= 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 <residue>
$$
\operatorname{Res}_{z=\pi i}F=\frac{e^{a\pi i}}{e^{\pi i}}=-e^{a\pi i}.
$$
On 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 gives
$$
(1-e^{2\pi ia})\int_{-\infty}^{\infty}\frac{e^{ax}}{1+e^x}\,dx=-2\pi i e^{a\pi i}.
$$
Since $1-e^{2\pi ia}=-2ie^{\pi ia}\sin(\pi a)$,
$$
\boxed{\int_{-\infty}^{\infty}\frac{e^{ax}}{1+e^x}\,dx=\frac\pi{\sin(\pi a)}.}
$$
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.
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