= Solution
For a <simple pole>, the <residue> is $\operatorname{Res}_{z_0}f=\lim_{z\to z_0}(z-z_0)f(z)$, the coefficient $a_{-1}$ of its <Laurent series>. On an anticlockwise circle $z=z_0+Re^{it}$, integration of each Laurent term is zero except $a_{-1}/(z-z_0)$, whose integral is $2\pi i a_{-1}$. This proves the stated single-pole <residue theorem>.
For $F(z)=e^{az}/(1+e^z)$, the only pole inside the rectangle of height $2\pi$ is $z=i\pi$, with <residue> $e^{ai\pi}/e^{i\pi}=-e^{ai\pi}$. The <residue theorem> therefore gives $I(R)=-2\pi i e^{ai\pi}$. The upper edge is traversed from right to left and satisfies $F(x+2\pi i)=e^{2\pi ia}F(x)$, so the two horizontal contributions sum to $(1-e^{2\pi ia})\int_{-R}^R e^{ax}/(1+e^x)dx$.
On the right edge the modulus is bounded by $e^{aR}/(e^R-1)$, and on the left by $e^{-aR}/(1-e^{-R})$. Multiplying by the fixed edge length $2\pi$ shows that both contributions tend to zero because $0<a<1$. Thus
$$
(1-e^{2\pi ia})J=-2\pi i e^{\pi ia}.
$$
Using $1-e^{2\pi ia}=-2ie^{\pi ia}\sin(\pi a)$ yields
$$
\boxed{J=\frac\pi{\sin(\pi a)}.}
$$
This <rectangular contour for an exponential beta integral> also gives the ordinary beta integral after the substitution $t=e^x$.
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