Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!


share|improve this question
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
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
show 2 more comments

1 Answer

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

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.