Integral over infinity of $F(x) = e^{ix}$ how to calculate integral of following function over infinity ?
$F(x) = e^{ix}$  ($i$ imaginary)
$$
\int\limits_{-\infty}^\infty e^{ix} \, dx
$$
 A: The integral does not exist in the usual sense obviously, so this depends on what meaning you put in it. Say, you want to understand $h(t) = \int_{\mathbb{R}} e^{i xt} dx$ as a (Schwartz) distribution. Then for any test function $f$,
$$
\left\langle h,f\right\rangle = \int_{\mathbb{R}}\int_{\mathbb{R}} e^{i xt} dx\, f(t)dt =\int_{\mathbb{R}}\int_{\mathbb{R}} e^{i xt} f(t)dt\, dx  \\
=\int_{\mathbb{R}} \hat f(x) dx = {\tau} f(0),
$$
by the Fourier inversion formula. This means that $h(t) = \tau\, \delta(t)$, in the sense of distributions. Therefore, "substituting $t=1$" (which is quite meaningless though) "gives" $\int_{\mathbb{R}} e^{ix}dx = 0$.
A: This is up to some factor the Fourier transform of the unit function $f(x) = 1$, thus the Dirac delta distribution $\delta(\xi)$.
And this evaluated for non-zero argument, thus resulting to $0$.
WA seems to be a bit more rigorous: 
It states $0$ only as Cauchy principal value (WA answer)
A: $$\int_0^{\infty}e^{ix}dx=\left[\frac{e^{ix}}{i}\right|_0^{\infty}=\left[-ie^{ix}\right|_0^{\infty}=\left[ie^{ix}\right|^0_{\infty}=i-e^{i\infty}$$
this result not converge, are you sure the sign in the exponential is correct?
