[VA] Short & Sweet Math Challenge #25 "San Valentin's Special: Weird Math"

[robve]

One more observation that may be of interest. Based on the visual clue that the integrant behaves as (x-1)/(phi-1), but is not the same, it is prudent to numerically verify that the function (x-1)/(phi-1) equals or closely approximates the integrand

$$ \frac{\Gamma\ln(\phi^2-x)}{\Gamma\ln x+\Gamma\ln(\phi^2-x)} $$

This can be done by listing the results of the two functions side-by-side for x between 1 and phi in increments of say 0.1 to compute the absolute or relative error of the difference between the two. We can also take the absolute difference of the two functions as the integrand and numerically evaluate the integral:

$$ \int_1^\phi \left|\frac{\Gamma\ln(\phi^2-x)}{\Gamma\ln x+\Gamma\ln(\phi^2-x)}-\frac{x-1}{\phi-1}\right| dx \approx 5.66308603939\times10^{-3}$$

The value is the result on the HP PRIME. My PC-1350 Romberg integrator gives the same 5.66308E-3. Therefore, the total area of the difference between the two functions is small, but not zero. Based on the numerical results, it does not appear convincing to state that

$$ \frac{\Gamma\ln(\phi^2-x)}{\Gamma\ln x+\Gamma\ln(\phi^2-x)} \stackrel{?}{=} \frac{x-1}{\phi-1} $$

They are approximate, but not equal. Function (x-1)/(phi-1) under estimates the integrand on the interval 1<=x<=phi/2 and overestimates the integrand on interval phi/2<=x<=phi by about 0.014 symmetrically:

   

Because of this symmetry, one could argue that

$$ \int_1^\phi \frac{\Gamma\ln(\phi^2-x)}{\Gamma\ln x+\Gamma\ln(\phi^2-x)} dx = \frac{\phi-1}{2} \approx 0.309016994375 $$

Furthermore, the integral of the difference is near zero

$$ \int_1^\phi \frac{\Gamma\ln(\phi^2-x)}{\Gamma\ln x+\Gamma\ln(\phi^2-x)}-\frac{x-1}{\phi-1} dx \approx -1.13498070108\times10^{-13} $$

Suggesting that the over and under errors of the (x-1)/(phi-1) curve nicely cancel out over 1<=x<=phi.

- Rob