What is the following expression equal to?

Question

What is the following expression equal to? $$z^{\alpha } \left(\, _2F_1\left(1,-\alpha ;1-\alpha ;\frac{1}{z}\right)+\, _2F_1(1,\alpha ;\alpha +1;z)-1\right)$$ The derivative of it with respect to z is zero. And the limit of it when z goes to zero is $$\pi \alpha (\cot (\pi \alpha )+i)$$, while the limit of it when z goes to 1 is $$\pi \alpha (\cot (\pi \alpha )-i)$$.

Answer 1

This is a partial answer. It only cover the case where $\alpha \ne 0$ and $0 \le \Re\alpha < 1$.

For any $\beta \in \mathbb{C}\setminus \mathbb{Z}$, let $G_\beta(z)$ be the shorthand for ${}_2F_1(1,\beta;\beta+1;z)$. The equality at hand can be rewritten as $$z^\alpha \left( G_{-\alpha}(1/z) + G_{\alpha}(z) - 1 \right) \stackrel{?}{=} \pi\alpha(\cot(\alpha) + i)\tag{*1}$$

For any nonzero $\beta$ with $|\Re\beta| < 1$ and $|z| < 1$, we have

$$\begin{align}G_\beta(z) &= \sum_{k=0}^\infty \frac{(1)_k (\beta)_k}{k!(\beta+1)_k} z^k = \sum_{k=0}^\infty \frac{\beta}{\beta+k} z^k = 1 + \beta\sum_{k=1}^\infty \left( \int_0^1 t^{\beta+k-1} dt \right) z^k\\ &= 1 + \beta \int_0^1 t^{\beta-1}\left(\sum_{k=1}^\infty (tz)^k \right) dt = 1 + \beta z \int_0^1 \frac{t^{\beta}}{1 - tz} dt \end{align}\tag{*2} $$ Please note that the integral in RHS$(*2)$ is well defined for any $z \in \mathbb{C}\setminus [1,\infty)$. Let $C$ be any closed contour in $\mathbb{C}\setminus [1,\infty)$ and consider following double integral:

$$\int_C \left( \int_0^1 \frac{t^\beta}{1-tz} dt \right) dz$$

Since $[0,1] \times C$ is compact and $1 - tz \ne 0$ for any $(t,z) \in [0,1] \times \mathbb{C}\setminus [1,\infty)$, the term $\displaystyle\;\left|\frac{1}{1-tz}\right|\;$ in the integrand is bounded from above over $[0,1] \times C$. As long as $|\Re\beta| < 1$, the double integral is absolutely integrable and we can exchange the order of integration. Together with Cauchy integral theorem, we find:

$$\int_C \left( \int_0^1 \frac{t^\beta}{1-tz} dt \right) dz = \int_0^1 \left( \int_C \frac{t^\beta}{1-tz} dz \right) dt = 0 $$ Since $C$ is arbitrary, by Morera's theorem, the integral $\int_0^1 \frac{t^\beta}{1-tz} dt$ defines an analytic function on $\mathbb{C}\setminus [1,\infty)$. We can use $(*1)$ to analytic continue $G_\beta(z)$ over $\mathbb{C}\setminus [1,\infty)$.

From now on, assume $z \in \mathbb{C} \setminus [0,\infty)$.

Let $\alpha \ne 0$ be any number in the strip $0 \le \Re\alpha < 1$.

Substitute $\beta$ by $\alpha$ in $(*2)$, we have

$$G_\alpha(z) = 1 + \alpha \int_0^1 t^{\alpha-1}\left(\frac{1}{1-tz} -1\right) dt = \alpha \int_0^1 \frac{t^{\alpha-1}}{1-tz}dt\tag{*3a}$$

Substitute $\beta$ by $-\alpha$ in $(*2)$ and change variable to $s = 1/t$, we find

$$G_{-\alpha}(1/z) = 1 - \frac{\alpha}{z}\int_1^\infty \frac{s^\alpha}{1 - \frac{1}{zs}} \frac{ds}{s^2} = 1 + \alpha \int_1^\infty \frac{s^{\alpha-1}}{1 - zs}ds\tag{*3b} $$ Combine $(*3a)$ and $(*3b)$, we have

$$z^\alpha \left(G_{-\alpha}(1/z) + G_{\alpha}(z) - 1\right) = \alpha z^\alpha \int_0^\infty \frac{t^{\alpha-1}}{1-zt} ds\tag{*4}$$ To evaluate the integral on $(*4)$, consider following contour $C_\epsilon$

$$+\infty - \epsilon i\quad\to\quad -\epsilon - \epsilon i \quad\to\quad -\epsilon + \epsilon i \quad\to\quad +\infty + \epsilon i$$

If we choose the branch cut of $s^{\alpha-1}$ along the positive real axis and pick the branch where argument of $t^{\alpha-1}$ is $0$ on the upper side of the cut. We have

$$(1 - e^{2\pi\alpha i})\int_0^\infty \frac{t^{\alpha-1}}{1-tz} dt = \lim_{\epsilon\to 0}\int_{C_\epsilon} \frac{t^{\alpha-1}}{1-tz} dt$$ When $\Re\alpha < 1$, the integrand fall off fast enough as $|t| \to \infty$. We can complete the contour $C_\epsilon$ by a circle at infinity and convert the contour integral over $C_\epsilon$ to evaluation of residues within the extended contour. We obtain

$$\alpha z^\alpha \int_0^\infty \frac{t^{\alpha-1}}{1-tz}dt = 2\pi i\left(\frac{\alpha z^\alpha}{1 - e^{2\pi\alpha i}}\right)\mathop{\text{Res}}_{t = 1/z}\left(\frac{t^{\alpha-1}}{1-tz}\right) = \frac{2\pi i\alpha}{e^{2\pi\alpha i} - 1} z^{\alpha-1} \left[ t^{\alpha-1} \right]_{t=1/z}$$

Since we are taking the branch cut for $t^{\alpha-1}$ along the positive real axis. When $z = re^{i\theta}$, the correct value of $1/z$ to put into $\left[ t^{\alpha-1} \right]_{t=1/z}$ should be $r^{-1} e^{(2\pi - \theta)i}$. This leads to

$$z^{\alpha-1} \left[ t^{\alpha-1} \right]_{t=1/z} = (re^{i\theta})^{\alpha-1} (r^{-1}e^{(2\pi - \theta)i})^{\alpha-1} = e^{2\pi\alpha i}$$ As a result,

$$\text{RHS}(*4) = 2\pi i\alpha \frac{e^{2\pi\alpha i}}{e^{2\pi\alpha i} - 1} = \pi \alpha i \left( \frac{e^{2\pi \alpha i} + 1}{e^{2\pi \alpha i} - 1} + 1\right) = \pi \alpha (\cot(\pi\alpha) + i ) $$

This justify $(*1)$ when

• $\alpha\ne 0$, falls inside the strip $0 \le \Re\alpha < 1$,
• $z \in \mathbb{C}\setminus [0,\infty)$,
• the branch cut of $z^{\alpha}$ is taking along the +ve real axis,
• $\arg z = 0$ on upper side of +ve real axis.