Principal value of hypergeometric function

[CAF123]

I am looking to write the hypergeometric function $${}_2F_1\left(1,1,2+\epsilon, -\frac{\alpha}{\beta}\right) = \int_0^1\,dt\,\frac{(1-t)^{\epsilon}}{1-tz + i\delta},$$ where $z=-\alpha/\beta$ and $0< \beta < - \alpha$, in terms of its real and imaginary part. The $i\delta$ prescription is to shift the denominator away from the pole at $t=1/z$. I know that $$\frac{1}{1-tz+i\delta} = \text{P.V} \frac{1}{1-tz} -i\pi \delta(1-tz)$$ so to compute the real part I am left with the problem with finding $$\text{P.V}\int_0^1\,dt\,\frac{(1-t)^{\epsilon}}{1-tz}$$ I tried writing this as $$\lim_{\tau \rightarrow 0} \left(\int_0^{1/z-\tau} + \int_{1/z+\tau}^1\right)\frac{(1-t)^{\epsilon}}{1-tz} dt$$ but I am not sure how to progress. I tried using the residue theorem and coming up with a closed contour but the limits do not extend to $\pm \infty$. Any help would be great, thanks!

 

[jvicens]

Thank you for your question. The hypergeometric function ${}_2F_1(a,b,c,z)$ is a special function that is defined as a power series in the variable $z$ and has many applications in mathematics and physics. In your case, you are interested in writing the hypergeometric function ${}_2F_1(1,1,2+\epsilon, -\frac{\alpha}{\beta})$ in terms of its real and imaginary part.

To do this, we can use the property of the hypergeometric function that it can be expressed as an integral. In particular, we can write ${}_2F_1(a,b,c,z)$ as an integral in the following form:

$${}_2F_1(a,b,c,z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1 t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a} dt$$

Using this property, we can rewrite your hypergeometric function as:

$${}_2F_1(1,1,2+\epsilon, -\frac{\alpha}{\beta}) = \frac{\Gamma(2+\epsilon)}{\Gamma(1)\Gamma(1+\epsilon)}\int_0^1 t^{0}(1-t)^{\epsilon}(1+\frac{\alpha}{\beta}t)^{-1} dt$$

Now, we can use the substitution $x = 1-t$ to rewrite the integral as:

$${}_2F_1(1,1,2+\epsilon, -\frac{\alpha}{\beta}) = \frac{\Gamma(2+\epsilon)}{\Gamma(1)\Gamma(1+\epsilon)}\int_1^0 (1-x)^{1+\epsilon} (1+\frac{\alpha}{\beta}(1-x))^{-1} (-dx)$$

Note that we have changed the limits of integration and introduced a negative sign due to the change in orientation of the integration. We can simplify this further to get:

$${}_2F_1(1,1,2+\epsilon, -\frac{\alpha}{\beta}) = \frac{\Gamma(2+\epsilon)}{\Gamma(1)\Gamma(1+\epsilon)}\int_0^1 (1-x)^{1+\epsilon} (1+\