Residue theorem

hbghlyj

complex.pdf page 62

Theorem 11.1. (Residue theorem): Suppose that $U$ is an open set in $\mathbb{C}$ and $\gamma$ is a path whose inside is contained in $U$, so that for all $z \notin U$ we have $I(\gamma, z)=0$. Then if $S \subset U$ is a finite set such that $S \cap \gamma^*=\emptyset$ and $f$ is a holomorphic function on $U \backslash S$ we have
\[
\frac{1}{2 \pi i} \int_\gamma f(z) d z=\sum_{a \in S} I(\gamma, a) \operatorname{Res}_a(f)
\]
Proof. For each $a \in S$ let $P_a(f)(z)=\sum_{n=-1}^{-\infty} c_n(a)(z-a)^n$ be the principal part of $f$ at $a$, a holomorphic function on $\mathbb{C} \backslash\{a\}$. Then by definition of $P_a(f)$, the difference $f-P_a(f)$ is holomorphic at $a \in S$, and thus $g(z)=$ $f(z)-\sum_{a \in S} P_a(f)$ is holomorphic on all of $U$. But then by Theorem 9.11 we see that $\int_\gamma g(z) d z=0$, so that
\[
\int_\gamma f(z) d z=\sum_{a \in S} \int_\gamma P_a(f)(z) d z
\]
But by the proof of Theorem 9.17, the series $P_a(f)$ converges uniformly on $\gamma^*$ so that
\[
\begin{aligned}
\int_\gamma P_a(f) d z &=\int_\gamma \sum_{n=-1}^{-\infty} c_n(a)(z-a)^n=\sum_{n=1}^{\infty} \int_\gamma \frac{c_{-n}(a) d z}{(z-a)^n} \\
&=\int_\gamma \frac{c_{-1}(a) d z}{z-a}=I(\gamma, a) \operatorname{Res}_a(f),
\end{aligned}
\]
since for $n>1$ the function $(z-a)^{-n}$ has a primitive on $\mathbb{C} \backslash\{a\}$. The result follows.

hbghlyj

\[=\int_\gamma \frac{c_{-1}(a) d z}{z-a}=\color{red}{2\pi i}I(\gamma, a) \operatorname{Res}_a(f),\]这里少一个$2πi$

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Example 11.3. Consider the integral $\int_0^{2 \pi} \frac{d t}{1+3 \cos ^2(t)}$. If we let $\gamma$ be the path $t \mapsto e^{i t}$ and let $z=e^{i t}$ then $\cos (t)=\Re(z)=\frac{1}{2}(z+\bar{z})=\frac{1}{2}(z+1 / z)$. Thus we have
\[
\frac{1}{1+3 \cos ^2(t)}=\frac{1}{1+3 / 4(z+1 / z)^2}=\frac{1}{1+\frac{3}{4} z^2+\frac{3}{2}+\frac{3}{4} z^{-2}}=\frac{4 z^2}{3+10 z^2+3 z^4},
\]
Finally, since $d z=i z d t$ it follows
\[
\int_0^{2 \pi} \frac{d t}{1+3 \cos ^2(t)}=\int_\gamma \frac{-4 i z}{3+10 z^2+3 z^4} d z
\]
Thus we have turned our real integral into a contour integral, and to evaluate the contour integral we just need to calculate the residues of the meromorphic function $g(z)=\frac{-4 i z}{3+10 z^2+3 z^4}$ at the poles it has inside the unit circle. Now the poles of $g(z)$ are the zeros of the polynomial $p(z)=3+10 z^2+3 z^4$, which are at $z^2 \in\{-3,-1 / 3\}$. Thus the poles inside the unit circle are at $\pm i / \sqrt{3}$. In particular, since $p$ has degree 4 and has four roots, they must all be simple zeros, and so $g$ has simple poles at these points. The residue at see (compare with Remark 8.14) that
\[
\begin{aligned}
\operatorname{Res}_{z=\pm i / \sqrt{3}}(g(z)) &=\lim _{z \rightarrow \pm i / \sqrt{3}} \frac{-4 i z(z-\pm i / \sqrt{3})}{3+10 z^2+3 z^4}=(\pm 4 / \sqrt{3}) \cdot \frac{1}{p^{\prime}(\pm i / \sqrt{3})} \\
&=(\pm 4 / \sqrt{3}) \cdot \frac{1}{20(\pm i / \sqrt{3})+12(\pm i / \sqrt{3})^3}=1 / 4 i
\end{aligned}
\]
It now follows from the Residue theorem that
\[
\int_0^{2 \pi} \frac{d t}{1+3 \cos ^2(t)}=2 \pi i\left(\operatorname{Res}_{z=i / \sqrt{3}}\left((g(z))+\operatorname{Res}_{z=-i / \sqrt{3}}(g(z))\right)=\pi .\right.
\]
Remark 11.4. Often we are interested in integrating along a path which is not closed or even finite, for example, we might wish to understand the integral of a function on the positive real axis. The residue theorem can still be a powerful tool in calculating these integrals, provided we complete the path to a closed one in such a way that we can control the extra contribution to the integral along the part of the path we add.

