Deformation of Gamma function integral contour

Terence Tao has described the gamma function as the inner product of a multiplicative and an additive character with respect to the Haar measure on $\Bbb R^+$. The gamma function is defined as follows:

$\Gamma(\omega) = \int_0^\infty \frac{x^{\omega}}{e^x} \frac{\text{d}x}{x}$

The functions $x^\omega$ and $e^x$ map from $\Bbb R^+$ to $\Bbb R^+$, hence they can be thought of as non-unitary characters.

More interesting to me is to replace them with the unitary characters $x^{i \omega}$ and $e^{ix}$, which map from $\Bbb R^+$ to $\Bbb C^\times$. That would yield the integral

$\int_0^\infty \frac{x^{i \omega}}{e^{ix}} \frac{\text{d}x}{x}$

I don't think the integral converges, but it seems likely to me that there is some method that can make something of it.

I am wondering if this integral is related to the ordinary Gamma function interval, or if it can be expressed in terms of the Gamma function, if contour deformation can be used to relate the two in some sense, etc.

• this is what you can do with contour deformation math.stackexchange.com/questions/645954/… – reuns Apr 27 '16 at 21:51
• Perfect, thank you! I had missed that while searching. – Mike Battaglia Apr 27 '16 at 21:57
• but no $\int_a^\infty x^{i \omega} e^{ix} dx$ doesn't converge, since integration by part gives $\displaystyle\int_a^b x^{i \omega} e^{ix} dx = x^{i \omega}\frac{e^ix}{i}|_a^b - \int_a^b i \omega x^{i \omega -1} \frac{e^{ix}}{i} dx$ where the second term converges when $b \to \infty$, but not the first one. – reuns Apr 27 '16 at 21:58
• hence you need to define some sort of regularization, for example $\displaystyle\lim_{s \to i \omega, Re(s) < 0} \int_0^\infty x^s e^{i x } dx$ and you can relate the result to $\Gamma(1+i \omega)$ with what is in the link above – reuns Apr 27 '16 at 22:00