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Let $$ \int_{\Gamma} dz \frac{e^{iz}}{1 + z^2} $$ be a contour integral. Now we have two cases. First $\Gamma$ is the real line $\mathbb{R}$ (i.e. the real axis), and second, where $\Gamma$ is the line $\mathbb{R} - 10i$. In both cases it is oriented from left to right. I computed the first part easily.

Let $$f(z) = \frac{e^{iz}}{1 + z^2} = \frac{e^{iz}}{(z+i)(z-i)} $$ Then $f(z)$ has simple poles at $z = \pm i$. Form the semicircular region in the upper halve of the plane that encloses the simple pole $ z = i$. The residue is \begin{align*} Res(f(z), z= i) = \lim_{z \to i} (z - i) f(z) = \lim_{z \to i} \frac{e^{iz}}{z + i } = \frac{1}{2 e i}. \end{align*} Hence by the residue theorem, we have that \begin{align*} \int_{\mathbb{R}} \frac{e^{iz}}{1 + z^2} dz = 2 \pi i ( \frac{1}{2ei}) = \frac{\pi}{e}. \end{align*} But now, for the second part, I'm not really sure what to do. If $\Gamma = \mathbb{R} - 10i$, what contour should I form? I know the answer is given as \begin{align*} \frac{\pi}{e} - \pi e, \end{align*} but I don't know how to find this. Should I let $z = x - 10 i$ and plug this in for $e^{iz}$ in $f(z)$?

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    $\begingroup$ Consider a rectangle with vertices $R-10 i, R, -R, -R-10i$. Let then $R\to +\infty$. $\endgroup$ – Daniel Fischer Jan 11 '16 at 21:47
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Well, you can use the residue theorem. Draw a rectangle with vertices $-R-i 10$, $R - i 10$, $R$, $-R$. The contour integral about this rectangle is

$$\int_{-R}^R dx \frac{e^{i (x-i 10)}}{1+(x-i 10)^2} + i \int_{-10}^0 dy \frac{e^{i (R+i y)}}{1+(R + i y)^2} \\ + \int_{R}^{-R} dx \frac{e^{i x}}{1+x^2} + i \int_0^{-10} dy \frac{e^{i (-R+i y)}}{1+(-R + i y)^2}$$

As $R \to \infty$, the second and fourth integrals vanish because their magnitudes vanish as $1/R$. Thus, by the residue theorem, we have, in this limit,

$$\int_{-\infty}^{\infty} dx \frac{e^{i (x-i 10)}}{1+(x-i 10)^2} - \int_{-\infty}^{\infty} dx \frac{e^{i x}}{1+x^2} = i 2 \pi \frac{e^{i (-i)}}{-i 2}$$

(The RHS is the residue at the pole $z=-i$.)

Thus,

$$\int_{-\infty}^{\infty} dx \frac{e^{i (x-i 10)}}{1+(x-i 10)^2} = \frac{\pi}{e} - \pi e = -2 \pi \sinh{1}$$

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    $\begingroup$ Thank you. That was very clear and helped me a lot! $\endgroup$ – Kamil Jan 11 '16 at 22:08

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