# Hypergeometric challenge # 2

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Prove the following

$$\displaystyle _2F_1 \left(a,1-a;c; \frac{1}{2} \right) = \frac{\Gamma \left(\frac{c}{2} \right)\Gamma \left(\frac{1+c}{2} \right) } {\Gamma \left(\frac{c+a}{2}\right)\Gamma \left(\frac{1+c-a}{2}\right)}.$$​

#### Random Variable

##### Well-known member
MHB Math Helper
This only took me a couple of hours to figure out.

Using Euler's integral representation of ${}_2F_{1}(a,b;c;z)$,

$$\displaystyle {}_2F_{1} \left(a,1-a;c;\frac{1}{2} \right) = \frac{\Gamma(c)}{\Gamma(1-a) \Gamma(a+c-1)} \int_{0}^{1} t^{-a} (1-t)^{a+c-2} \left(1- \frac{t}{2} \right)^{-a} \ dt$$

$$= \frac{\Gamma(c)}{\Gamma(1-a) \Gamma(a+c-1)} \int_{0}^{1} (1-u)^{-a} u^{a+c-2} \left(\frac{1}{2} + \frac{u}{2} \right)^{-a} \ du$$

$$= \frac{2^{a} \ \Gamma(c)}{\Gamma(1-a) \Gamma(a+c-1)} \int_{0}^{1} u^{a+c-2} (1-u^{2})^{-a} \ du$$

$$= \frac{2^{a-1} \ \Gamma(c)}{\Gamma(1-a) \Gamma(a+c-1)} \int_{0}^{1} w^{\frac{a}{2} + \frac{c}{2} -\frac{3}{2}} (1-w)^{-a} \ dw$$

$$= \frac{2^{a-1} \ \Gamma(c)}{\Gamma(1-a) \Gamma(c+a-1)} \frac{\Gamma(\frac{a}{2} + \frac{c}{2}- \frac{1}{2}) \Gamma(1-a)}{\Gamma(\frac{c}{2}- \frac{a}{2}+ \frac{1}{2})} = \frac{2^{a-1} \ \Gamma(c)}{\Gamma(a+c-1)} \frac{\Gamma(\frac{a}{2} + \frac{c}{2}- \frac{1}{2}) }{\Gamma(\frac{c}{2}- \frac{a}{2}+ \frac{1}{2})}$$

$$= \frac{2^{a-1} \ \Gamma(c)}{\Gamma(a+c-1)} \frac{\Gamma\big(\frac{1}{2} (a + c- 1)\big)}{\Gamma(\frac{c}{2}- \frac{a}{2}+ \frac{1}{2}) }$$

And using the duplication formula twice,

$$= 2^{a-1} \frac{2^{c-1} \Gamma(\frac{c}{2}) \Gamma (\frac{c}{2}+ \frac{1}{2})}{\sqrt{\pi}} \frac{\sqrt{\pi}}{2^{a+c-2}\Gamma(\frac{a}{2} + \frac{c}{2})} \frac{1}{\Gamma(\frac{c}{2} - \frac{a}{2} + \frac{1}{2})}$$

$$= \frac{\Gamma(\frac{c}{2}) \Gamma(\frac{c}{2} + \frac{1}{2})}{\Gamma(\frac{a}{2} + \frac{c}{2}) \Gamma(\frac{c}{2} - \frac{a}{2} + \frac{1}{2})}$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Hey Greg , I am pleased to have you back . Using the duplication formula was the tricky part and using it twice is surely tiresome !

#### Random Variable

##### Well-known member
MHB Math Helper
What seems even more tricky is to use Euler's integral representation to show that $$\displaystyle _{2}F_{1}\left( a,b;\frac{a+b+1}{2},\frac{1}{2} \right) = \frac{\sqrt{\pi} \ \Gamma (\frac{a+b+1}{2})}{\Gamma(\frac{a+1}{2}) \Gamma (\frac{b+1}{2})}$$

Wikipedia refers to this as Gauss' second summation theorem.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
I would use a combination of euler transformation and kummer equation . Staring from scratch is sureley cumbersome!

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
By the way greg have you worked th mellin transform of the gauss hypergeometric function ,it is quite intresting.
See here

#### Random Variable

##### Well-known member
MHB Math Helper
You can use Ramanujan's master formula to evaluate the Mellin transform of $${}_pF_{q} (a_{1},a_{2}, \ldots, a_{p};b_{1},b_{2}, \ldots, b_{q}; -x) = \sum_{k=0}^{\infty} \frac{\Gamma(a_{1}+k) \Gamma(a_{2}+k) \cdots \Gamma(a_{p}+k) \Gamma(b_{1}) \Gamma(b_{2}) \cdots \Gamma(b_{q})}{\Gamma(a_{1}) \Gamma(a_{2}) \cdots \Gamma(a_{p}) \Gamma (b_{1}+k) \Gamma(b_{2}+k) \cdots \Gamma(b_{q}+k)} \frac{(-x)^{k}}{k!}$$

Specifically, $$\int_{0}^{\infty} x^{s-1} {}_pF_{q} (a_{1},a_{2}, \ldots, a_{p};b_{1},b_{2}, \ldots, b_{q}; -x) \ dx = \Gamma(s) \ \frac{\Gamma(a_{1}-s) \Gamma(a_{2}-s) \cdots \Gamma(a_{p}-s) \Gamma(b_{1}) \Gamma(b_{2}) \cdots \Gamma(b_{q})}{\Gamma(a_{1}) \Gamma(a_{2}) \cdots \Gamma(a_{p}) \Gamma (b_{1}-s) \Gamma(b_{2}-s) \cdots \Gamma(b_{q}-s)}$$

I've used the formula to evaluate the Mellin transforms of functions with simple hypergeometric representations of that form.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Yes , quite remarkable theorem , thanks to Ramnujan .