Integral of a power combined with a Gaussian hypergeometric function

Question

I think the following is true for $k \ge 3$, $$ \int_0^{\infty } (w+1)^{\frac{2}{k}-2} \, _2F_1\left(\frac{2}{k},\frac{1}{k};1+\frac{1}{k};-w\right) \, dw = \frac{\pi \cot \left(\frac{\pi }{k}\right)}{k-2} . $$ I have checked Table of Integrals, Series, and Products without any hit. Has anyone seen any formula that resembles the left hand side?

Edit: This came up in my research when I tried to show another integral is equals the RHS. I was able to transform the integral into the LHS and it looked promising. But eventually I used another proof. Anyway, it's still nice to see this direction also works.

Answer 1

Using the integral representation 15.6.6 $$ \, _2F_1\left(\frac{2}{k},\frac{1}{k};1+\frac{1}{k};-w\right)=\frac{\Gamma\left(1+1/k\right)}{2\pi i\Gamma\left(1/k\right)\Gamma\left(2/k\right)}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(1/k+t\right)\Gamma\left(2/k+t\right )\Gamma\left(-t\right)}{\Gamma\left(1+1/k+t\right)}w^{t}dt, $$ where the contour of integration separates the poles of $\Gamma\left(1/k+t\right)\Gamma\left(2/k+t\right)$ from the poles of $\Gamma(-t)$, interchanging the order of integration and calculating the integral over $w$ $$ \int_0^{\infty } (w+1)^{\frac{2}{k}-2} w^tdw=B(t+1,1-2/k-t) $$ we get \begin{align} &\int_0^{\infty } (w+1)^{\frac{2}{k}-2} \, _2F_1\left(\frac{2}{k},\frac{1}{k};1+\frac{1}{k};-w\right) \, dw\\ &=\frac{1/k}{2\pi i\Gamma\left(2/k\right)}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(1/k+t\right)\Gamma\left(2/k+t\right )\Gamma\left(-t\right)\Gamma(t+1)\Gamma\left(1-2/k-t\right)}{\Gamma\left(1+1/k+t\right)\Gamma\left(2-2/k\right)}dt\\ &=\frac{\sin \left(\frac{2 \pi }{k}\right)}{2i(k-2)} \int_{-i\infty}^{i\infty}\frac{dt}{t\sin (\pi/k-\pi t) \sin \left(\pi/k+\pi t\right)}. \end{align} The contour of integration here is so that the origin lies to the left. We see that the integrand is odd. Therefore if we choose the contour as $(-i\infty,-ir)\cup \Gamma_r\cup(ir,i\infty)$,$r>0$, where $\Gamma_r=\left\{|z|=r,\text{Re}~z>0\right\}$, the integrals along $(-i\infty,-ir)$ and $(ir,i\infty)$ cancel out. The integral along $\Gamma_r$ can be calculated in the limit $r\to +0$ and equals $$ \frac{\pi i}{\sin^2(\pi/k)}. $$ Thus we get $$ \frac{\sin \left(\frac{2 \pi }{k}\right)}{2i(k-2)} \cdot \frac{\pi i}{\sin^2(\pi/k)}= \frac{\pi \cot \left(\frac{\pi }{k}\right)}{k-2}. $$

Answer 2

Expanding on my comment, we can alternatively use Euler's integral representation of the hypergeometric function, along with a well-known integral representation of the digamma function.

We also need the reflection formula for the digamma function.

$$\begin{align} I(k) &= \int_{0}^{\infty}(1+w)^{2/k-2} \, _2F_1 \left(\frac{2}{k}, \frac{1}{k}; 1 + \frac{1}{k};-w \right) \, dw \\ &= \int_{0}^{\infty}(1+w)^{2/k-2} \frac{1}{B \left(\frac{1}{k},1 \right)} \int_{0}^{1} x^{1/k-1}(1+wx)^{-2/k} \, dx \, dw \\ &= \frac{1}{k} \int_{0}^{1} x^{1/k-1}\int_{0}^{\infty} (1+w)^{2/k-2} (1+xw)^{-2/k} \, dw \, dx \\ &=\frac{1}{k} \int_{0}^{1} x^{1/k-1} \int_{1}^{\infty} \left(\frac{u-1}{x}+1 \right)^{2/k-2}u^{-2/k} \, \frac{du}{x} \, dx \\ &= \frac{1}{k} \int_{0}^{1} x^{-1/k} \int_{1}^{\infty} \frac{1}{(u-1+x)^{2}} \, \left(\frac{u}{u-1+x} \right)^{-2/k} \, du \, dx \\ &= \frac{1}{k} \int_{0}^{1} \frac{x^{-1/k}}{x-1} \int_{1/x}^{1} v^{-2/k} \, dv \, dx \tag{1} \\ &= \frac{1}{k-2} \int_{0}^{1} \frac{x^{1/k-1}-x^{-1/k}}{1-x} \, dx \\ &= \frac{1}{k-2} \left(\psi \left(1- \frac{1}{k} \right)-\psi \left(\frac{1}{k} \right) \right) \\ &= \frac{\pi}{k-2} \, \cot \left(\frac{\pi}{k} \right) \end{align} $$

This should hold for $k>1$, excluding $k=2$.

If $k=2$, then $(1)$ becomes $$ \frac{1}{2} \int_{0}^{1} \frac{x^{-1/2}\log(x)}{x-1} \, dx = \frac{1}{2} \, \psi_{1} \left(\frac{1}{2} \right) = \frac{\pi^{2}}{4},$$ which is the limit of $\frac{\pi}{k-2} \cot \left(\frac{\pi}{k} \right)$ as $k \to 2$.

https://en.wikipedia.org/wiki/Polygamma_function#Integral_representation