Is this :$\int_{-\infty}^{+\infty} \frac{\exp(1-ix)}{x^2+1}=\pi$? I have tried to evaluate this integral:$$\int_{-\infty}^{+\infty} \frac{\exp(1-ix)}{x^2+1}$$  playing by denominator $x^n+1$ , I have accrossed for $n=2$  a value is close to $\pi$ as shwon here in wolfram alpha , Now my question is to know what is the exact value of that integral ? is it $\pi$ ?
Note: $i$ is the unit imaginary part 
 A: This integral can be found using the residue theorem. We can close the contour in the lower half-plane, since $\exp(-iz)$ tends to zero as the imaginary part of $z$ tends to $-\infty$, giving
$$ I = -2\pi i \, \left.\mathrm{Res}\frac{\exp(1-iz)}{z^2+1}\right|_{-i} = \pi \,.$$
A: For the lower half plane, pole $z = -i$, the longer method is:
$$0 = \int_{\text{low}} \frac{e^{1 - i z}}{z^2 + 1} \, dz = \left(\int_{R} + \int_{-\infty}^{0} + \int_{0}^{-i} + \int_{\gamma} + \int_{-i}^{0} + \int_{0}^{\infty} \right) \frac{e^{1 - i z}}{z^2 + 1} \, dz.$$
The integral for $R$ is the outer radius as $R \to \infty$ and is zero. The integral for $\gamma$ is the residue, the integrals $\int_{0}^{-i} + \int_{-i}^{0}$ sum to zero. From this 
\begin{align}
0 &= \int_{- \infty}^{\infty} \frac{e^{1 - ix}}{x^2 + 1} \, dx + 2\pi \, i \lim_{z \to -i} \left\{ \frac{(z + i)}{z^2 + 1} \, e^{1 - i z} \right\} \\
\int_{- \infty}^{\infty} \frac{e^{1 - ix}}{x^2 + 1} \, dx &= - 2\pi \, i \lim_{z \to -i} \left\{ \frac{1}{z - i} \, e^{1 - i z} \right\} 
\end{align}
or 
$$\int_{- \infty}^{\infty} \frac{e^{1 - ix}}{x^2 + 1} \, dx = \pi.$$
A: 
I thought it would be instructive to present a way forward that uses real analysis only and not complex analysis.  To that end, we now proceed.


First, we write
$$\int_{-\infty}^\infty \frac{e^{1-ix}}{1+x^2}\,dx=2e\int_0^\infty \frac{\cos(x)}{1+x^2}\,dx\tag1$$
In THIS ANSWER and in Methodology $2$ of THIS ONE, I showed directly using real analysis only that the Fourier Cosine Transform of $\frac1{1+x^2}$ is given by
$$\int_0^\infty \frac{\cos(\omega x)}{1+x^2}\,dx=\frac\pi2e^{-|\omega|}\tag2$$
Setting $\omega=1$ in $(2)$ and substituting the result into $(1)$ yields the result
$$\int_{-\infty}^\infty \frac{e^{1-ix}}{1+x^2}\,dx=\pi $$
