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Suppose f is a continuous function on an open set Ω which has an anti-derivative on Ω. if γ, γ' are two paths in Ω with the same beginning point and the same end point, are below statement true? $$\int_{\gamma} f(z) \, dz =\int_{\gamma'} f(z) \, dz $$

My approach: I think it should be the same, since it has two curves ends in the same points.

Also by the fundamental theorem of the calculus, I know that using F as derivative of f. $$\int_{\gamma} f(z) \, dz = F(\gamma(b)) - F(\gamma(a))$$

$$\int_{\gamma'} f(z) \, dz = F(\gamma'(b)) - F(\gamma'(a))$$

But I am not sure how to approach this problem to show that these are equal. I think I need to use the fact $$\int_{\gamma} f(z) \, dz = 0$$ But not sure how to prove this.

mathrock1453
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  • @saulspatz: $f(z) = 1/z$ doesn't have antiderivative on your $\Omega$. – xyzzyz Sep 30 '19 at 23:15
  • Yes, of course. Don't know where my mind went. – saulspatz Sep 30 '19 at 23:16
  • @saulspatz, can you help me with this how having the same end points results γ(a)=γ′(a) and γ(b)=γ′(b). I just cannot seem to find the proof for this. – mathrock1453 Sep 30 '19 at 23:21
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    What are beginning and end points according to your definitions? If $\gamma:[a,b]\to\mathbb{C}$ is a curve then as I know its beginning point is $\gamma(a)$ while the end point is $\gamma(b)$. (simply by definition) – Mark Sep 30 '19 at 23:24
  • @Mark, my beginning point is "a" and ending point is "b". Other than that, I don't have any other information. – mathrock1453 Sep 30 '19 at 23:25
  • No, $a$ and $b$ are not end points of $\gamma$, they are just end points of the parametrization interval (i.e. usually $a = 0$ and $b = 1$). The end points are $\gamma(a)$ and $\gamma(b)$. If $\gamma(a)$ is not equal to $\gamma'(a)$, then the curves $\gamma, \gamma'$ have different end points, and there is no reason to expect that the integrals will be equal. – xyzzyz Sep 30 '19 at 23:29

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If $\gamma$ and $\gamma'$ have the same end points, that is, $\gamma(a) = \gamma'(a)$ and $\gamma(b) = \gamma'(b)$, then obviously $F(\gamma(a)) = F(\gamma'(a))$,$F(\gamma(b)) = F(\gamma'(b))$, so $F(\gamma(b)) - F(\gamma(a)) = F(\gamma'(b)) - F(\gamma'(a))$.

xyzzyz
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  • want to know how having the same end points result γ(a)=γ′(a) and γ(b)=γ′(b) – mathrock1453 Sep 30 '19 at 23:19
  • This is literally what "same end points" means. What else would "same end points" mean? – xyzzyz Sep 30 '19 at 23:28
  • That is what I wrote on the exam, but got my mark off for not showing the proof for this. – mathrock1453 Sep 30 '19 at 23:32
  • There's no proof of this, this is the definition of what "same end point" means. The end points of $\gamma: [a, b] \to \mathbb{C}$ are $\gamma(a)$ and $\gamma(b)$. There isn't anything else to it. – xyzzyz Oct 01 '19 at 00:06