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I am working on evaluating the following equation:

$\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)$

If I'm understanding correctly, the above is an increasing function which can be demonstrated by the following argument using the digamma function $\frac{\Gamma'}{\Gamma}(x) = \int_0^\infty(\frac{e^{-t}}{t} - \frac{e^{-xt}}{1-e^{-t}})$:

$\frac{\Gamma'}{\Gamma}(\frac{1}{2}x) - \frac{\Gamma'}{\Gamma'}(\frac{1}{3}x) = \int_0^\infty\frac{1}{1-e^{-t}}(e^{-\frac{1}{3}xt} - e^{-\frac{1}{2}xt})dt > 0 (x > 1)$

Please let me know if this reasoning is incorrect or if you have any corrections.

Thanks very much!

-Larry

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1  
The series for $\psi$ might be more expedient to use... –  J. M. Apr 4 '13 at 6:33
    
Thanks very much! I'll check it out. –  Larry Freeman Apr 4 '13 at 6:51
    
I reviewed the series for $\psi$. Thanks. Is this correct: Using $\psi(x) = -\gamma+\sum_{k=0}^\infty(\frac{1}{k+1}-\frac{1}{k+z})$ gets me to the derivative of $\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)$ to be: $\sum_{k=0}^{\infty}(\frac{1}{k+\frac{x}{3}} - \frac{1}{k+\frac{x}{2}})$ which shows an increasing function. –  Larry Freeman Apr 4 '13 at 7:04
1  
Well, the terms of your resulting series are all positive for positive argument, so... –  J. M. Apr 4 '13 at 11:05
    
Great. Thanks for the help. –  Larry Freeman Apr 4 '13 at 14:21

1 Answer 1

up vote 1 down vote accepted

This answer is provided with help from J.M.

$\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)$ is an increasing function. This can be shown using this series for $\psi$:

The function is increasing if we can show: $\frac{d}{dx}(\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)) > 0$

We can show this using the digamma function $\psi(x)$:

$$\frac{d}{dx}(\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)) = \frac{\psi(\frac{1}{2}x)}{2} - \frac{\psi(\frac{1}{3}x)}{3}$$

$$\frac{\psi(\frac{1}{2}x)}{2} - \frac{\psi(\frac{1}{3}x)}{3} = -\gamma + \sum_{k=0}^\infty(\frac{1}{k+1} - \frac{1}{k + {\frac{1}{2}}}) + \gamma - \sum_{k=0}^\infty(\frac{1}{k+1} - \frac{1}{k+\frac{1}{3}})$$

$$= \sum_{k=0}^\infty(\frac{1}{k+\frac{1}{3}} - \frac{1}{k+\frac{1}{2}})$$

Since for all $k\ge 0$: $k + \frac{1}{3} < k + \frac{1}{2}$, it follows that for all $k\ge0$: $\frac{1}{k+\frac{1}{3}} > \frac{1}{k+\frac{1}{2}}$ and therefore: $$\sum_{k=0}^\infty(\frac{1}{k+\frac{1}{3}} - \frac{1}{k+\frac{1}{2}}) > 0.$$

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