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what will be the value of this integral?

$$\int_{|z|=1}\frac{1}{z(z-1)}$$

Well, I dont know what how to tackle when there is a singularity (here is a pole at $1$) at the boundary of domain. Thank you.

$Res(0)=-1$, $Res(1)=1$

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The integral is not properly defined when there is a singularity on the contour. –  Ragib Zaman Dec 16 '12 at 12:43
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up vote 2 down vote accepted

The integral $$\oint_{|z|=1}\frac{1}{z(z-1)}$$ is not defined as $f(z)=\frac{1}{z(z-1)}$ is not defined at $z=1$.

If we consider the integral $$\oint_{|z|=r}\frac{1}{z(z-1)}$$

$0<r\neq 1$ then, $$\oint_{|z|=r}\frac{1}{z(z-1)}=-2\pi i$$ if $r<1$ and $$\oint_{|z|=r}\frac{1}{z(z-1)}=(-1+1)2\pi i=0$$ if $r>1$.

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Are you sure you are not missing a $2\pi i$ or something? –  GEdgar Dec 16 '12 at 17:58
    
@GEdgar Oh yes I am... –  Nameless Dec 16 '12 at 18:22
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There is a singularity on the path of integration at $z=1$ that causes difficulty.

Let us parametrize the path of integration as $z=e^{2it}$ where $0\le t\le\pi$. Then your integral becomes $$ \begin{align} \int_{|z|=1}\frac{\mathrm{d}z}{z(z-1)} &=\int_0^{\pi}\frac{2i\,e^{2it}\,\mathrm{d}t}{e^{2it}(e^{2it}-1)}\\ &=\int_0^{\pi}\frac{2i\,e^{-it}\,\mathrm{d}t}{e^{it}-e^{-it}}\\ &=\int_0^{\pi}\frac{(\cos(t)-i\sin(t))\,\mathrm{d}t}{\sin(t)}\\ &=\int_0^{\pi}(\cot(t)-i)\,\mathrm{d}t\\ &=\int_0^{\pi}\cot(t)\,\mathrm{d}t-\pi i \end{align} $$ $\cot(t)$ blowsup to $+\infty$ near $t=0$ and $-\infty$ near $t=\pi$, just as $\frac1{z(z-1)}$ does near $z=1$.

Cauchy Principal Value

One standard method used to deal with this situation is called the Cauchy Principal Value. To get the Cauchy Principal Value for the integral of $\cot(t)$ above we take out a piece of the path on either side of the singularity of size $\epsilon$ and let $\epsilon\to0^+$: $$ \begin{align} \mathrm{PV}\int_0^\pi\cot(t)\,\mathrm{d}t &=\lim_{\epsilon\to0^+}\int_\epsilon^{\pi-\epsilon}\cot(t)\,\mathrm{d}t\\ &=\lim_{\epsilon\to0^+}\int_\epsilon^{\pi-\epsilon}\frac12(\cot(t)-\cot(\pi-t))\,\mathrm{d}t\\ &=\lim_{\epsilon\to0^+}\left(\frac12\int_\epsilon^{\pi-\epsilon}\cot(t)\,\mathrm{d}t -\frac12\int_\epsilon^{\pi-\epsilon}\cot(\pi-t)\,\mathrm{d}t\right)\\ &=\lim_{\epsilon\to0^+}\left(\frac12\int_\epsilon^{\pi-\epsilon}\cot(t)\,\mathrm{d}t -\frac12\int_\epsilon^{\pi-\epsilon}\cot(s)\,\mathrm{d}s\right)\\ &=\lim_{\epsilon\to0^+}0\\ &=0 \end{align} $$ where we have taken advantage of the fact that $\cot(\pi-t)=-\cot(t)$ and used the substitution $s=\pi-t$.

Therefore, in this sense $$ \mathrm{PV}\int_{|z|=1}\frac{\mathrm{d}z}{z(z-1)}=-\pi i $$ Contour Integration

Another way to compute the Cauchy Principal Value is using Contour Integration.

To avoid the singularity, we can make a semicircular arc on either side of the singularity, as below in red and green.

$\hspace{4.5cm}$enter image description here

The residue of $\frac1{z(z-1)}=\frac1{z-1}-\frac1z$ at $z=0$ is $-1$ and the residue at $z=1$ is $+1$.

Since the green semicircle (of radius $\epsilon$) goes half-way counterclockwise around the singularity at $z=1$, the limit of the integral as $\epsilon\to0$ is only $\pi i$ times the residue, not $2\pi i$. That is, the integral over the green semicircle is $\pi i$.

Since the red semicircle (of radius $\epsilon$) goes half-way clockwise around the singularity at $z=1$, the limit of the integral as $\epsilon\to0$ is $-\pi i$ times the residue. That is, the integral over the red semicircle is $-\pi i$.

The contour consisting of the black circle and green arc contains both singularities, so the integral over that contour is $2\pi i$ times the sum of the residues, which is $0$. Taking out the integral over the green semicircle leaves that the integral over the black circle is $-\pi i$.

The contour consisting of the black circle and red arc contains only the singularity at $z=0$, so the integral over that contour is $2\pi i$ times this residue, which is $-2\pi i$. Taking out the integral over the red semicircle leaves that the integral over the black circle is $-\pi i$.

Thus, both contours give that the integral over the black circle is $-\pi i$. This is the same as the Cauchy Principal Value since we took out equal parts of the contour on either side of $z=1$.

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Using this lemma's corollary, we get ($\,0<\epsilon<<1\,$):

$$\gamma_\epsilon:=\left\{z\in\Bbb C\;\;;\;\;\,\,,\,z=1+\epsilon e^{it}\,\,,\,\frac{\pi}{2}\leq t\leq\frac{3\pi}{2}\,\right\}\Longrightarrow $$

$$\lim_{\epsilon\to 0}\int_{\gamma_\epsilon}\frac{dz}{z(z-1)}=i\pi\,Res_{z=1}(f)=i\pi\cdot 1=\pi i$$

And since

$$Res_{z=0}(f)=\lim_{z\to 0}\frac{1}{z-1}=-1$$

we get using Cauchy's Residue Theorem:

$$\int_{|z|=1}\frac{dz}{z(z-1)}=\int_{|z|=1\,,\,z\notin \stackrel{\circ}\gamma_\epsilon}\frac{dz}{z(z-1)}-\int_{\gamma_\epsilon}\frac{d}{z(z-1)}=-2\pi i+2\pi^2=2\pi(\pi-i)$$

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Yes, but be sure to point out that the integral is divergent. You've computed the principal value of the integral. –  mrf Dec 16 '12 at 15:00
    
Well, yes...but I really don't think the term divergent applies here. What I did is more like stating that $\,\sin x/x\,$ is a continuous function, whereas the correct thing to say is that the only non-definition point is a removable discontinuity and we do actually remove it and...etc. In the present case, I think it'd be more accurate, perhaps, to say the integral isn't even defined if we go the basic definitions, but that...etc. –  DonAntonio Dec 16 '12 at 15:02
    
The integral is most certainly divergent. $|f| \approx 1/|z-1|$ near $z=1$. It's not just a matter of $f$ not being defined at $z=1$. (Compare the real integral of $1/x$ from $-1$ to $1$. Would you say that its value is $0$?) –  mrf Dec 16 '12 at 15:05
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