Prove an identity involving Gamma function and Gaussian hypergeometric function The question asks to prove the following identity:
$$\int_0^{2\pi}\!\!\frac{e^{-in\Psi}}{\left[\left(\frac{\rho}{\rho_0}\right)^2+1-2\left(\frac{\rho}{\rho_0}\right)\cos\Psi\right]^{\frac{1+u}{2}}}\mathbb{d}\Psi=\frac{2\pi\Gamma\left(n+\frac{1+u}{2}\right)}{\Gamma\left(\frac{1+u}{2}\right)\Gamma(n+1)}\left(\frac{\rho}{\rho_0}\right)^n F\left(\frac{1+u}{2},n+\frac{1+u}{2},n+1,\left(\frac{\rho}{\rho_0}\right)^2\right)
$$
where $i$ is the unit imaginary number, $n$ is an arbitrary integer, $\Gamma$ denotes the Gamma function, $F$ denotes the Gaussian hypergeometric function.
I have not even the slightest clue on how to deal with it. Can anyone help?

The left side is an even function of $n$, so let us assume (as you implicitly do on the right) that $n\geq0$. Also, denote $t=\frac{\rho}{\rho_0}$ and $a=\frac{1+u}{2}$. We want to compute
$$ I= \int_0^{2\pi}\frac{e^{-in\Psi}}{\left(1-2t\cos\Psi+t^2\right)^{a}}d\Psi.\qquad\qquad (\mathrm{A})$$
Write
$$ 1-2t\cos\Psi+t^2=1-t(e^{i\Psi}+e^{-i\Psi})+t^2= (1-te^{i\Psi})(1-te^{-i\Psi}).$$
Recall that $(1-x)^{-a}=\sum_{j=0}^{\infty}\frac{\Gamma(a+j)}{\Gamma(a)\Gamma(j+1)}x^j$. Multiplying two such series, we can write
$$ \left(1-2t\cos\Psi+t^2\right)^{-a}=\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\frac{\Gamma(a+j)\Gamma(a+k)}{\Gamma(a)\Gamma(j+1)\Gamma(a)\Gamma(k+1)}t^{j+k}e^{i(j-k)\Psi}.$$
Now we want to substitute this double series into (A) and integrate $e^{i(j-k-n)\Psi}$ w.r.t. $\Psi$. This is an easy task since almost all integrals are equal to zero. The only non-zero terms come from $j-k-n=0$, where the integral is equal to $2\pi$. Therefore (leaving only the sum over $k$ and replacing $j$ by $n+k$) we find
\begin{align*} I =2\pi\sum_{k=0}^{\infty}\frac{\Gamma(a+n+k)\Gamma(a+k)}{\Gamma(a)\Gamma(n+k+1)\Gamma(a)\Gamma(k+1)}t^{n+2k}=\\
=2\pi t^n\times\frac{\Gamma(a+n)}{\Gamma(a)\Gamma(n+1)}\times\sum_{k=0}^{\infty}\frac{\Gamma(n+a+k)\Gamma(n+1)\Gamma(a+k)}{\Gamma(n+a)\Gamma(n+k+1)\Gamma(a)}\frac{\left(t^2\right)^k}{k!}.
\end{align*}
The last sum is the standard series representation of $_2F_1(a,n+a,n+1,t^2)$, and we obtain the necessary identity. $\blacksquare$
There are several other ways to prove the same formula. For example, if you are ok with complex analysis, you can consider (A) as an integral over unit circle in the complex plane $z=e^{-i\Psi}$ and then shrink it to the branch cut $[0,t]$. This will give Euler integral representation of $_2F_1$.
