# How to prove that if $int_L vec{F}(vec{x})cdot dvec{x}$ exists, then integral on separate curve exists?

From wiki I found the very basic definition of line integral on vector:

For any partition $$P={vec{x}_1,…,vec{x}_{m+1}}$$ separate $$L$$, any pick of points $$vec{xi}_iin vec{x}_ivec{x}_{i+1}$$, the limit $$lim_{lambdato0}sum_{i=1}^n vec{F}(vec{xi}_i)cdotDelta vec{x}_i$$ always converge to the same value, then we can say
$$lim_{lambdato0}sum_{i=1}^n vec{F}(vec{xi}_i)cdotDelta vec{x}_i=int_L vec{F}(vec{x})cdot dvec{x}$$

Based on that I tried to prove the existence of $$int_{L_1} vec{F}(vec{x})cdot dvec{x},int_{L_2} vec{F}(vec{x})cdot dvec{x}$$ (where $$L_1cap L_2=varnothing land L_1cup L_2=L$$). However, it is a little bit hard for me and I only can prove it based on $$vec{F}(vec{x})$$ is a continues function.

So I would like to ask for a prove that do not base on that condition. Or if the condition is needed, an example that do not satisfy.

Thanks a lot!!

The following is a list of some definitions and assumptions, in order to better describe my question:

1. $$L$$ is a curve in Euclid Space $$mathbb{R}^n$$ and is rectifiable.
2. $$vec{F}(vec{x})=begin{cases}f_1(vec{x})\…\f_n(vec{x})end{cases},vec{x}in R^n$$ is a $$mathbb{R}^nto mathbb{R}^n$$ function
3. $$P={vec{x}_1,…,vec{x}_{m+1}}$$ saperate $$L$$ into $$n$$ parts
4. $$Delta vec{x}_i=vec{x}_{i+1}-vec{x}_i=begin{cases}Delta x_{i(1)}\…\Delta x_{i(n)}end{cases}$$
5. $$|Delta vec{x}_i|=sqrt{sum_{j=1}^nleft(Delta x_{i(j)}right)^2}$$ which is the chord length of $$vec{x}_ivec{x}_{i+1}$$
6. $$Delta s_i$$ donate the arc length of $$vec{x}_ivec{x}_{i+1}$$
7. $$vec{xi}_iin vec{x}_ivec{x}_{i+1}$$
8. $$lambda=max(Delta s_i)$$

Mathematics Asked by Shore on November 21, 2021

I'll leave the proof that persuaded myself here, since I'm a beginner, anyone who would willing to look into it and find any mistake are very welcome.

Part 1

Since $$int_Lvec{F}(vec{x})cdot dvec{x}$$ exists, we have the following conditions:

1. $$vec{F}(vec{x})$$ have definition on $$L$$ since $$L$$ is close set, $$vec{F}(vec{x})$$ is limited
2. $$forallepsilon>0,existsdelta,foralllambda for any partion $$P$$ and pick of points $$vec{xi}_i$$, we have $$left|sum_{i=1}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Iright|
3. $$lim_{lambdato0}sum_{i=1}^m vec{F}(vec{x}_i)cdotDelta vec{x}_i=I$$

Obviously there will include partitions that separates $$L_1,L_2$$ on $$vec{x}_kin P$$.

Therefore we have $$left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i+sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Iright|

and also $$lim_{lambdato0}sum_{i=1}^n vec{F}(vec{x}_i)cdotDelta vec{x}_i=lim_{lambdato0}left{sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i+sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_iright}=I$$

Where we define the first part of the sum is from $$L_1$$ and the second part is from $$L_2$$

Part 2

Now we assume that $$lim_{lambdato0}sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i$$ does not converge.

List the assumption as $$existsepsilon_1>0,foralldelta>0,existslambda_1 exists partition and pick of points $$exists(P^1,vec{xi}^1_i)$$

That for all $$Ainmathbb{R}$$ we have $$left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Aright|>epsilon_1$$ for $$L_1$$

Part 3

Then we use the condition from Part 1 and set $$epsilon=frac{epsilon_1}{2}$$

Therefore we get $$delta>$$ that $$foralllambda we have $$left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i+sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Iright|

According to Part 2 there is a partition and pick of points that makes $$forall ARightarrowleft|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Aright|>epsilon_1$$

We make $$L_1$$ follow the above partition and pick of points, $$L_2$$ can be any partition and pick of points as long as $$lambda

Thus we derive:

begin{aligned} because&frac{epsilon_1}{2}>left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i+sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Iright|\ =&left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-A+sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-(I-A)right|\ geq&left|left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Aright|-left|sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-(I-A)right|right|\ therefore&-frac{epsilon_1}{2}epsilon_1\ therefore&left|sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-(I-A)right|>frac{epsilon_1}{2} end{aligned}

Note that the above condition only used the specific partition and pick of points of $$L_1$$. For $$L_2$$, despite of the partition and pick of points the result holds. Also dispute the value of $$A$$, the result holds.

Part 4

Using condition 1 of Part 1, we define $$left|vec{F}(vec{x})right|leq M$$ thus we have $$left|vec{F}(vec{x})cdotDeltavec{x}right|leq mleft|vec{F}(vec{x})right|left|Deltavec{x}right|leq nMleft|Deltavec{x}right|$$

While $$n$$ is the dimension of $$vec{F}(vec{x})$$

Therefore, we have $$left|sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_iright|leqsum_{i=k}^{m}left|vec{F}(vec{xi}_i)cdotDeltavec{x}_iright| where $$l_2$$ is the length of $$L_2$$ since $$L$$ is rectifiable.

We construct a partition of $$P={vec{x}_k,...,vec{x}_{m+1}}$$ separate $$L_2$$ equally on length into $$m-k$$ parts and pick of points $$vec{xi}_i$$ are the middle points of arc $$vec{x}_ivec{x}_{i+1}$$ on length.

Then donate $$S_m=sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i,m>k$$, with the previous proof we know that S_m is limited series.

Using Bolzano-Weierstrass theorem that there must be a sub-series that converges.

Donate $$S_{m_u}$$ that have $$lim_{utoinfty}S_{m_u}=B$$

This means that $$forallepsilon>0,exists U,forall u>URightarrowleft|sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Bright|

Part 5

From Part 4 we set $$epsilon=frac{epsilon_1}{2}$$ then we exists $$U_1$$

we set $$u=maxleft(U_1,left[frac{l}{delta}right]+1right)$$ then obviously we have $$left|sum_{i=k}^{n}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Bright|

From Part 3 we know that for any partition, pick of points, and $$A$$,

as long as $$lambda=ldiv uleq ldiv left(left[frac{l}{delta}right]+1right), for $$L_2$$ we have $$left|sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-(I-A)right|>frac{epsilon_1}{2}$$

We pick $$A=I-B$$ then have $$lambda, for $$L_2$$ we have $$left|sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Bright|>frac{epsilon_1}{2}$$

Thus $$lim_{lambdato0}sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i,lim_{lambdato0}sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i$$ both converge despite of partition and pick of points.

It is also easy to prove that they would converge into the same value by for any partition and pick of points $$lim_{lambdato0}left{sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i+sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_iright}$$ always converge to $$I$$.

Thus it persuaded me that the existence are proved.

Answered by Shore on November 21, 2021

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