Units inside a logarithm I have troubles understanding a seemingly simple integral in a physical context. Take a look at $\int_{V_1}^{V_2} \frac{\mathrm{d}V}{V}$ which appears in isothermal expansions (V being the volume of a gas). Now of course the result is $\ln{V_2}-\ln{V_1}$ or using log laws $\ln{\frac{V_2}{V_1}}$. 
The first expression would require you to evaluate the logarithm of a unit of volume which to my unterstanding is impossible. The second expression makes sense, though. How can you account for this apparent discrepancy?
 A: 
How can you account for this apparent discrepancy?

There is no discrepancy in the value of the difference of the logs 
$$
\log(V_1)-\log(V_2)
$$
and the value of the log of the ratio
$$
\log{V_1/V_2}\;.
$$
And neither of these depends on the choice of units. This is because, for both the difference and the ratio, the units cancel. For example, suppose I represent the "units" as a number "U" such that 
$$
V=\alpha U\;,
$$
where $\alpha$ is unitless.
Then
$$
\log(V_1)-\log(V_2)=\log(\alpha_1 U)-\log(\alpha_2 U)=\log(\alpha_1)+\log(U)-\log(\alpha_2)-\log(U)
$$
$$
=\log(\alpha_1)-\log(\alpha_2)\;.
$$
And, similarly,
$$
\log(V_1/V_2)=\log(\frac{\alpha_1 U}{\alpha_2 U})=\log(\frac{\alpha_1}{\alpha_2})\;.
$$
Not much else to say but that... Clearly, trying to interpret $\log(V_1)$ on its own doesn't make sense since changing the unit changes the numerical value... but the interpretation and sensibleness of the difference of the logs certainly makes sense and is independent of the choice of units.
A: If you introduce any unit $v$ in terms of which you express the numerical values of $V_1,V_2$, then you see
$\ln\left(\frac{V_2}{V_1}\right)=\ln\left(\frac{V_2/v}{V_1/v}\right)=\ln\left(V_2/v\right)-\ln\left(V_1/v\right)$
So if $V_1$ and $V_2$ were the integral bounds of a single integral sign, it's only important that make sure you express the numerical value of $V_1,V_2$, in the difference of the two log-expressions, as multiples of the same arbitrary scale. 
Writing $\ln\left(V_2\right)-\ln\left(V_1\right)$ implies you fixed a unit for them and it wasn't explicitly emphasised. People don't get in trouble here because the result is really independent of the unit, as it being equal to $\ln\left(\frac{V_2}{V_1}\right)$ demonstrates.
A: In mathematics when we put a number x in ln x, we assume x belong to a field, so x must be a dimensionless number.
So it is possible to put a pure number of x in ln x.
However, many people seem not to know the following expression of lnx as an infinite series expansion.
In fact, lnx=2{(x-1)/(x+1)+(1/3)[(x-1)(x-1)(x-1)]/[(x+1)(x+1)(x+1)]+(1/5)[(x-1)(x-1)(x-1)(x-1)(x-1)]/[(x+1)(x+1)(x+1)(x+1)(x+1)]+,,,,,,,} .
So now it is clear that we cannot put x in lnx if x has a dimension.
I have published these issues in the three papers:


*

*Mayumi, K. and Giampietro, M. 2010. “Dimensions and Logarithmic Function: A Short Critical Analysis” , Ecological Economics,Vol.69, pp. 1604-1609;

*Mayumi, K., and Giampietro, M. 2012. “Response to “Dimensions and logarithmic function in economics: A comment””, Ecological Economics Vol. 75, pp. 12-14;

*Mayumi, K., Giampietro, M., And Ramos-Martin, J. 2012. “Reconsideration of Dimensions and Curve Fitting Practice in View of Georgescu-Roegen’s Epistemology in Economics”, 
Romanian Journal of Economic Forecasting, Vol.15, No.4, pp.17-35.

