i!

Define the Factorial of a Complex number.

In usual sense, factorial is defined as,
$$n! = \prod_{k=1}^{n} k=1 \cdot2\cdot3\ldots(n-2)\cdot(n-1)\cdot n \tag*{(1)}$$
Now, the not so usual definition is based on the famous Gamma Function,
$$\Gamma(t) = \int_0^\infty e^{-x}x^{t-1}\mathrm{d}x \tag*{(2)}$$
There is an unique and very useful property,
$$\Gamma(n+1)=n! \tag*{(3)}$$
To extend $(3)$ into the complex domain, we will first have to go through Analytic Continuation, please read about Analytic Continuation here.
Therefore, after analytic continuation, we can write it as,
$$z! \overset{\text{def}}{=} \Gamma(z+1) \tag*{(4)}$$
For, $z \in \mathcal C-{-1,-2...}$
So, now, clearly,
$$i!= \Gamma(i+1) \tag*{(5)}$$
By $(3)$
$$\Gamma(i+1)=\int_0^\infty e^{-x} x^{(i+1)-1}\mathrm{d}x \tag*{(6)}$$
Clearly,
$$\Gamma(i+1)=\int_0^\infty e^{-x} x^{i}\mathrm{d}x \tag*{(6)}$$
For easier computation, please catch that,
$$x^i = e^{i \ln x} = \cos ( \ln x)+i \sin( \ln x)$$
Let’s break it down,
$$i! = \int_0^\infty e^{-x} (\cos ( \ln x)+i \sin( \ln x))\mathrm{d}x$$
$$\implies i! = \int_0^\infty e^{-x} \cos ( \ln x)\mathrm{d}x+i\int_0^\infty e^{-x} \sin ( \ln x)\mathrm{d}x$$
If you have reached this far, you obviously know how to solve the above integral.
Cheers!