Integration of hypergeometric function on complex plane

Question

I have come across an integral that involves a hypergeometric function, which can be expressed as follows:

$$I = \int_0^1 x^{1/2}(1-x)^{\epsilon-1} {_{2}F_1}(\frac{1}{2}+\epsilon,1+\epsilon;\frac{3}{2};x) dx.$$

Here, $\epsilon$ is a small complex quantity where $|\epsilon|\ll1$.

I found an integral formula in "Table of Integrals, Series, and Products," ET II 399(4), as follows: $$\int_0^1 x^{\gamma-1}(1-x)^{\rho-1}F(\alpha,\beta,;\gamma;x)dx=\frac{\Gamma(\gamma)\Gamma(\rho)\Gamma(\gamma+\rho-\alpha-\beta)}{\Gamma(\gamma+\rho-\alpha)\Gamma(\gamma+\rho-\beta)}$$, for $Re \ \gamma\gt0, Re\ \rho\gt0, Re\ (\gamma +\rho -\alpha - \beta)\gt0$.

As one can see, for my case, $\alpha=1/2+\epsilon, \beta=1+\epsilon, \gamma=3/2, \rho=\epsilon$, and if assume $Re\ (\epsilon) \gt 0$(this is not necessarily true), the third condition can not be satisfied as $Re\ (\gamma +\rho -\alpha - \beta)= Re\ (-\epsilon)\lt0$.

I have a question regarding the integral $I$. Does the given case imply that $I$ diverges, or is there any other way to determine the result? If possible, I would appreciate it if someone could guide me on how to solve this integral. Thank you all for taking the time to read my post!

---

EDIT: A little background for the problem set:

I intend to compute this complex integral along the path $C$ on the complex plane as follows: $$K=\int_C\frac{z^2}{1+z^2} {_2F_1(\frac{1}{2}+\epsilon,\frac{1}{2}-\epsilon;\frac{3}{2};-z^2)}dz,$$ where $z=x/\delta$, $x$ is real, $\delta=\sqrt{(4\epsilon^2-1)\eta}$, here, $\epsilon$ is a small complex quantity ($|\epsilon|\ll1$)(so $Re (\epsilon)$ can be positive or negative, whatever helps, we can assume that) and $0<\eta\ll1$ is a real parameter; the integration path $C$ is determined by $z=x/\delta$, where x is taken from $(0,+\infty)$.

My attempt is first taking a transform of variable, $t=z^2$, and the integral $K$ is converted into the following form: $$K=\frac{1}{2}\int_{C'} t^{1/2}(1+t)^{-1} {_2F_1(\frac{1}{2}+\epsilon,\frac{1}{2}-\epsilon;\frac{3}{2};-t)}dt$$, then making use of the relation: $$F(a,b;c;z)=(1-z)^{-a}F(a,c-b;c;\frac{z}{z-1})$$, yielding $$K=\frac{1}{2}\int_{C''} t^{1/2}(1+t)^{-3/2-\epsilon} {_2F_1(\frac{1}{2}+\epsilon,1+\epsilon;\frac{3}{2};\frac{t}{t+1})}dt$$. At this point, I took a transform of the variable as $l=t/(t+1)$ and ended up something like this: $$K=\frac{1}{2}\int_{C'''}l^{1/2}(1-l)^{\epsilon-1} {_2F_1(\frac{1}{2}+\epsilon,1+\epsilon;\frac{3}{2};l)dl},$$ where the paths of integration, $C'$, $C''$ and $C'''$ change accordingly with the transformation of variables.

Finally, if $x: 0 \to +\infty$, $l$ varies from $0 \to 1$ along the path $C'''$, therefore I thought it is safe to say(I think I made a serious mistake here, even though I am not quiet sure why): $$K=\frac{1}{2}\int_0^1l^{1/2}(1-l)^{\epsilon-1} {_2F_1(\frac{1}{2}+\epsilon,1+\epsilon;\frac{3}{2};l)dl}.$$ Obviously, from the comment by @Conrad, at $l=1$, the integrant is singular, so I need to be careful by changing the integral path from $C'''$ to the real axis.

What should I do at this point?

I would appreciate it so much if someone could tell me how to deal with this problem; thank you all for your time reading my post.

Answer 1

For the antiderivative $$I = \int\sqrt x\,\,(1-x)^{\epsilon-1} \, _2F_1\left(\frac{1}{2}+\epsilon,1+\epsilon;\frac{3}{2};x\right)\, dx$$ $$ _2F_1\left(\frac{1}{2}+\epsilon,1+\epsilon;\frac{3}{2};x\right)=\frac{\left(1+\sqrt{x}\right)^{2 \epsilon }-\left(1-\sqrt{x}\right)^{2 \epsilon } } {4 \,\epsilon\, \sqrt{x}\, (1-x)^{2 \epsilon } }$$ $$I=\frac 1{4 \,\epsilon}\int \frac{\left(1+\sqrt{x}\right)^{2 \epsilon }-\left(1-\sqrt{x}\right)^{2 \epsilon } } {(1-x)^{\epsilon +1}}\,dx$$ $$I=\frac{x}{4 \epsilon }\Big(F_1\left(2;1+\epsilon ,1-\epsilon ;3;\sqrt{x},-\sqrt{x}\right)-F_1\left(2;1-\epsilon ,1+\epsilon ;3;\sqrt{x},-\sqrt{x}\right)\Big)$$ where appear Appell hypergeometric functions of two variables.

The definite integral tends to infinity when $x\to 1^-$

Answer 2

If you use Eulers integral representation of hypergeometric series, you get a double integral over the unit square

$$\text{Assuming}\left[0<\epsilon <\frac{1}{2},\text{Simplify}\left[\frac{\Gamma \left(\frac{3}{2}\right) \int_0^1 \left(\int_0^1 \sqrt{x} (t x+1) (1-t)^{\frac{1}{2}-\epsilon } t^{\epsilon +1} (1-x)^{\epsilon -1} \, dt\right) \, dx}{\Gamma (\epsilon +1) \Gamma \left(\frac{1}{2}-\epsilon \right)}\right]\right]$$ $$=\frac{\sqrt{\pi } (20 \epsilon +33) \Gamma \left(\frac{3}{2}-\epsilon \right) \Gamma (\epsilon ) \Gamma (\epsilon +2)}{105 \Gamma \left(\frac{1}{2}-\epsilon \right) \Gamma (\epsilon +1) \Gamma \left(\epsilon +\frac{5}{2}\right)}$$

using Mathematica.

A numerical test confirms

   With[{\[Epsilon] = 1/3}, 
   NIntegrate[ Gamma[3/2]/
    (Gamma[1 + \[Epsilon]] Gamma[1/2 - \[Epsilon]])* 
    ((1 - t)^(1/2 - \[Epsilon])*  t^(1 + \[Epsilon])*
     (1 - x)^(-1 + \[Epsilon]) *  Sqrt[x] (1 + t x)) , 
    {t, 0, 1}, {x, 0, 1}]]

      0.231148



     (Sqrt[\[Pi]] (33 + 20 \[Epsilon]) Gamma[
    3/2 - \[Epsilon]] Gamma[\[Epsilon]] Gamma[2 + \[Epsilon]])/(
  105 Gamma[1/2 - \[Epsilon]] Gamma[1 + \[Epsilon]] Gamma[
    5/2 + \[Epsilon]]) /. {\[Epsilon] -> 1/3} // N

    0.231148