What is the limit as $x\to\infty$ of the incomplete elliptic integral of the third kind with imaginary argument $\phi=-ix$?

Question

I am trying to evaluate the limit as $x\to\infty$ of the incomplete integral of the third kind,

$$\Pi(n;\phi|m), \quad \mathrm{(using~the~Wolfram~language~convention)}$$

with $\phi=-ix$.

In particular I need to consider

$$\lim\limits_{x\to\infty}\Pi(a+ib;-ix|(-1)^{1/3}).$$

I've been unable to find anything in Abramovitz and Stegun, but experimenting on Wolfram Alpha (WA) shows that the limit can be calculated for a given $a+ib$. With $a+ib=1+i$, WA calculates the variation of the argument of the limit with $x$ as

$\Pi(1+i;-ix|(-1)^{1/3})$ against $x$" />,

showing that, for $a+ib=1+i$ at least, the limit is approached quickly. WA calculates the limit to be $-0.24-0.69i$ to 2s.f. in this case.

Is there a closed form for the limit in terms of $a$ and $b$? I would like to compute the limit over a range of $a$ and $b$, this would be much simpler if I didn't have to take the limit each time. Thanks in advance for your help.

Answer 1

Have a look at my Elliptic Integrals and Functions monograph. The relevant part is in section 2.3:

Let $m^*=1-m$ and $\tan\psi=-i\sin\varphi$, then (BF 161.01, .02, DLMF 19.7.7, 22.6(iv)) $$\Pi(n,\varphi,m)=\frac i{1-n}(F(\psi,m^*)-n\Pi(1-n,\psi,m^*))$$

If $\varphi=-ix$ with $x\ge0$, $\psi=-\tan^{-1}\sinh x$; as $x\to\infty$, $\psi\to-\frac\pi2$. Thus the integrals become complete in the limit: $$\lim_{x\to\infty}\Pi(n,-ix,m)=\frac i{1-n}(F(-\pi/2,1-m)-n\Pi(1-n,-\pi/2,1-m))$$ $$=\frac i{1-n}(n\Pi(1-n,1-m)-K(1-m))$$