Integral of a power combined with a Gaussian hypergeometric function 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.

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
