Closed from of $\int_0^{\infty} \frac{e^{iax}}{x^{n}+1}dx$?

Question

I've been trying to find the general form of a certain group of integrals of the form$$I(a,n)=\int_0^{\infty} \frac{e^{iax}}{x^{n}+1}dx$$

I know that the real part of $I(a,2)$ can be calculated using Fourier Transform or residues, and $I(a,1)$ reduces to a form of the exponential integral.

I thought about approaching the integral via Fourier Transform but I did not know how to apply it to this integral. It might be able to be calculated with residues but I am not that great at complex analysis. I'm very interested in a closed form for this integral so any help would be appreciated.

Answer 1

Since the integrand is a product of two Meijer G-functions and the integration range is $[0, \infty)$, there is a closed form, but it involves the Fox H-function: $$\int_0^\infty \frac {e^{i a x}} {x^n + 1} dx = \int_0^\infty G_{0, 1}^{1, 0} {\left(- i a x \middle| { - \atop 0} \right)} G_{1, 1}^{1, 1} {\left(x^n \middle| { 0 \atop 0} \right)} dx = \frac i a H_{2, 1}^{1, 2} {\left( \left( \frac i a \right)^{\!n} \middle| {(0, 1), (0, n) \atop (0, 1)} \right)}.$$ This becomes a G-function if $n$ is rational, but a rational $n$ produces an infinite number of double poles. This gives an infinite sum of polygamma terms instead of gamma terms when the H-function is evaluated by applying the residue theorem. Such a sum may have a closed form in terms of simpler functions in some special cases, which happens for $n = 1$ and $n = 2$.