Conjecture on an asymptotic expansion $\int_0^\infty \frac{e^{-\mu x}}{x} (e^{\pm ix^n}-1) dx$ from Mathematica For $n\in \mathbb N$, I conjecture the asymptotic expansion
$$I_n(\mu) := \int_0^\infty \frac{e^{-\mu x}}{x} (e^{\pm ix^n}-1) dx = \log\mu + \left( \frac{n-1}{n} \gamma \pm \frac{i\pi}{2n} \right) + O(\mu),$$
as $\mu \to 0^+$. Here, $\gamma$ is the Euler gamma constant. Is this true?
For $n=2,3,4$, the conjecture is verified by Mathematica. Explicit closed form for $I_n(\mu)$ can be obtained (in terms of complicated special functions), and taking the series expansion gives the above result.
Also, for $n=2$, I previously asked a question Asymptotic evaluation of $\int_0^\infty \frac{e^{-\lambda x}}{x} (e^{i\alpha \lambda x^2}-1) dx$ , where the answer involves explicit computation of $I_2(\mu)$. Hence, this question is a generalization to the previous question.
I appreciate if one only finds the leading behavior $I_n(\mu) = \log \mu + O(1)$.
 A: I looked at
$$I_n(\mu) = \int_0^\infty \frac{e^{-\mu x}}{x} (e^{ ix^n}-1)\, dx$$
Using $x^n=t$ leads to
$$I_n= \frac 1n \int_0^\infty \frac{ e^{-\mu \,   t^{\frac{1}{n}}}}{t}\left(e^{i t}-1\right)\,dt$$ Mathematica provides solutions in terms of generalized hypergeometric functions.
Defining
$$J_n=\frac 1 \mu \Big[I_n -\log(\mu)-\frac{  n-1}{n}\gamma-\frac{i \pi }{2 n}  \Big]$$ (this is obviously the correct conjecture) and
$$\color{blue}{\large J_n=-\frac{ \Gamma
   \left(\frac{1}{n}\right)}{n}\exp\left(i \frac{\pi }{2 n}\right)}$$
A: You have that
$$
\frac{dI_n}{d\mu} = \int_0^\infty e^{-\mu x}(1-e^{\pm i x^n}) dx = \frac{1}{\mu} - \int_0^\infty e^{-\mu x + ix^n} dx.
$$
The latter term is (after a substitution) a stationary phase type integral one could show is $O(\mu)$, so you have the leading order behaviour.  The fact that this latter integral is $O(\mu)$ is also essentially what you'd need to show that @Gary's remainder term is constant + $O(\mu)$.  I guess what remains is to show that constant is what you think it is.
