# 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?

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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
yeah, sure. The complex absolute value is nowhere complex differentiable. This makes it pretty much impossible. This was the only hope I had coming a some closed form computation somewhat close. I cannot just say the integral exists. I have to provide a way to get a real number, hence "really". – user13655 Sep 5 '11 at 17:50