 Solution

Let $F(z)=e^{az}/(1+e^z)$ and integrate counterclockwise around the rectangle with vertices $-R,R,R+2\pi i,-R+2\pi i$. It contains a single simple pole at $z=\pi i$, with residueOn the top edge $F(x+2\pi i)=e^{2\pi ia}F(x)$, and its orientation reverses that of the bottom edge. On the right edge $|F|\le e^{aR}/(e^R-1)$, so its integral tends to zero for $a<1$. On the left $|F|\le e^{-aR}/(1-e^{-R})$, which tends to zero for $a>0$. The residue theorem therefore givesSince $1-e^{2\pi ia}=-2i{e}^{\pi ia}\sin (\pi a)$,The restriction is also necessary for ordinary convergence: at $-\infty$ the integrand behaves as $e^{ax}$, requiring $a>0$, while at $+\infty$ it behaves as $e^{(a-1)x}$, requiring $a<1$. At either endpoint one tail approaches a nonzero constant; outside this range one tail grows. There is no cancellation because the real integrand is positive.