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

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

• What happens to the integrand in the lower half-plane? Specifically, this: $\left | \int_{\gamma}^{ }\frac{e^{-i z}}{z^2+1}dz \right |\overset{R\rightarrow \infty}{\rightarrow}?$ – user36196 May 30 '18 at 10:08

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.
• Why does it goes to $\infty$? $\left | \int_{\gamma}^{ }\frac{e^{-i z}}{z^2+1}dz \right |\leq \pi R\sup_{z\in \gamma} \left | \frac{e^{-i z}}{z^2+1} \right |\leq \pi R \frac{1}{R^2-1} \overset{R\rightarrow \infty}{\rightarrow}0$ is not true? – bp7070 May 30 '18 at 10:24
• @bp7070 You seem to assume that $|e^{-iz}|\leqslant1$. This is not true. Note that$$|e^{-iz}|=e^{\operatorname{Re}(-iz)}=e^{\operatorname{Im}(z)}.$$ – José Carlos Santos May 30 '18 at 10:27
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 $b=\Im(x) <0$).