# true for every closed path/line/contour integral

$5^{* *}$. Given these two statements:

i) $\forall \gamma \in \mathbb{C}\backslash \{0,1\}$ where $\gamma$ is a closed path, it holds that: $\displaystyle{\int_{\gamma} \frac{d\eta}{\eta(\eta-1)}=0}$

ii) $\forall \gamma \in \mathbb{C}\backslash[0,1]$, where $\gamma$ is a closed path, it holds that: $\displaystyle{\int_{\gamma} \frac{d\eta}{\eta(\eta-1)}=0}$

which one is true ?

My thoughts:

In i) also points between 0 and 1 are allowed, only 0 and 1 aren't. In ii) the piece with between 0 and 1 is not allowed. I guess that the first statement is true, but the second isn't, because there are more points not allowed in the second one.

How does one prove i) and contradict ii)? Does anybody see a way? Please do tell!

Proof attempt:

i)

Let $\gamma = 0.5e^{it}+1 , t\in [0,2\pi]$, this is a closed path with the point 1 in it but not 0. From the Cauchy Goursat Theorem and Cauchy Integral formula it follows that with:

$\eta_{0}=0 , \eta_{1}=1$ only $\eta_{1}$ lies within this disc, so one has:

$\displaystyle{\int_{\gamma} \frac{\frac{1}{(\eta)}}{\eta-1}}= 2\pi i (\frac{1}{\eta|_{\eta=1}})= 2\pi i$

So i) can not be true.

ii)

Only both singularities can be within the circle or outside. If they are outside, then the Integral is 0. If they are both inside, then with the Cauchy Goursat theorem and the Cauchy integral formula: we can write them as the sums in the form:

$\displaystyle{\int_{\gamma}\frac{1}{\eta(\eta-1)}d\eta = \int_{\gamma} \frac{\frac{1}{\eta-1}}{\eta}}d\eta + \int_{\gamma}\frac{\frac{1}{\eta}}{\eta-1}d\eta = -2\pi i + 2\pi i = 0$

Is this proof correct so far, I believe one also needs to show that only these two cases are possible and not any other. How to do that?

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What happens if you have a closed path with $1$ inside and $0$ outside? Can you have that in (i)? In (ii)? –  Arturo Magidin Nov 15 '11 at 5:20
In ii) no, i) yes. What is the meaning of this?? –  VVV Nov 15 '11 at 5:23
Now, can you figure out what the integral will be if one of the poles of the function is inside the path and the other one is outside, e.g., a loop like I suggested? –  Arturo Magidin Nov 15 '11 at 5:25
it will be : $2\pi i$ ? How to prove this? –  VVV Nov 15 '11 at 5:33
So, it's not zero, then. That means that (i) cannot be true. So if one of them is true, then it's (ii). How to prove it? Show what any such $\gamma$ will either have both poles outside or both poles inside the closed path. Go from there. –  Arturo Magidin Nov 15 '11 at 5:34

Let me note first that I think your reasoning is exactly backwards. Because in statement (i) you are excluding fewer points than in (ii), that means that there are more paths $\gamma$ that satisfy the condition in (i) than those that satisfy the conditions in (ii): any path that satisfies (ii) will also satisfy (i), but there are $\gamma$s that satisfy (i) but not (ii).

That means that (i) is a stronger statement than (ii), not a weaker one (as you seem to imply). In fact, as I mentioned, if (i) is true, then (ii) has to be true, because any path that satisfies the conditions in (ii) will also satisfy the conditions in (i). That means that it is logically impossible for (i) to be true and (ii) to be false. So either they are both false, they are both true, or (i) is false and (ii) and is true.

So the key is to take a single path that satisfies the condition in (i) but not the condition in (ii) and see what happens. That was my suggestion in comments: take a look that has, say, 1 in the interior and 0 outside of it, and see what the integral is. Cauchy's Residue Theorem will come into play.

Now, in (ii), every $\gamma$ will either have both $0$ and $1$ inside, or will have neither $0$ nor $1$ inside; if it has neither inside, then by the Residue Theorem the integral is $0$ and you are done. So you just need to figure out what happens if both $0$ and $1$ are inside the closed path.

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The residue theorem was not proven…… Is there a way to show without using it? –  VVV Nov 15 '11 at 5:55
@VVV: I don't know what tools you have on hand, so I don't know how you would go about proving it; in any case, note that your reasoning was definitely backwards: in this case, "excluding more points" means fewer paths, means weaker conclusion, not stronger one. –  Arturo Magidin Nov 15 '11 at 5:57
Is the reasoning correct now?? –  VVV Nov 15 '11 at 12:10
@VVV: Modulo explaining why in (ii) the only two possibilities are that both $0$ and $1$ are inside or both outside the loop, it looks correct to me. I do wish you would not over-use the boldface, though... –  Arturo Magidin Nov 15 '11 at 18:13
Arturo, Thank You. –  VVV Nov 15 '11 at 19:18