# Hypergeometric function

1. Jun 7, 2006

### Wishbone

I am having trouble with a problem that asks me to show that if I change the variable of integration of the following equation from t to t-1 the following

(disregard that z in the denominator, that should not be there)

will equal the following

(this one uses z's instead of t's)

I have tried simply changing the all the t's to t-1, and the only differentiating the t-1 term. That of course didn't work. I also tried a U substitution and that gave me an integral that looked like it couldn't be solved analytically (if you want me to post that code I can). I am not sure whether I even need to integrate, or whether this is one of those problems with a really quick shortuct and I can avoid doing any integration. Anyways, any help would be appreicated, thanks.

Last edited: Jun 7, 2006
2. Jun 8, 2006

### AKG

Why do you think it should not be there? Without it, the right side does not depend on z at all, although the left side being 2F1(a,b;c;z) suggests it should.

3. Jun 12, 2006

### benorin

a Hypergeometric transformation

$$_2F_1(a,b;c;z)=\int_{t=0}^{1}\frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^{a}}dt$$

now substitute t=1-u which gives dt=-du to get

$$_2F_1(a,b;c;z) = -\int_{u=1}^{0}\frac{(1-u)^{b-1}u^{c-b-1}}{(1-z+uz)^{a}}du = \int_{u=0}^{1}\frac{u^{c-b-1}(1-u)^{b-1}}{(1-z)^a\left( 1+u\frac{z}{1-z}}\right) ^{a}}du$$
$$= (1-z)^{-a} \int_{u=0}^{1}\frac{u^{c-b-1}(1-u)^{b-1}}{\left( 1+u\frac{z}{1-z}}\right) ^{a}}du = (1-z)^{-a} \, _2F_1 \left( a,c-b;c ;\frac{z}{z-1}\right)$$

Last edited: Jun 12, 2006