# when integrating a Laurent series $f(z)=\sum\limits_{j=-\infty}^{\infty}a_j(z-z_0)^j$, why do all terms for $j\neq-1$ dissappear?

In my complex analysis book we are looking at a Laurent series expression for $f(z)$ around a singularity $z_0$ that converges to $f(z)$ for all $z\in C, z\neq z_0$. The Laurent series looks like $$f(z)=\sum\limits_{j=-\infty}^{\infty}a_j(z-z_0)^j$$ We are integrating this series along some simply connected closed contour $\Gamma$ lying in $C$ which goes around $z_0$ with a positive orientation. (So f(z) is analytic everywhere on and within $\Gamma$ except at $z_0$). It says that we can integrate termwise which makes a lot of sense but somehow the integral of all terms for $j\neq-1$ is said to dissappear. This is strange to me because these terms are of the form $\frac{a_{-2}}{(z-z_0)^2}, \frac{a_{-3}}{(z-z_0)^3}\ldots$ meaning that each of these terms has a singularity at $z_0$ and thus the integral along $\Gamma$ should not vanish? It might be that i am missing something obvious but I can not seem to figure it out. Any hints or suggestions towards an answer would be greatly appreciated!

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If $f(z)=\dfrac{1}{(z-z_0)^k}$ and $k\neq 1$, then $F(z)=\dfrac{-1}{k-1}\dfrac{1}{(z-z_0)^{k-1}}$ satisfies $F'(z)=f(z)$ for all $z\neq z_0$. If $\Gamma$ is a path from $a$ to $b$ that avoids $z_0$, then $\int_{\Gamma}f(z)dz =F(b)-F(a)$. Hence if $\Gamma$ is closed, $\int_{\Gamma}f(z)dz = 0$.

This argument doesn't apply when $k=1$, because $\dfrac{1}{z-z_0}$ has no antiderivative defined everywhere on $\mathbb C\setminus\{z_0\}$.

Incidentally, term-by-term integration is justified in this case by uniform convergence on compact sets.

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Awesome this makes a lot of sense, thanks! So would it be correct to say that the problem is that the antiderivative for $\frac{a_{-1}}{z-z_0}$ is $\log{z-z_0}$ which is undefined for the point on $\Gamma$ directly 'to the left of' $z_0$ where $z-z_0$ is simply a negative real number? – Slugger Jan 21 '13 at 21:34
@TeunVerstraaten: $\log$ can be used to define an antiderivative of $\frac{1}{z-z_0}$ on any simply connected subset of $\mathbb C\setminus\{z_0\}$; it is not necessary to choose to remove those points to the left of $z_0$, but that would be one option, obtained by translating the principal branch of $\log$. Using the fact that $\log(z-z_0)$ cannot be extended to the whole punctured plane continuously is one way to prove that $\frac{1}{z-z_0}$ has no antiderivative on the punctured plane. – Jonas Meyer Jan 21 '13 at 21:37

The integral around such a "singularity" is indeed zero (assuming $z_0=0$ for simplicity):

$$\int_\gamma \frac{a_{-k}}{z^k}dt=\int\limits_0^1\frac{a_{-k}}{e^{2k\pi it}}2\pi ie^{2\pi it}dt=2\pi i\int\limits_0^1 \frac{a_{-k}}{e^{2(k-1)\pi it}}dt$$

The last integral is zero for $k\neq 1$ because that function has no singularities -> is holomorphic in a unit sphere. Therefore, indeed, every other laurent series term vanishes. This is represented by the residue theorem aswell, If you already handled that in your reading.

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This is a good way to see it in the case where $\Gamma$ is a circle centered at $z_0$, where the direct computation is easier, and if the OP knows something about homotopy invariance of the integral then they can generalize. – Jonas Meyer Jan 21 '13 at 21:32

Except for $k=-1$, we have that $$(z-z_0)^{k}=\frac{\mathrm{d}}{\mathrm{d}z}\frac{(z-z_0)^{k+1}}{k+1}$$ Therefore, $$\int_\gamma (z-z_0)^k\,\mathrm{d}z=\left.\frac{(z-z_0)^{k+1}}{k+1}\right]_{\gamma(0)}^{\gamma(1)}$$ where $\gamma(0)$ is the beginning of the curve and $\gamma(1)$ is the end of the curve. Since $\gamma$ is a closed curve, $\gamma(0)=\gamma(1)$. Therefore, $$\int_\gamma (z-z_0)^k\,\mathrm{d}z=\left.\frac{(z-z_0)^{k+1}}{k+1}\right]_{\gamma(0)}^{\gamma(1)}=0$$ Therefore, $$\int_\gamma\left(\sum_{k}a_k(z-z_0)^k\right)\,\mathrm{d}z=\int_\gamma a_{-1}(z-z_0)^{-1}\,\mathrm{d}z$$

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