The rate of increase of the Gamma Function over real numbers

Question

If

$$ x_1 > x_2 > 0$$ and $$\Delta{x}>0$$

does it follow that: $$\ln\Gamma(x_1 + \Delta{x}) - \ln\Gamma(x_1) \ge \ln\Gamma(x_2 + \Delta{x}) - \ln\Gamma(x_2)$$

Would it be enough to show that the Gamma Function is a strictly increasing function for $x > 0$?

Thanks,

-Larry

Answer 1

$$f(x)=\ln{\Gamma{x}}$$ $$f'(x_1) \approx \frac{\ln{\Gamma{(x_1+dx_1)}}-\ln{\Gamma{x_1}}}{dx_1}$$ $$f''(x_1) \approx \frac{f'{(x_1+dx_2)}-f'{x_1}}{dx_2}$$ Your question is essentially about when $dx_2=x_2-x_1$.

If $f''(x)>0$ for all x in your range, the statement is correct.

Unfortunately $f''(x)=\psi^{(1)}(x)$ and i don't know if that is ever less than 0. http://www.wolframalpha.com/input/?i=polygamma%281%2C+x%29&lk=1&a=ClashPrefs_*Math-