Consider the integral
\frac{\int_{b-i \infty }^{b+i
\infty } \frac{\exp (a
z)}{\sin (\pi z)} \, dz}{2
\pi i}
where 0 < b <1 and |Im a| < 1. Below is Mathematica code which evaluates this integral 1) directtly and 2) evaluating the indefinite integral between the two limits.
The Mathematica command:
value1=Assuming[0 < b < 1 && Abs[Im[a]] < Pi,1/(2 Pi I) Integrate[Exp[a z]/Sin[Pi z],{z,b- I Infinity,b+I infinity}]]
produces the result
-\frac{1}{\pi
\left(e^a+1\right)}
If we calculate the indefinite integral
indef=1/(2 Pi I) Integrate[Exp[a z]/Sin[Pi z],z]
we obtain
\frac{i e^{(a+i \pi ) z} \,
_2F_1\left(1,\frac{\pi -i
a}{2 \pi
};\frac{3}{2}-\frac{i a}{2
\pi };e^{2 i \pi
z}\right)}{\pi (\pi -i a)}
Next, we set
temp = FullSimplify[(indef /. {z -> b + I R}) - (indef /. {z ->
b - I R})]
obtaining
-\frac{e^{(a+i \pi ) (b+i R)}
\, _2F_1\left(1,\frac{\pi
-i a}{2 \pi
};\frac{3}{2}-\frac{i a}{2
\pi };e^{2 i \pi (b+i
R)}\right)-e^{(a+i \pi )
(b-i R)} \,
_2F_1\left(1,\frac{\pi -i
a}{2 \pi
};\frac{3}{2}-\frac{i a}{2
\pi };e^{2 \pi (i
b+R)}\right)}{\pi (a+i \pi
)}
value2 = Assuming[0 < b < 1 && Abs[Im[a]] < Pi && R > 0,
Limit[temp, R -> Infinity]]]
gives
\frac{e^{a/2} \Gamma
\left(\frac{3}{2}-\frac{i
a}{2 \pi }\right) \Gamma
\left(\frac{i a+\pi }{2 \pi
}\right)}{\pi (\pi -i a)}
value2 = Factor[FullSimplify[value2]]
gives
\frac{e^a}{\pi
\left(e^a+1\right)}
{value1, value2}
\left\{-\frac{1}{\pi
\left(e^a+1\right)},\frac{e
^a}{\pi
\left(e^a+1\right)}\right\}
It can be shown by contour integration that value2 is the correct value.