Yesterday I posted a question about the Gamma function (see here: Lower bound for combined Gamma functions)

It helped me very much in solving all my remaining exercises concerning the Gamma function, except one last. It is the following. Let $0<a<1$ and $b\ge 2, b\in\mathbb{N}$. Prove that there exists a positive constant $c$ which does not depend on $a$ (but it would be fine it would depend on $b$) such that $$\frac{\Gamma\big((b+2a)/2\big)}{\Gamma(2-a)}\le c.$$

Here $\Gamma$ is defined, for $Re(z)>0$, as $$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt$$ (here $(b+2a)/2$ and $2-a$ are all real and nonnegative numbers).

I tried by using the same strategy in the linked question. It worked with all the other exercise, but in this case I can not get rid of the dependence on $a$.

Anyone could help?


1 Answer 1


This has nothing to do with the Gamma function. You only need that $\Gamma$ is continuous and does not have any zeros. Then for every fixed $b$, the function $$f : [0,1] \to \mathbb{R}, \quad f(a) = \frac{\Gamma((b+2a)/2)}{\Gamma(2-a)}$$ is continuous and hence bounded.


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