- Thread starter
- #1

- Jan 17, 2013

- 1,667

\(\displaystyle \beta(a,b,c;s) = \int^1_0 \frac{1}{(1-x)^a(s-x)^b x^c}\, dx \,\,\,\,\,\,-1<a,b,c<1 \,\,\,s\geq 0 \)

This is NOT a tutorial , all suggestions are encouraged.

- Thread starter ZaidAlyafey
- Start date

- Thread starter
- #1

- Jan 17, 2013

- 1,667

\(\displaystyle \beta(a,b,c;s) = \int^1_0 \frac{1}{(1-x)^a(s-x)^b x^c}\, dx \,\,\,\,\,\,-1<a,b,c<1 \,\,\,s\geq 0 \)

This is NOT a tutorial , all suggestions are encouraged.

- Thread starter
- #2

- Jan 17, 2013

- 1,667

Assume \(\displaystyle \Re(a+b)>0 , \Re(1-c)>0\)

\(\displaystyle \beta(a,b,c;\,1) = \int^1_0 (1-x)^{-(a+b)} x^{-c}dx= \beta(1-c,1-(a+b))=\frac{\Gamma(1-c)\Gamma(1-(a+b))}{\Gamma(2-(a+b+c))}\)

\(\displaystyle \tag{1} \beta(a,b,c;\,1) = \frac{\Gamma(1-c)\Gamma(1-(a+b))}{\Gamma(2-(a+b+c))} \,\,\,\, \Re(a+b)>0 , \Re(1-c)>0\)

- Sep 16, 2013

- 337

\(\displaystyle \beta(a,b,c;s) = \int^1_0 \frac{1}{(1-x)^a(s-x)^b x^c}\, dx \,\,\,\,\,\,-1<a,b,c<1 \,\,\,s\geq 0 \)

This is NOT a tutorial , all suggestions are encouraged.

You're on tip-top form, Zaid! I like it!

A few things jump out at me, which might be worth exploring...

Firstly, the reflection substitution \(\displaystyle x \to 1-x\) might be worth a look... Also, being the logarithmic fiend that I am, I think it might be worth considering partial derivatives wrt any/all of the parameters.

I'll definitely come back to this topic when it's not so close to bed time. Very interesting! (heart)