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$$ f(z) = \frac{e^{-iz}}{z^2+9}, \:\:\: z \in \mathbb{C} $$

I am trying to compute

$$ I = \int_\limits{-\infty}^{\infty} f(z)\:dz $$ The zeros of the denominator give that $f$ is analytic in $\mathbb{C}\backslash\left\{-3i,3i\right\}$. The residues of these simple poles are: $$ \text{Res}\left(f,-3i\right) = \frac{i}{6 e^3} \:\wedge\: \text{Res}\left(f,3i\right) = -\frac{ie^3}{6} $$

Let $\gamma_1$ be the contour going from $-R$ to $R$ on the real axis, and let $\gamma_2$ be the positively oriented semicircle of radius R, centred at the origin. Let $\Gamma = \gamma_1+\gamma_2$.

$3i$ is interior to $\Gamma$

By definition, and by the residue theorem applied to $\Gamma$, and Jordan's lemma applied to $\gamma_2$:

$$ \lim_{R \to \infty} \int_\Gamma f(z)\:dz = \lim_{R \to \infty} \left(\int_{\gamma_1}+\int_{\gamma_2}\right) f(z)\:dz = 2\pi i\left(-\frac{ie^3}{6}\right) = \frac{\pi e^3}{3} = I+0 $$

This is, however, incorrect. Wolfram Alpha gives the solution as $I = \frac{\pi}{3e^3}$, which agrees with the alternative choice of performing the computation with the negatively oriented semicircle in the lower half plane, and the residue of other pole. What is my mistake?

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    $\begingroup$ $e^{-iz}$ blows up (exponentially) in the upper half-plane, so the premises of Jordan's lemma aren't given. You must take the semicircle in the lower half-plane (and thus get a negatively oriented contour). $\endgroup$ – Daniel Fischer Jan 31 '17 at 21:54

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