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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

For a simple pole, the residue is ${\mathrm{Res}}_{z_0}f={\lim }_{z\to z_0}(z-z_0)f(z)$, the coefficient $a_{-1}$ of its Laurent series. On an anticlockwise circle $z=z_0+R{e}^{it}$, 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 $0<a<1$. ThusUsing $1-e^{2\pi ia}=-2i{e}^{\pi ia}\sin (\pi a)$ yieldsThis rectangular contour for an exponential beta integral also gives the ordinary beta integral after the substitution $t=e^x$.