What's the best way to symbolically evaluate this integral?

$$\frac{1}{\hbar}\int_{-\infty}^\infty e^{iux/\hbar}\Psi^{*}_n(p-u/2)\Psi_n(p+u/2)\,du$$


$$\Psi_n(p)=\frac{1}{(1+\alpha p^2)^{\sqrt{q+r_n}}}{}_2F_1\left(a_n,-n;c_n;\frac{1}{2}+i\frac{\sqrt{\alpha}}{2}p\right)$$






${}_2F_1(a,b;c;z)$ is the Gauss Hypergeometric Function



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?


  • $\begingroup$ Finding a good table of integrals usually helps. Though that seems over the top for a problem like this (which I would guess is a quantum problem?) Also note that when one of the parameters is an integer the function possibly simplifies. (See WolframAlpha for the $n=2$ case). $\endgroup$ Jul 29, 2014 at 19:33
  • $\begingroup$ To expand on that last point a little: if either upper parameter is an integer, then the hypergeometric series terminates after that many terms. $\endgroup$ Jul 29, 2014 at 20:08
  • $\begingroup$ Ah, good point. This will greatly simplify my integral. $\endgroup$
    – user85503
    Jul 29, 2014 at 21:31
  • $\begingroup$ Actually, take a look at the Jacobi polynomials page at Wikipedia. If I'm looking at that right, your hypergeometrics are cases of the definition they give there. (Note that the order of the first two parameters is irrelevant.) So if you translate your problem into Jacobi polys, then you may be able to find a more specific result in an integral table. $\endgroup$ Jul 29, 2014 at 21:50

1 Answer 1


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.

$\int_{-\infty}^{+\infty} (\alpha+ix)^{-2\mu} (\beta-ix)^{-2\nu} e^{-ixy} dx = \begin{cases} I_{-}, & y < 0\\ I_{+}, & y > 0\end{cases}$.


$I_{-} = +2\pi(\alpha+\beta)^{-\mu-\nu}[\Gamma(2\mu)]^{-1} e^{y(\alpha-\beta)/2} (-y)^{\mu+\nu-1} W_{\mu-\nu,1/2-\mu-\nu}(-(\alpha+\beta)y)$ $I_{+} = -2\pi(\alpha+\beta)^{-\nu-\mu}[\Gamma(2\nu)]^{-1} e^{y(\beta-\alpha)/2} (+y)^{\nu+\mu-1} W_{\nu-\mu,1/2-\nu-\mu}(+(\alpha+\beta)y)$

and $\mathcal{Re}(\mu+\nu) > 1$, $\mathcal{Re}(\alpha) > 0$, $\mathcal{Re}(\beta) > 0$. I think the $W$-functions are Whittaker-functions.


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