Adding imaginary number to exponential of Euler Gamma function This is gamma function:
$\Gamma (n) = \int_0^\infty x^{n-1}e^{-x}\,dx$
What will be Result if I add Imaginary Number to Exponential of Euler Gamma Function?
$$? = \int_0^\infty x^{n-1}e^{-ix}\,dx$$
where the $i^2=-1$
isn't it a new function!?
it will and will not converge?
 A: Looks related to Fourier Transform: 
$$
\hat{f}(\xi)=\mathcal{F}f(x) = \int_{-\infty}^{\infty} f(x)\ e^{- 2\pi i x \xi}\,dx
$$
A scaled version of your integral with limits $]-\infty,\infty[$
is given here:

308 | $\mathcal{F}x^n \rightarrow  \left(\frac{i}{2\pi}\right)^n \delta^{(n)} (\xi)\,$| Here, $n$ is a natural number and $\textstyle \delta^{(n)}(\xi)$ is the $n$-th distribution derivative of the Dirac delta function. This rule follows from rules 107 and 301. Combining this rule with 101, we can transform all polynomials.

A: No this is not a "new function", because the integral diverges for every $n$ (as $x\to0$ if $n\leqslant0$ and as $x\to+\infty$ if $n\geqslant0$).
A: It converges by contour integration (quarter cake slice in the lower right quadrant, no residues):
$\int_0^\infty dz\, z^{n-1} e^{-iz} + \int_{-i \infty}^0 dz\, z^{n-1} e^{-iz} +$ 
$+ \lim_{a \rightarrow 0^-}\lim_{R \rightarrow \infty}\int_a^{-\pi/2}d\phi \,i R e^{i \phi} R^{n-1} e^{i(n-1) \phi} e^{-i R \cos \phi} e^{R \sin \phi} = 0$
so
$\int_0^\infty dz\, z^{n-1} e^{-iz} =  \int_0^{-i \infty} dz\, z^{n-1} e^{-iz} = (-i)^n \int_0^\infty dy\, y^{n-1} e^{-y} = (-i)^n \Gamma(n)$ .
