Calculate $\int_{-\infty}^{\infty}\frac{e^{-i x}}{x^2+1}dx$ using residue theorem

Question

I need to calculate $\int_{-\infty}^{\infty}\frac{e^{-i x}}{x^2+1}dx$ using the Residue theorem.

I know that I should choose an upper semicircle so that for

$\gamma=\left \{ z\in\mathbb{C}: z=Re^{it}, t\in[0,\pi] \right \}$

$\left | \int_{\gamma}^{ }\frac{e^{-i z}}{z^2+1}dz \right |\overset{R\rightarrow \infty}{\rightarrow}0$

Then I'm left with the original integral which now equals to

$2\pi i Res(\frac{e^{-i z}}{z^2+1},i)=2\pi i (-\frac{i e}{2})=\pi e$

But if I choose a lower semicircle the answer is

$2\pi i (-Res(\frac{e^{-i z}}{z^2+1},-i))=2\pi i (-\frac{i}{2e})=\frac{\pi}{e}$

Why do I get 2 different answers for the same integral? Where is my mistake?

Thanks

Answer 1

It's not as if you had an option here. The integral along the top half of the semicircle centered at $0$ with radius $R$ goes to $\infty$ as $R\to\infty$. This does not occur with the bottom half; in that case, it goes to $0$.

If the numerator was $e^{ix}$, then you would have to choose the top half and not the bottom half.

Answer 2

If you let $x=a+bi$, then $\exp(-ix)=\exp(-i(a+bi))=\exp(-ai)\exp(b)$, so it will only go to $0$ if you choose the bottom half (Where $b=\Im(x) <0$).