2\pi.}
$$
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