Confusion in Maxwell's derivation of Ampere's Force Law - Part III

Note: This is not a duplicate of my previous questions with same title asked last year.

I am reading Maxwell's a treatise on electricity and magnetism, Volume 2, page 156 about "Ampere's Force Law".

Edit:

I apologise for not being able to write my whole analysis of the Treatise. It is because it is too long. However I would write down the important equations within the question

$$\begin{matrix} P'Q'=ds',\\ PQ=ds\\ r=\xi\hat{i}+\eta\hat{j}+\zeta\hat{k}\\=(x-x')\hat{i}+(y-y')\hat{j}+(z-z')\hat{k}\\ \\ \end{matrix}$$

\begin{align} &\dfrac{d\xi}{ds}=\dfrac{dx}{ds}=l,\\ & \dfrac{d\eta}{ds}=\dfrac{dy}{ds}=m,\\ & \dfrac{d\zeta}{ds}=\dfrac{dz}{ds}=n,\\ & -\dfrac{d\xi}{ds'}=\dfrac{dx'}{ds'}=l',\\ & -\dfrac{d\eta}{ds'}=\dfrac{dy'}{ds'}=m',\\ & -\dfrac{d\zeta}{ds'}=\dfrac{dz'}{ds'}=n',\\ \end{align}

$P, B$ and $C$ are unknown functions of $r$. $Q=-\int_{0}^{r} C dr$ is also a function of $r$.

By making use of the definition of $l'$ equation (19) in book becomes:

\begin{align} \dfrac{dX}{ds} &= \biggl[P \xi^{2}-Q \biggl]_{0}^{s^{\prime}}-\int_{0}^{s^{\prime}} (2Pr-B-C)\frac{l' \xi}{r}ds'\\ &=\biggl[P \xi^{2}-Q \biggl]_{0}^{s^{\prime}}+\int_{0}^{s^{\prime}} (2Pr-B-C)\frac{ \xi}{r}d\xi'\\ \end{align}

Now since $P,\xi$ and $Q$ are functions of $r$, they disappear after closed integration around $s'$ and the first term is zero. However if we look carefully, in the second term, the integrand and the variable are all functions of $r$. Therefore this term should also disappear.

But its written$-$below equation (19)$-$that "the second term will not in general disappear". Why is this so even after the second term is a function of $r$?

• Maxwell's original notation has not aged well, and digging through the book to connect his notation to the modern usage is maybe a job that you should do, and leave the question to just the physics. – Emilio Pisanty Feb 23 '17 at 7:53
• I have already deciphered the whole chapter. This is the only place where I am confused. Also I can't find any modern textbooks on this. Maxwell's treatise is the only place where I have found this derivation. – N.G.Tyson Feb 23 '17 at 8:05
• If you have already deciphered the whole chapter, why don't you post the deciphered version and phrase the question in the modern notation? If you're having trouble with the physics, we can help, but if you're expecting people to reconstruct all of the notation from scratch (particularly if that means redoing work that you've already done) then that's rather different, and not particularly kind to the people you're asking for help. – Emilio Pisanty Feb 23 '17 at 8:07
• I thought experts can easily understand and decipher the treatise. I have not posted my deciphered version because it is too long. One page in Maxwell's treatise makes about 5 pages on mine. If nobody can understand the treatise easily I think I have to write it down here. – N.G.Tyson Feb 23 '17 at 8:17
• Experts can understand the treatise given enough time, but it is a laborious process since the notation has changed significantly in the intervening 150 years. It is not an impossible proposition but it is a much more labour-intensive task than if you just ask about the physics. – Emilio Pisanty Feb 23 '17 at 14:30

It is true that the integrand is function of $r$, but the variable is an infinitesimal length on the $x$ axis. So it need not necessarily be a function of $r$. Also, the path of integration is not along $r$. It is along the curve $s'$. Therefore the variable is $s'$ and not $r$. Therefore $\dfrac{dX}{ds}$ remains a function of $s'$.
In your $\dfrac{dX}{ds}$ equation, integration w.r.t $d\xi'$ means integration along projection of path of circuit $s'$ on $x$-axis and not along projection of $\vec{r}$ on $x$ axis.
Therefore it is better to write integration w.r.t $dx'$ than $d\xi'$