Integrals 3.384 from Gradshteyn and Ryzhik

I'm interested in understanding the computation of

$$\int_{-\infty}^\infty\frac{e^{-ip x}}{(1 + ix)^{2u}(1-ix)^{2v}}\mathrm{d}x,$$

which is evaluated in 3.384.9 of Gradshteyn and Rhysik for sufficiently large $u, v$ as

$$(2\pi) 2^{-u-v} \frac{p^{u+v-1}}{\Gamma(2v)}W_{v-u, \frac 12 - v - u}(2p).$$

(I'm pretty sure I didn't mess up my transcription of that). The reference given in Gradshteyn and Rhysik is Erdelyi's Table of Integrals, volume I, pg 119 equation 12 (actually, my copy of Gradshteyn and Rhysik has a typo and says pg 19 equation 12 - so it goes). But in Erdelyi's Table, there are no proofs nor references.

By $W$, I mean Whittaker's $W$ function. Since I don't quite know the most convenient characterization to give as I don't know how to do this integral (yet), I might be misleading you. But I think the convenient characterization will be

$$W_{\lambda, \mu}(z) = \frac{z^{\mu + \frac 12} e^{-z/2}}{\Gamma(\mu - \lambda + \frac 12)} \int_0^\infty t^{\mu - \lambda - \frac 12}e^{-t}\left(1 + \frac{t}{z}\right)^{\mu + \lambda - \frac 12} \mathrm{d}t,$$

for real part of $\mu - \lambda > -\frac 12$ and $|\arg z| < \pi$.

Do you know how to compute this integral, or alternatively have a reference for it?

• Since I don't have Gradshteyn and Ryzhik in my midst, what is the $W$ function? – Igor Rivin Dec 10 '13 at 1:25
• @Igor: I'd like to say sorry again! I've added what I think will be the most convenient form of the Whittaker function. Hopefully that helps. On a completely different note, I hope you're enjoying ICERM (I say as a Brown student just up the hill from you). – davidlowryduda Dec 10 '13 at 8:54

With $\ds{t \equiv -\ic x\quad\imp\quad x = \ic t}$: \begin{align}&\color{#c00000}{\int_{-\infty}^{\infty} {\expo{-\ic px} \over \pars{1 + \ic x}^{2\mu}\pars{1 - \ic x}^{2\nu}}\,\dd x} =-\ic\int_{-\infty\ic}^{\infty\ic} {\expo{pt} \over \pars{1 - t}^{2\mu}\pars{1 + t}^{2\nu}}\,\dd t \end{align} Now, it seems 'somehow' related to Laplace Transforms and we'll take advantage of that fact: If $\ds{p < 0}$ we'll close the integral with a contour 'to the right' $\ds{\pars{~\Re\pars{t} > 0~}}$ and 'to the left' $\ds{\pars{~\Re\pars{t} < 0~}}$ when $\ds{p > 0}$. In both cases we take into account the branch cuts of the denominator.
However, under the change $\ds{p \to -p}$ the integral is symmetric with the exchange of $\ds{\mu}$ and $\ds{\nu}$ such that we need to performe the integration for just one of the cases mentioned above.
Let's perform the calculation when $\ds{\large p > 0}$: \begin{align}&\color{#c00000}{\int_{-\infty}^{\infty} {\expo{-\ic px} \over \pars{1 + \ic x}^{2\mu}\pars{1 - \ic x}^{2\nu}}\,\dd x} \\[3mm]&=\ic\int_{-\infty}^{-1} {\expo{pt} \over \pars{1 - t}^{2\mu}\verts{1 + t}^{2\nu}\expo{2\pi\nu\ic}}\,\dd t +\ic\int_{-1}^{-\infty} {\expo{pt} \over \pars{1 - t}^{2\mu}\verts{1 + t}^{2\nu}\expo{-2\pi\nu\ic}}\,\dd t \\[3mm]&=\ic\expo{-2\pi\nu\ic}\int_{1}^{\infty} {\expo{-pt} \over \pars{1 + t}^{2\mu}\pars{t - 1}^{2\nu}}\,\dd t -\ic\expo{2\pi\nu\ic}\int_{1}^{\infty} {\expo{-pt} \over \pars{1 + t}^{2\mu}\pars{t - 1}^{2\nu}}\,\dd t \\[3mm]&=2\sin\pars{2\pi\nu}\int_{1}^{\infty} {\expo{-pt} \over \pars{1 + t}^{2\mu}\pars{t - 1}^{2\nu}}\,\dd t \\[3mm]&=2\sin\pars{2\pi\nu}\expo{-p}2^{-2\mu}\int_{0}^{\infty} t^{-2\nu}\expo{-pt}\pars{1 + {t \over 2}}^{-2\mu}\,\dd t \\[3mm]&=2\sin\pars{2\pi\nu}\expo{-p}2^{-2\mu}\int_{0}^{\infty} p^{2\nu}\pars{pt}^{-2\nu}\expo{-pt}\pars{1 + {pt \over 2p}}^{-2\mu} \,{p\,\dd t \over p} \\[3mm]&=\color{#44f}{\large% 2^{1 - 2\nu}\sin\pars{2\pi\nu}\expo{-p}p^{2\nu - 1}\int_{0}^{\infty} t^{-2\nu}\expo{-t}\pars{1 + {t \over 2p}}^{-2\mu}\,\dd t} \end{align}
$\tt\mbox{At this point, I guess the OP is able to see the relation to the Whittaker}$ $\tt\mbox{function !!!.}$