# Computing complex principal value integral - sgn-function?

I currently face a less appealing integral which emerged computing the expectation of some random variable. It reads as (omitting all unnecessary constants except $\alpha\in(0,1)$)

$$PV \int_{-\infty}^\infty \frac{\exp[1-\exp[iu-|u|^\alpha(1-i\text{sgn}(u)]]}{u}du.$$

It converges as a principal value integral. However, I really have to compute it. Numerically this is not a trivial task, because it does not satisfy the Hoelder condition around zero.

First thought: Contour integration. Could I replace $\text{sgn}(u)$ with $\frac{u}{|u|}\mathbb{1}_{u\neq 0}$? Due to the $\text{sgn}$-function we don't have a simple pole: $z\frac{\exp[1-\exp[iz-|z|^\alpha(1-i\frac{z}{|z|}\mathbb{1}_{z\neq 0}]]}{z}$ is not holomorphic in zero. Computing the residue will be a pain, I guess. What is the best way to attack this problem?

Idea: Compute $a_{-1} = \int_{\gamma}f(z)dz$, where $\gamma$ is a suitable contour and use series expansion of the exponential function. Or is it possible to obtain the Laurent series?

If contour integration is to be of any help, you'll have to get rid of the $|u|$ somehow -- since $|z|$ is not differentiable, the integrand does not satisfy Cauchy's theorem at all. The signum and the absolute value will annihilate each other in the $|u|^\alpha i\mathrm{sgn}u$ term, but the pure $-|u|^\alpha$ term will be more trouble. Perhaps restrict your interest to rational $\alpha$s of the form $2p/q$ in lowest terms, and hope you can recover the rest by continuity at the end of the day? (Getting the branches to line up correctly might still be a problem, though). – Henning Makholm Sep 5 '11 at 17:31
Reside theorem and most of complex analysis simply doesn't apply where a function isn't holomorphic (which this isn't at $0$ obviously). You can rearrange to get the integral as e\int_0^\infty\frac{1}{u}\left(e^{-e^{iu}}-e^{-e^{-iu-2u^{\alpha}}}\right)du.$‌​$ I'm pretty doubtful you'll get a closed-form for this. What makes you say you "really have to compute it"? – anon Sep 5 '11 at 17:38