Integral of a hypergeometric function

Question

Does anyone know if there is a closed form expression for

$I(a,v,\tau_2) = \int_{-1/2}^{1/2}d\tau_1 \;{_2}F_1(a,a;1;q^v)$

where $q=e^{2\pi i \tau}$ and $\tau=\tau_1+i\tau_2$? Also $a$ is some real number probably $\leq 1$ and $v\in[0,1]$.

Answer 1

I'll assume $v \in (0, 1)$. If $z = e^{2 \pi i v \tau_1}$, then $$I = \frac 1 {2 \pi i v} \int_{|z| = 1 \land \left| \arg z \right| < \pi v} \frac {{_2 F_1}(a, a; 1; \omega z)} z dz,$$ where $\omega = e^{-2 \pi v \tau_2}$. An antiderivative of the integrand is $$F(z) = a^2 \omega z \,{_4 F_3}(1, 1, a + 1, a + 1; 2, 2, 2; \omega z) + \ln z.$$ Assume also that the branch cuts of the hypergeometric functions are along $[1, \infty)$ and that $\ln z$ is the principal branch of the logarithm. If $\tau_2 > 0$, then $\omega z$ is inside the unit circle and $F$ is continuous on the integration contour. If $\tau_2 < 0$, then there is a discontinuity at $z = 1$. Therefore $$I = \frac 1 {2 \pi i v} \cases { F(e^{\pi i v}) - F(e^{-\pi i v}) & $\tau_2 > 0$ \\ F(e^{\pi i v}) - F(1 + i0) + F(1 - i0) - F(e^{-\pi i v}) & $\tau_2 < 0$}.$$