I was looking at the Product Integral. I'm defining it here with the Gamma symbol like this:
$\Gamma _a^bf(x)dx=e^{\int_a^b \log (f(x)) \, dx}$
It occured to me that the product integral of $f(x) = x$ in the range $[\frac {1} {a}, a]$ "should" be 1, because every element in that range has an inverse in that range. But Mathematica argues otherwise:
E^Integrate[Log[x], {x, 1/2, 2}]
$e^{\frac{1}{2} (\log (32)-3)}$
Which is about 1.2622. The general solution for a > 1 is:
$e^{\frac{1}{a}-a+\left(\frac{1}{a}+a\right) \log(a)}$
This is definitely not 1, so my assumption is probably naive. The first explanation that comes to mind (besides the integral not being $0$) is the size difference. Although every element in the range $[\frac {1}{a}, a]$ does have an inverse in that set, $[\frac {1}{a},1]$ is a smaller range than $[1,a]$. So perhaps in some way this imbalance comes out in the product integral.
I have two questions then:
1) Is the product integral telling us anything here? If $\Gamma _{\frac{1}{a}}^{a}x\,dx \approx 1.2622$, does that say anything about the range $[\frac {1}{a},a]$?
2) Is there an obvious product integral which would yield 1 for the function $f(x)=x$ in the range $[\frac {1}{a},a]$?