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I got help (Thanks to Mako) to show this result:

\begin{equation} \begin{gathered} \frac{\Gamma(\frac{1}{2}+\frac{1}{q-1})}{\Gamma(\frac{1}{q-1})}\frac{\Gamma(\frac{q}{q-1})}{ \Gamma(\frac{q}{q-1}+\frac{1}{2})} = \frac{2}{(q+1)} \end{gathered} \end{equation}

The Gammas can be rewritten as:

$$\frac{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}\right)\Gamma\left(\frac{1}{q-1}+1\right)}{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}+1\right)\Gamma\left(\frac{1}{q-1}\right)}$$

Using $\Gamma(x+1) = x\Gamma(x)$ We can simplify the expressions above by considering $$x = \frac{1}{2}+\frac{1}{q-1}$$ $$y = \frac{1}{q-1}$$

So that we have the following,

$$\frac{\Gamma(x)\Gamma(y+1)}{\Gamma(x+1)\Gamma(y)}$$

Now using the identity $\Gamma(x+1) = x\Gamma(x)$, we can write out

$$ = \frac{y\Gamma(x)\Gamma(y)}{x\Gamma(x)\Gamma(y)}$$

and now simplifying by cancelling $\Gamma(x)$ and $\Gamma(y)$ we get

$$ = \frac{y}{x}$$ which when plugging expressions for $x$ and $y$ back in yield

$$ = \frac{1}{q-1}\frac{1}{\frac{1}{2}+\frac{1}{q-1}}$$ $$ = \frac{2(q-1)}{(q-1)(q+1)}$$ $$ = \frac{2}{q+1}$$

However, I´m trying to obtain the following results unsuccessfully:

\begin{equation} \begin{gathered} \frac{\Gamma(\frac{1}{2}+\frac{1}{q-1})}{\Gamma(\frac{1}{q-1})}\frac{\Gamma(\frac{q}{q-1})}{ \Gamma(\frac{q}{q-1}+\frac{3}{2})} = \frac{4(q-1)}{(q+1)(3q-1)} \end{gathered} \end{equation} And: \begin{equation} \begin{gathered} \frac{\Gamma(\frac{1}{2}+\frac{1}{q-1})}{\Gamma(\frac{1}{q-1})}\frac{\Gamma(\frac{q}{q-1})}{ \Gamma(\frac{q}{q-1}+\frac{5}{2})} = \frac{8(q-1^2)}{(q+1)(3q-1)(5q-3)} \end{gathered} \end{equation}

I´ve tried to rewrite the $\frac{3}{2}$ as:

$$\frac{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}\right)\Gamma\left(\frac{1}{q-1}+1\right)}{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}+1+1\right)\Gamma\left(\frac{1}{q-1}\right)}$$

And the $\frac{5}{2}$ as:

$$\frac{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}\right)\Gamma\left(\frac{1}{q-1}+1\right)}{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}+1+1+1\right)\Gamma\left(\frac{1}{q-1}\right)}$$

Repeating the same steps above for the first Gamma Function , I can use $\Gamma(x+1) = x\Gamma(x)$ and achieve $\frac{y}{x^2}$ and $\frac{y}{x^3}$ respectivelly for these Gamma Functions with $\frac{3}{2}$ and $\frac{5}{2}$ arguments. However this path is not leading me to the desired solutions which were obtainned by MAPLE software.

JC94
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1 Answers1

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I'm not entirely sure what mistake you're making. Both reductions should indeed follow immediately from the identity $\Gamma(x+1)=x\,\Gamma(x)$.

For the first one, we have

$$\begin{align*} \frac{\color{red}{\Gamma\left(\frac1{q-1} + \frac12\right)} \color{blue}{\Gamma\left(\frac q{q-1}\right)}}{\color{blue}{\Gamma\left(\frac1{q-1}\right)} \color{red}{\Gamma\left(\frac q{q-1} + \frac32\right)}} &= \frac{\color{red}{\Gamma\left(\frac1{q-1} + \frac12\right)} \color{blue}{\Gamma\left(\frac1{q-1}+1\right)}}{\color{blue}{\Gamma\left(\frac1{q-1}\right)} \color{red}{\Gamma\left(\frac 1{q-1} + \frac52\right)}} \\ &= \frac{\color{red}{\Gamma\left(\frac1{q-1} + \frac12\right)} \color{blue}{\frac1{q-1} \Gamma\left(\frac1{q-1}\right)}}{\color{blue}{\Gamma\left(\frac1{q-1}\right)} \color{red}{\left(\frac 1{q-1} + \frac32\right) \left(\frac 1{q-1} + \frac12\right) \Gamma\left(\frac 1{q-1} + \frac12\right)}} \\ &= \frac{\frac1{q-1}}{\left(\frac 1{q-1} + \frac32\right) \left(\frac 1{q-1} + \frac12\right)} = \frac{4(q-1)}{(q+1)(3q-1)} \end{align*}$$

and the second is nearly identical except for an additional factor of $\dfrac1{q-1}+\dfrac52$ in the denominator.

user170231
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  • Thanks man, I found my mistake. It does not leads to x^2 and x^3 because the arguments are different. – JC94 Jan 10 '24 at 20:00