 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 $z=\pi i$, withThe 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 givesSince $1-e^{2\pi ia}=-2i{e}^{\pi ia}\sin (\pi a)$,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