Evaluate the integral p.v. $\int_{-\infty}^\infty\frac{\mathrm{e}^{-2\mathrm{i}x}}{x^2+1}\,\mathrm{d}x$ using residues

Question

I have to evaluate the principal value of the integral $\displaystyle\int_{-\infty}^\infty\frac{\mathrm{e}^{-2\mathrm{i}x}}{x^2+1}\,\mathrm{d}x$ and I have to use residues.

First, I thought that because of the exponent, there is an infinite number of poles, so I cannot use a semicircle as a parametrization from $-R$ to $R$, where $R\to\infty$ in the limit. This would imply that the residues would become an infinite series.

Next, I considered a rectangular contour shifted by $\frac12\pi$. I thought this meant that the $\gamma_3$ parameter was just the $\gamma_1$ parameter multiplied by a factor $\mathrm{e}^\pi$. The $\gamma_2$ and $\gamma_4$ go to zero when $R\to\infty$. However, $\gamma_1$ and $\gamma_3$ do not. So then, I have to calculate the residue of the function, which is the residue of $\frac14\pi$ and of $\mathrm{i}$. The residue of $\frac14\pi$ gives me trouble. If I use that $\textrm{Res}=\frac{P(x)}{Q'(x)}$ defined at $x=\frac14\pi$, I do not get a value.

Answer 1

The poles should only be at x = i, x = - i. Recalling that $e^0 = 1$

Form the usual semi-circle and then isolate those points in the usual way. Then use the residue theorem.

Answer 2

The poles are at $x=i$ and $x=-i$. By using a contour with a semi-circle:

$$\Gamma=\mathbb{R}\cup \{Re^{i\phi}\}_{\phi=\pi}^{2\pi}$$

and letting $R\to\infty$, we have

$$\oint_\Gamma f(z)\;\mathrm{d}z=2\pi i \mathrm{Res}\left(f, z=-i\right)=-\int_{-\infty}^\infty f(x)\;\mathrm{d}x+\lim_{R\to\infty}\int_0^\pi f(Re^{j\phi})iRe^{i\phi}\;\mathrm{d}\phi$$

where $f(z)=\frac{e^{-2iz}}{z^2+1}$. The integral along the semi-circle satisfies

$$\left|\int_{\pi}^{2\pi} \frac{e^{-2iRe^{i\phi}}}{R^2 e^{2i\phi}+1}iRe^{i\phi}\;\mathrm{d}\phi\right|\le \int_{\pi}^{2\pi} \frac{Re^{2R\sin\phi}}{\sqrt{1+R^4+2R^2\cos(2\phi)}}\;\mathrm{d}\phi\le \int_{\pi}^{2\pi}\frac{R}{|1-R^2|}\;\mathrm{d}\phi=\frac{\pi R}{|1-R^2|}\to0$$

i.e., it goes to $0$ as $R\to\infty$. Therefore we have

$$\int_{-\infty}^\infty f(x)\;\mathrm{d}x=-2\pi i\mathrm{Res}\left(f, z=-i\right)=-2\pi i\frac{e^{-2}}{-2i}=\pi e^{-2}.$$