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**Haven** This question is similar to another one I've posted

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$\displaystyle \int_{-\infty}^{\infty} \frac{\sin{x}}{x^2+4x+5} = \frac{-\pi\sin{2}}{e}$

so I let $\displaystyle f(z) = \frac{\sin{z}}{z^2+4z+5}$

And I used a closed semicircle which contains $\displaystyle z = -2+i$ of radius R.

Let $\displaystyle C_{R} = \{z=Re^{i\theta}: 0\leq\theta\leq\pi\, R<\infty\} $

$\displaystyle C_{+} = \{z=r: -R<r<R, R<\infty\}$

I let: $\displaystyle I = \int_{C_{+}} f(z)dz$ and $\displaystyle I_{R} = \int_{C_R} f(z)dz$

By Jordan's Lemma, $\displaystyle I_{R} = 0$ as $\displaystyle R\to\infty$

By the Cauchy Residue Theorem

$\displaystyle

I + I_{R} = I = 2\pi i {Res}_{z=-2+i} f(z)$

Now I use the following theorem to evaluate the theorem

If $\displaystyle p(z)$ is entire and $\displaystyle q(z)$ is analytic everywhere except $\displaystyle z_{0}$ but $\displaystyle q'(z)$ is entire.

The $\displaystyle {Res}_{z=z_0} \frac{p(z)}{q(z)} = \frac{p(z_0)}{q'(z_0)}$

However, I can't get the following to evaluate correctly

$\displaystyle I = 2\pi i \frac{\sin{(-2+i)}}{2(-2+i) + 4(-2+i)}$

I am having troubles with $\displaystyle \sin{(-2+i)}$. None of the identities in my text book seem to help.