Validity of Jordan's lemma I'm trying to calculate the following integral
$$ \int_0^\infty dx \, e^{ix}$$
with contour integration. By choosing a quarter circle contour $\gamma$, by the residue theorem we get
\begin{align} 
0 =\oint_\gamma dz \, e^{iz} =& \int_0^\infty dx\, e^{ix} + \int_{\mathrm{Arc}} dz \, e^{iz} -i\int_0^\infty dy \, e^{-y} \\
=& \int_0^\infty dx\, e^{ix} + \int_{\mathrm{Arc}} dz \, e^{iz} -i
\end{align}
therefore, if the integral over the arc is zero, the original integral becomes $i$.
Can Jordan's lemma, of a variation thereof, be applied in this case?
Edit: the parametrization would be $z=R e^{i\theta}$ with $\theta \in [0, \frac{\pi}{2}]$.
Aside: it seems to me that the original integral can be interpreted as a distribution as the Fourier transform of the theta function at $k=1$, which would give $i$ as well.
 A: That integral is not well-defined because for any horizontal line $z=x+iy,e^{iz}$ follows the circle of radius $e^{-y}$ in the complex plane. Therefore, symmetry shows us that the integral $\int_0^{2n\pi}e^{ix}dx=0$ but $\int_0^{(2n+1)\pi} e^{ix}dx=\left. \frac{1}{i}e^{ix}\right|_0^\pi=-i(-1-1)=2i$. Jordan's lemma will not allow you to solve the improper integral, because no solution exists. The reason for this is that the proof of Jordan's lemma relies on the symmetry $\sin\theta=\sin(\pi-\theta)$, so only the half-circular arc in the upper half-plane can apply it. However, if you want to apply a similar result, you could instead consider the maximum modulus of the function over that interval (in this case, that approach fails because $|e^{ix}|=1$ for all real $x$).
A: The answer to my question is that, as pointed out by Nico Terry, the integral does not converge. However, it can be assigned a value by analytic continuation as follows
$$ f(z)=\int_0^\infty dt \, e^{zt} = -\frac{1}{z}, \qquad \mathrm{for\, Re}(z) < 0 $$
so we can analytically continue $f(z)$ over all $\mathbb C -\{0\}$. Hence $f(i)=i$ means that we can assign the value $i$ to the original integral.