Example 11.5. If we have a function $f$ which we wish to integrate over the whole real line (so we have to treat it as an improper Riemann integral) then we may consider the contours $\Gamma_R$ given as the concatenation of the paths $\gamma_1:[-R, R] \rightarrow \mathbb{C}$ and $\gamma_2:[0,1] \rightarrow \mathbb{C}$ where
\[
\gamma_1(t)=-R+t ; \quad \gamma_2(t)=R e^{i \pi t}
\]
(so that $\Gamma_R=\gamma_2 \star \gamma_1$ traces out the boundary of a half-disk). In many cases one can show that $\int_{\gamma_2} f(z) d z$ tends to 0 as $R \rightarrow \infty$, and by calculating the residues inside the contours $\Gamma_R$ deduce the integral of $f$ on $(-\infty, \infty)$. To see this strategy in action, consider the integral
\[
\int_0^{\infty} \frac{d x}{1+x^2+x^4} .
\]
It is easy to check that this integral exists as an improper Riemann integral, and since the integrand is even, it is equal to
\[
\frac{1}{2} \lim _{R \rightarrow \infty} \int_{-R}^R \frac{d x}{1+x^2+x^4} d x .
\]
If $f(z)=1 /\left(1+z^2+z^4\right)$, then $\int_{\Gamma_R} f(z) d z$ is equal to $2 \pi i$ times the sum of the residues inside the path $\Gamma_R$. The function $f(z)=1 /\left(1+z^2+z^4\right)$ has poles at $z^2=\pm e^{2 \pi i / 3}$ and hence at $\left\{e^{\pi i / 3}, e^{2 \pi i / 3}, e^{4 \pi i / 3}, e^{5 \pi i / 3}\right\}$. They are all simple poles and of these only $\left\{\omega, \omega^2\right\}$ are in the upper-half plane, where $\omega=e^{i \pi / 3}$. Thus by the residue theorem, for all $R>1$ we have
\[
\int_{\Gamma_R} f(z) d z=2 \pi i\left(\operatorname{Res}_\omega(f(z))+\operatorname{Res}_{\omega^2}(f(z))\right)
\]
and we may calculate the residues using the limit formula as above (and the fact that it evaluates to the reciprocal of the derivative of $1+z^2+z^4$ ): Indeed since $\omega^3=-1$ we have $\operatorname{Res}_\omega(f(z))=\frac{1}{2 \omega+4 \omega^3}=\frac{1}{2 \omega-4}$, while $\operatorname{Res}_{\omega^2}(f(z))=\frac{1}{2 \omega^2+4 \omega^6}=\frac{1}{4+2 \omega^2}$. Thus we obtain:
\[
\begin{aligned}
\int_{\Gamma_R} f(z) d z &=2 \pi i\left(\frac{1}{2 \omega-4}+\frac{1}{2 \omega^2+4}\right) \\
&=\pi i\left(\frac{1}{\omega-2}+\frac{1}{\omega^2+2}\right) \\
&=\pi i\left(\frac{\omega^2+\omega}{2\left(\omega-\omega^2\right)-5}\right)=-\sqrt{3} \pi /(-3)=\pi / \sqrt{3},
\end{aligned}
\]
(where we used the fact that $\omega^2+\omega=i \sqrt{3}$ and $\omega-\omega^2=1$ ). Now clearly
\[
\int_{\Gamma_R} f(z) d z=\int_{-R}^R \frac{d t}{1+t^2+t^4}+\int_{\gamma_2} f(z) d z,
\]
and by the estimation lemma we have
\[
\left|\int_{\gamma_2} f(z) d z\right| \leq \color{red}{\sup _{z \in \gamma_2^*}|f(z)|} \cdot \color{green}{\ell\left(\gamma_2\right) }\leq \frac{\color{green}{\pi R}}{\color{red}{R^4-R^2-1}} \rightarrow 0,
\]
as $R \rightarrow \infty$, it follows that
\[
\pi / \sqrt{3}=\lim _{R \rightarrow \infty} \int_{\Gamma_R} f(z) d z=\int_{-\infty}^{\infty} \frac{d t}{1+t^2+t^4}
\]