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Hypergeometric challenge # 2

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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
Jan 31, 2012
253
This only took me a couple of hours to figure out.(Whew)


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
Jan 17, 2013
1,667
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
Jan 31, 2012
253
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
Jan 17, 2013
1,667
I would use a combination of euler transformation and kummer equation . Staring from scratch is sureley cumbersome!
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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
Jan 31, 2012
253
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
Jan 17, 2013
1,667
Yes , quite remarkable theorem , thanks to Ramnujan .