I'm reading The Variational Principles of Mechanics by Cornelius Lanczos; here is the concerned excerpt:
A second landmark is the geometry of Riemann, which grew out of the ingenious investigations of Gauss concerning the intrinsic geometry of the curved surfaces. Riemann's geometry is based on one single differential quantity called the "line-element" $\bar{ds}^1\;.$ .... Let us consider for example the infinitesimal distance $\bar{ds}$ between two points $P$ and $P^1$ with the coordinates $x,y,z$ and $x+dx,y+dy,z+dz$. According to the Pythagorean theorem we have $$\bar{ds}^2= dx^2 + dy^2 + dz^2\,.$$ [...]
At the footnote
$^1$ Here we adopt the custom of denoting non-integrable differentials, which cannot be considered as the infinitesimal change of something, by putting a dash above the symbol.Thus $dq_k$ or $dt$ means the "$d$ of $q_k$" or the " $d$ of $t$", while $\bar{ds}$, for example, has to conceived as an infinitesimal expression and not as the "$d$ of $s$".
He further adds:
if $\bar{ds}$ were integrable, it would be impossible to find the shortest distance between two points because length of any curve between these two points - depending on the initial and end positions only - would be the same.
I couldn't get the point why "line-element" is non-integrable; he did add an explanation at the footnote, but I couldn't comprehend his reasoning: so according to him, a differential displacement between two points is non-integrable, is it so?
How could we integrate then using a differential displacement like $\mathrm dx$ in one-dimension or $\mathrm d\mathbf r$ in three dimensions or so?
Sorry, if I sound silly; but I'm a bit confused.
Can anyone please clarify to me why "line-element" is non-integrable?
"Non-integrable" is not the word I would use here, because it has other, non-related meanings. I think that is the source of the confusion.
What he is saying is that the result of integrating $ds$ along a curve is path-dependent, not just a function of the endpoints. If $f$ is any smooth function defined on a simply-connected region (in particular the $x$ coordinate), integrating $df$ on any curve from $p$ to $q$ in this region will give you $f(q) - f(p)$, so that is not path-dependent. So $dx$ is "integrable" in his sense, but $ds$ is not.