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?

• Here is one way to derive the expressions for the upper/lower limits. – Maxim Apr 15 at 11:22

1 Answer

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 $$\frac{\sin(\pi(c-b-a))}{\pi}\,\Re \mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};x+i0\right) = \frac{1}{\Gamma(c-a)\Gamma(c-b)} \,\mathbf F\left(\begin{array}{} \ \ \ \ \ \ \ \ a,b\\a+b-c+1\end{array};1-x\right) - \frac{\cos(\pi(a+b-c))(x-1)^{c-b-a}}{\Gamma(a)\Gamma(b)} \,\mathbf F\left(\begin{array}{} \ \ c-a,c-b\\c-a-b+1\end{array};1-x\right)\,.$$