What's the best way to symbolically evaluate this integral?
$$\frac{1}{\hbar}\int_{-\infty}^\infty e^{iux/\hbar}\Psi^{*}_n(p-u/2)\Psi_n(p+u/2)\,du$$
where:
$$\Psi_n(p)=\frac{1}{(1+\alpha p^2)^{\sqrt{q+r_n}}}{}_2F_1\left(a_n,-n;c_n;\frac{1}{2}+i\frac{\sqrt{\alpha}}{2}p\right)$$
$$\sqrt{q+r_n}=\frac{1}{2}\left(n+\frac{1}{2}\right)+\frac{1}{4}\sqrt{1+16r}$$
$$a_n=-n-\sqrt{1+16r}$$
$$c_n=1-2\sqrt{q+r_n}$$
$$r=\frac{\eta^2}{4\alpha^2}$$
$$\eta=\frac{1}{m\hbar\omega}$$
${}_2F_1(a,b;c;z)$ is the Gauss Hypergeometric Function
$$n\in\mathbb{Z}$$
$$\alpha,p,x,m,\omega,\hbar\in\mathbb{R}$$
My first attempt was with $\eta=\hbar=\alpha=1$ and $n=0$, but that yielded:
$$\int_{-\infty}^\infty \frac{e^{iux}}{\left(1+0.15\left(p-\frac{u}{2}\right)^2\right)^{3.5927}\left(1+0.15\left(p+\frac{u}{2}\right)^2\right)^{3.5927}}\,du$$
I'm not sure how to evaluate this comparatively simple integral, though. In general, I'm looking for the solution to the top integral to be a function of $n$, $\alpha$, $x$, and $p$. Any ideas?
Thanks.
I think I have found something for you.
See http://authors.library.caltech.edu/43489/ for the Integral Tables of the Bateman Project.
In Volume 1, p.119 (PDF p. 135) Formula (12), which I hope should be applicable for your $n=0$ case.
$\int_{-\infty}^{+\infty} (\alpha+ix)^{-2\mu} (\beta-ix)^{-2\nu} e^{-ixy} dx = \begin{cases} I_{-}, & y < 0\\ I_{+}, & y > 0\end{cases}$.
with
$I_{-} = +2\pi(\alpha+\beta)^{-\mu-\nu}[\Gamma(2\mu)]^{-1} e^{y(\alpha-\beta)/2} (-y)^{\mu+\nu-1} W_{\mu-\nu,1/2-\mu-\nu}(-(\alpha+\beta)y)$ $I_{+} = -2\pi(\alpha+\beta)^{-\nu-\mu}[\Gamma(2\nu)]^{-1} e^{y(\beta-\alpha)/2} (+y)^{\nu+\mu-1} W_{\nu-\mu,1/2-\nu-\mu}(+(\alpha+\beta)y)$
and $\mathcal{Re}(\mu+\nu) > 1$, $\mathcal{Re}(\alpha) > 0$, $\mathcal{Re}(\beta) > 0$. I think the $W$-functions are Whittaker-functions.