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?
On
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.
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 $n\geqslant0$).