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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 $[\frac {1}{a},a]$?

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Why "should" it be $1$, according to the definition? –  Siminore Aug 12 '12 at 8:44
The measure $dx$ is a Haar measure corresponding to the addition. That is, this measure weighs in such a way that it is compatible with the addition. Thus in order for your idea to be realized, it seems more natural to choose a Haar measure corresponding to the multiplication, namely $dx/x$. Thus we may define $$ \Gamma_{a}^{b} f(x) \; dx = \exp\left( \int_{a}^{b} \log f(x) \; \frac{dx}{x} \right). $$ Then we have correctly $\Gamma_{1/a}^{a} x \; dx = 1$. –  sos440 Aug 12 '12 at 9:44
@sos440 But you are defining a different product integral, according to Wikipedia. In the standard definition, the measure is not weighted. –  Siminore Aug 12 '12 at 9:50
@Siminore: Yes, I agree. I just wanted to point out where the problem arises and how it can be ameliorated. But clearly this integral is not a genuine product integral. –  sos440 Aug 12 '12 at 9:56
@sos440 Thanks, that's what I was looking for. It's a lot prettier than the standard definition, in my opinion. :) –  amr Aug 12 '12 at 19:34
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1 Answer

up vote 1 down vote accepted

The fact that the sizes of the intervals $[1/a,1]$ and $[1,a]$ are different indeed points to the right direction. To see this, let us try to show that the integral is zero because the contributions of $f(x)=\log(x)$ and $f(1/x)=\log(1/x)=-\log(x)$ would compensate each other (they do not but let us see what happens when we try to show they do).

A natural idea is to consider a subdivision $(x_k)_{0\leqslant k\leqslant n}$ of the interval $[1,a]$, since the integral of $\log f$ on $[1,a]$ is roughly the Riemann sum $$ S=\sum_{k=1}^n(x_k-x_{k-1})\log f(x_k). $$ The symmetry of $\log f$ suggests to consider the subdivision $(z_k)_{0\leqslant k\leqslant n}$ made with the inverses of the points of the first subdivision, more precisely, $z_k=x_{n-k}^{-1}$ for every $0\leqslant k\leqslant n$. The integral of $\log f$ on $[1/a,1]$ is roughly the Riemann sum $$ S'=\sum_{k=1}^n(z_k-z_{k-1})\log f(z_{k-1})=\sum_{k=1}^n(x_{n-k}^{-1}-x_{n-k+1}^{-1})\log f(x_{n-k+1}^{-1}), $$ that is $$ S'=-\sum_{k=1}^n\frac{x_k-x_{k-1}}{x_{k}x_{k-1}}\log f(x_{k}). $$ Since the coefficient of $\log f(x_k)$ in $S'$ is not the length of the interval $x_k-x_{k-1}$ as in $S$, but a biased version of this length, the argument does not lead a priori to $S'=-S$.

To sum up, the function $\log f$ is indeed symmetric with respect to the transformation $x\to1/x$ but the lengthes of the subintervals of $[1,a]$ and $[1/a,1]$ are not (and in particular, but this is only the tip of the iceberg, the lengthes of $[1,a]$ and $[1/a,1]$ themselves do not coincide). Still in other words, Area = Width $\times$ Height and Height does not change but Width changes, hence Area changes as well.

To save the day, the transformation $x\to1/x$ should leave Width unchanged. As explained in comments, the only way to do this is that the length is measured locally at $x$ by $\dfrac{\mathrm dx}x$. The biased Riemann sums become $$ \tilde S=\sum_{k=1}^n\frac{x_k-x_{k-1}}{x_k}\log f(x_k),\qquad \tilde S'=\sum_{k=1}^n\frac{z_k-z_{k-1}}{z_{k}}\log f(z_{k-1})=-\tilde S. $$

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