Evaluate $\frac{\Gamma(n/2)}{\Gamma(n/2+h)},$ $n,h \in \mathbb{N}$

Question

Suppose $n,h \in \mathbb{N}$. How can I evaluate $$\frac{\Gamma(n/2)}{\Gamma(n/2+h)},$$ where $\Gamma$ is the Euler gamma function.

All I know is that $$\Gamma(n+1) = n!$$

Answer 1

In fact you have the relationship $\Gamma(x+1) = x\Gamma(x)$ holds for any $x \in \mathbb{R^{+}}$, which is in fact one of the reason why mathematician came up with the Gamme function, i.e. a function that generalizes the factorial over the positive real numbers.

Now by succesivelly applying it you should get:

$$\Gamma\left(\frac n2 + h\right) = \left(\frac n2 + h - 1\right)\Gamma\left(\frac n2 + (h-1)\right) = \dots $$ $$= \left(\frac n2 + h - 1\right)\left(\frac n2 + h - 2\right) \cdots \left(\frac n2 \right)\Gamma\left(\frac n2\right)$$

Answer 2

The well known recurrence

$$n!=n(n-1)!$$ extends to the Gamma function, so that

$$\Gamma(n)=(n-1)\Gamma(n-1).$$

Then

$$\frac{\Gamma(n/2)}{\Gamma(n/2+h)}=\frac{\Gamma(n/2)}{(n/2+h-1)\Gamma(n/2+h-1)}=\frac{\Gamma(n/2)}{(n/2+h-2)(n/2+h-1)\Gamma(n/2+h-2)}\\=\cdots=\frac{\Gamma(n/2)}{(n/2+h-2)(n/2+h-1)\cdots(n/2+h-h)\Gamma(n/2+h-h)}.$$

Answer 3

If you consider the case where $n$ is large (and much larger than $h$), take logarithms and use Stirling approximation. This would give $$\log \left(\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n}{2}+h\right)}\right)=\log \left(2^h\right)-h \log ({n})+\frac{h-h^2}{n}+\frac{2 h^3-3 h^2+h}{3 n^2}+O\left(\frac{1}{n^{3}}\right)$$

For example, using $n=100$ and $h=10$, the exact value would be $\approx 4.38613\times 10^{-18}$ while the truncated expression would lead to $\approx 4.40747\times 10^{-18}$.

Edit

Continuing from the previous expansion and Taylor series, usind $A=e^{\log(A)}$, we can avoid exponentiation and get (as a slightly worse approximation) $$\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n}{2}+h\right)}=\left(\frac{2}{n}\right)^h\left(1-\frac{(h-1) h}{n}+\frac{(h-1) h (h+1) (3 h-2)}{6 n^2}+O\left(\frac{1}{n^3}\right)\right)$$

Answer 4

H.M. Edwards gives the expression in the opening of Riemann's Zeta Function (1974, reprint 2001)

$$ \Gamma(s) = \prod_{n=1}^\infty\bigg(1+\frac{s-1}{n}\bigg)^{-1}\bigg(1+\frac{1}{n}\bigg)^{s-1}, $$

which implies $$\frac{\Gamma(s/2)}{\Gamma(s/2+h)}= \prod_{n=1}^\infty\frac{\bigg(1+\frac{(s/2)-1}{n}\bigg)^{-1}\bigg(1+\frac{1}{n}\bigg)^{(s/2)-1}}{\bigg(1+\frac{(s/2)+h-1}{n}\bigg)^{-1}\bigg(1+\frac{1}{n}\bigg)^{(s/2)+h-1}} = $$

$$ = \prod_{n=1}^\infty \frac{\bigg(1+\frac{(s/2)-1}{n}\bigg)^{-1}}{\bigg(1+\frac{(s/2)+h-1}{n}\bigg)^{-1}\bigg(1+\frac{1}{n}\bigg)^{h}} = \prod_{n=1}^\infty \frac{\bigg(1+\frac{(s/2)+h-1}{n}\bigg)}{\bigg(1+\frac{(s/2)-1}{n}\bigg)\bigg(1+\frac{1}{n}\bigg)^{h}}$$

Another expression is given by Titchmarsh for when Re$(s)$ > 1.

$$ \Gamma(s) = \frac{1}{\zeta(s)} \int_0^{\infty}\frac{x^{s-1}}{e^x-1}dx $$

which also implies an expression

$$\frac{\Gamma(s/2)}{\Gamma(s/2+h)} = \frac{\zeta(s/2 + h)\int_0^{\infty}\frac{n^{(s/2)-1}}{e^n-1}dn}{\zeta(s/2)\int_0^{\infty}\frac{m^{(s/2)+h-1}}{e^m-1}dm} $$

And the integral over the first quadrant $(m,n) \in Q_1 = \bigg(\mathbb{R} \cap [0,\infty) \bigg)^2 $

$$ \frac{\int_0^{\infty}\frac{n^{(s/2)-1}}{e^n-1}dn}{\int_0^{\infty}\frac{m^{(s/2)+h-1}}{e^m-1}dm} = \int_{Q_1}\bigg( \frac{e^m -1}{e^n-1} \bigg) \bigg( \frac{1}{m^h} \bigg) \bigg( \frac{n}{m} \bigg)^{(s/2)-1} {dm} \ {dn}, $$

implying

$$ \frac{\Gamma(s/2)}{\Gamma(s/2+h)} = \frac{\zeta( (s/2) + h)}{\zeta(s/2)} \int_{Q_1}\bigg( \frac{e^m -1}{e^n-1} \bigg) \bigg( \frac{1}{m^h} \bigg) \bigg( \frac{n}{m} \bigg)^{(s/2)-1} {dm} \ {dn} $$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} {\Gamma\pars{n/2} \over \Gamma\pars{n/2 + h}} & = {1 \over \Gamma\pars{n/2 + h}/\Gamma\pars{n/2}} = {1 \over \pars{n/2}^{\large\overline{h}}} = {1 \over \prod_{k = 0}^{h - 1}\pars{n/2 + k}} \\[5mm] & = \bbx{1 \over \pars{n/2}\pars{n/2 + 1}\cdots\pars{n/2 + h - 1}} \end{align}

Answer 6

This is more a comment than an answer, though it can give some possibly useful asymptotics.

$\lim_{x \to \infty} \dfrac{(x+(n-1)/2)^n-x(x+1)...(x+n-1)}{(x+(n-1)/2)^{n-2}} =\dfrac{n^3-n}{24} $.

As a consequence, $x(x+1)...(x+n-1) = (x+(n-1)/2)^n-(x+(n-1)/2)^{n-2}\dfrac{n^3-n}{24} +O(x^{n-3}) $.