<s>[INST] Discontinuity of Hypergeometric function along the branch cut I am trying to evaluate an expression involving the hypergeometric function evaluated near its (principal) branch cut discontinuity, which is placed on the real line from $1$ to infinity.
For $x>1$, DLMF gives the following expression for the difference (or ''imaginary part'') of the regularized hypergeometric function $\mathbf F(a,b;c;z)={}_2F_1(a,b;c;z)/\Gamma(c)$ across the branch cut,
$$
\mathbf F(a,b;c;x+i0) - \mathbf F(a,b;c;x-i0)$$
$$ = \tfrac{i2\pi}{\Gamma(a)\Gamma(b)}(x-1)^{c-a-b}\mathbf F(c-a,c-b;c-a-b+1;1-x)\,.
$$
However, I am interested in the analogous expression for the sum (or, the ''real part''), again for $x>1$,
$$
\mathbf F(a,b;c;x+i0)+\mathbf F(a,b;c;x-i0)=\,?
$$
I tried plugging specific values of this into mathematica, in particular for integer $D>2$, which yields
$$
\mathbf F\left(\tfrac{D-1}{2},\tfrac{D}{2};\tfrac{D+1}{2};x+i0\right)
+
\mathbf F\left(\tfrac{D-1}{2},\tfrac{D}{2};\tfrac{D+1}{2};x-i0\right)$$
$$=
(1+(-1)^D)\mathbf F\left(\tfrac{D-1}{2},\tfrac{D}{2};\tfrac{D+1}{2};x\right)
+
\tfrac{2i^{D+1}\pi^{3/2}x^{-\tfrac{D-1}{2}}}{\cos\left(\tfrac{D\pi}{2}\right)\Gamma\left(\frac{D-3}{2}\right)\Gamma\left(\tfrac{D-1}{2}\right)\Gamma\left(\tfrac{D}{2}\right)}\,.
$$
This makes perfect sense when $D$ is odd, since then the right-hand side reads
$$
2\sqrt{\pi}(-1)^{\frac{D-1}{2}} \frac{x^{-\frac{D-1}{2}}}{\Gamma\left(\frac{D-1}{2}\right)\Gamma\left(\frac{D}{2}\right)}\,,
$$
however, it doesn't make much sense to me for even $D$, since then the right-hand side seems to still have a nonzero imaginary part.
Is there a general expression analogous to that provided for the difference by DLMF, but expressing the sum across the branch cut?
 [/INST] It turns out that to answer the question it is sufficient to employ in a suitable way some other formulas already present in the DLMF. 
A possibility is to use the ''inversion formula'' 
$$
\frac{\sin(\pi(b-a))}{\pi}\,
\mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};z\right)
=
\frac{(-z)^{-a}}{\Gamma(b)\Gamma(c-a)}
\,\mathbf F\left(\begin{array}{} a,a-c+1\\ \ \ a-b+1\end{array};\frac{1}{z}\right)
-
\frac{(-z)^{-b}}{\Gamma(a)\Gamma(c-b)}
\,\mathbf F\left(\begin{array}{} b,b-c+1\\ \ \ b-a+1\end{array};\frac{1}{z}\right)\,.
$$
Substituting $z=x+i0$ with $x>1$ and noting that, if we place the branch cut of the logarithm along the negative real axis, $(-(x+i0))^{-a}=e^{i\pi a}\,x^{-a}$, we have
$$
\frac{\sin(\pi(b-a))}{\pi}\,
\mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};x+i0\right)
=
\frac{e^{i\pi a}\,x^{-a}}{\Gamma(b)\Gamma(c-a)}
\,\mathbf F\left(\begin{array}{} a,a-c+1\\ \ \ a-b+1\end{array};\frac{1}{x}\right)
-
\frac{e^{i\pi b}\,x^{-b}}{\Gamma(a)\Gamma(c-b)}
\,\mathbf F\left(\begin{array}{} b,b-c+1\\ \ \ b-a+1\end{array};\frac{1}{x}\right)\,.
$$
This gives both real and imaginary parts, provided $b-a$ is not an integer (otherwise, the left-hand side vanishes).
For the specific values of $a$, $b$ and $c$ chosen in the example, $b-a=1/2$, so this formula applies and
$$
\Re \mathbf F\left(\begin{array}{} \frac{D-1}{2},\frac{D}{2}\\\ \ \frac{D+1}{2}\end{array};x+i0\right)
=\begin{cases}
(-)^{\frac{D}{2}+1}\sqrt{\pi} x^{-\frac{D}{2}}/\Gamma(\frac{D-1}{2}) &\text{even }D\\
(-)^{\frac{D-1}{2}}\pi x^{\frac{1-D}{2}}/\Gamma(\frac{D}{2}) &\text{odd }D\,.
\end{cases}
$$
Another workaround is provided by the ''reflection formula''
$$
\frac{\sin(\pi(c-b-a))}{\pi}\,
\mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};z\right)
=
\frac{1}{\Gamma(c-a)\Gamma(c-b)}
\,\mathbf F\left(\begin{array}{} \ \ \ \ \ \ \ \  a,b\\a+b-c+1\end{array};1-z\right)
-
\frac{(1-z)^{c-b-a}}{\Gamma(a)\Gamma(b)}
\,\mathbf F\left(\begin{array}{} \ \ c-a,c-b\\c-a-b+1\end{array};1-z\right)\,,
$$
hence
$$
\frac{\sin(\pi(c-b-a))}{\pi}\,
\mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};x+i0\right)
=
\frac{1}{\Gamma(c-a)\Gamma(c-b)}
\,\mathbf F\left(\begin{array}{} \ \ \ \ \ \ \ \  a,b\\a+b-c+1\end{array};1-x\right)
-
\frac{e^{i\pi(a+b-c)}(x-1)^{c-b-a}}{\Gamma(a)\Gamma(b)}
\,\mathbf F\left(\begin{array}{} \ \ c-a,c-b\\c-a-b+1\end{array};1-x\right)\,,
$$
for $x>1$. As a byproduct, we get the formula for the imaginary part quoted in the question,
$$
\Im \mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};x+i0\right) = \frac{\pi(x-1)^{x-a-b}}{\Gamma(a)\Gamma(b)}\,\mathbf F\left(\begin{array}{} \ \ c-a,c-b\\c-a-b+1\end{array};1-x\right)\,,
$$ 
together with
$$
\frac{\sin(\pi(c-b-a))}{\pi}\,\Re \mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};x+i0\right)
=
\frac{1}{\Gamma(c-a)\Gamma(c-b)}
\,\mathbf F\left(\begin{array}{} \ \ \ \ \ \ \ \  a,b\\a+b-c+1\end{array};1-x\right)
-
\frac{\cos(\pi(a+b-c))(x-1)^{c-b-a}}{\Gamma(a)\Gamma(b)}
\,\mathbf F\left(\begin{array}{} \ \ c-a,c-b\\c-a-b+1\end{array};1-x\right)\,.
$$
</s>