数学代写|MTH829 Complex Analysis

Recall the following Taylor series:
$$
\begin{aligned}
& e^z-1=z+\frac{z^2}{2 !}+\frac{z^3}{3 !}+\cdots \
& \sin ^3 z=z^3-\frac{z^5}{2}+\cdots \
& \frac{e^z-1}{\sin ^3 z}=\frac{z+\frac{z^2}{2 !}+\frac{z^3}{3 !}+\cdots}{z^3-\frac{z^5}{2}+\cdots}
\end{aligned}
$$
By performing long division
the coefficient of $\frac{1}{z}$ is seen to be $\frac{1}{2}$ so that
$$
\operatorname{Res}\left(\frac{e^z-1}{\sin ^3 z}, 0\right)=\frac{1}{2} \text {. }
$$

(a)
$$
\begin{aligned}
& \int_0^{2 \pi} \frac{d \theta}{1-2 a \cos \theta+a^2}=\oint_{|z|=1} \frac{d z}{i z\left(1-2 a\left(\frac{z+z^{-1}}{2}\right)+a^2\right)}=\oint_{|z|=1} \frac{d z}{-a i(z-a)\left(z-\frac{1}{a}\right)} \
= & \begin{cases}2 \pi i \operatorname{Res}\left(\frac{1}{-a i(z-a)\left(z-\frac{1}{a}\right)}, a\right), & |a|>1 \
2 \pi i \operatorname{Res}\left(\frac{1}{-a i(z-a)\left(z-\frac{1}{a}\right)}, \frac{1}{a}\right), & |a|<1\end{cases} \
= & \begin{cases}\frac{2 \pi}{a} \lim {z \rightarrow a}\left(\frac{1}{z-\frac{1}{a}}\right)=\frac{2 \pi}{a^2-1}, & |a|>1 \ \frac{2 \pi}{a} \lim {z \rightarrow \frac{1}{a}}\left(\frac{1}{z-a}\right)=\frac{2 \pi}{1-a^2}, & |a|<1\end{cases}
\end{aligned}
$$
(b) Consider the closed contour $C$ as depicted in the following figure
$$
\oint_C \frac{z^2}{z^4+1} d z=\int_{C_R} \frac{z^2}{z^4+1} d z+\int_{-R}^R \frac{x^2}{x^4+1} d x .
$$
By Residue calculus
$$
\oint_C \frac{z^2}{z^4+1} d z=2 \pi i\left[\operatorname{Res}\left(\frac{z^2}{z^4+1}, e^{i \pi / 4}\right)+\operatorname{Res}\left(\frac{z^2}{z^4+1}, e^{i 3 \pi / 4}\right)\right]
$$
where the isolated singularities of $\frac{z^2}{z^4+1}$ enclosed inside $C$ are $z=e^{i \pi / 4}$ and $z=$ $e^{i 3 \pi / 4}$. Both are simple poles so that
$$
\begin{gathered}
\operatorname{Res}\left(\frac{z^2}{z^4+1}, e^{i \pi / 4}\right)=\left.\frac{z^2}{4 z^3}\right|{z=e^{i \pi / 4}}=\frac{1}{4} e^{-i \pi / 4} \ \operatorname{Res}\left(\frac{z^2}{z^4+1}, e^{i 3 \pi / 4}\right)=\frac{1}{4} e^{-i 3 \pi / 4} . \ \int{C_R} \frac{z^2}{z^4+1} d z=O\left(\frac{R^2}{R^4}\right) R \rightarrow 0 \text { as } R \rightarrow \infty .
\end{gathered}
$$
By taking the limit $R \rightarrow \infty$, we obtain
$$
\begin{aligned}
\int_{-\infty}^{\infty} \frac{x^2}{x^4+1} d x & =2 \pi i\left[\operatorname{Res}\left(\frac{z^2}{z^4+1}, e^{i \pi / 4}\right)+\operatorname{Res}\left(\frac{z^2}{z^4+1}, e^{i 3 \pi / 4}\right)\right] \
& =\frac{2 \pi i}{4}\left[e^{-i \pi / 4}+e^{-i 3 \pi / 4}\right]=\frac{\pi}{\sqrt{2}} .
\end{aligned}
$$
(c) Consider
$$
\oint_C \frac{z e^{i a z}}{z^2+b^2} d z=\int_{-R}^R \frac{x e^{i a x}}{x^2+b^2} d x+\int_{C_R} \frac{z e^{i a z}}{z^2+b^2} d z
$$
Letting $R \rightarrow \infty$, then the integral over $C_R$ vanishes by Jordan’s lemma. This is because $\left|\frac{z}{z^2+b^2}\right| \rightarrow 0$ as $R \rightarrow \infty$. The integrand has a singularity at $z=b i$ which is enclosed inside the closed contour. Since $b>0$, we have
$$
2 \pi i \operatorname{Res}\left(\frac{z e^{i a z}}{z^2+b^2}, b i\right)=\left.2 \pi i\left(\frac{-z e^{i a z}}{z+i b}\right)\right|{z=b i}=\frac{\pi}{b} b i e^{-a b} $$ so that $$ \operatorname{Im} \oint_C \frac{z e^{i a z}}{z^2+b^2} d z=\int{-\infty}^{\infty} \frac{x \sin a x}{x^2+b^2} d x=\frac{\pi}{e^{a b}}
$$

One may be tempted to say that the given integral equals the imaginary part of
$$
\text { PV } \int_{-\infty}^{\infty} \frac{e^{i x}}{x+i} d x
$$
This is wrong! (Why?) Moreover, we cannot use $(\sin z) /(z+i)$ either, because it is unbounded in both the upper and lower half-planes. We try the substitution
$$
\sin x=\frac{e^{i x}-e^{-i x}}{2 i}
$$
which lead to the representation
$$
I=\frac{1}{2 i}\left(\mathrm{PV} \int_{-\infty}^{\infty} \frac{e^{i x}}{x+i} d x-\mathrm{PV} \int_{-\infty}^{\infty} \frac{e^{-i x}}{x+i} d x\right) .
$$
Now we deal with each integral separately. For
$$
I_1:=\mathrm{PV} \int_{-\infty}^{\infty} \frac{e^{i e}}{x+i} d x
$$
we close the contour $[-\rho, \rho]$ with the half-circle $C_\rho^{+}$in the upper half-plane. Then, by Jordan’s lemma
$$
\lim {\rho \rightarrow \infty} \int{C_\rho^{+}} \frac{e^{i z}}{z+i} d z=0
$$
and since the only singularity of the integrand is in the lower half-plane at $z=-i$, we deduce that $I_1=0$.
Now the second integral
$$
I_2:=\mathrm{PV} \int_{-\infty}^{\infty} \frac{e^{-i x}}{x+i} d x
$$
involves the function $e^{-i z}$, which is unbounded in the upper half-plane, so we close the contour $[-\rho, \rho]$ in the lower half-plane with the semicircle $C_\rho^{-}: z=\rho e^{-i t}, 0 \leq t \leq \pi$ (see the above figure). Then by the analogue of Jordan’s lemma for the case when $m<0$, we deduce that
$$
\lim {\rho \rightarrow \infty} \int{C_\rho^{-}} \frac{e^{-i z}}{z+i} d z=0 .
$$
Observing that the closed contour in the figure is negatively oriented, we obtain
$$
\begin{aligned}
I_2 & =-2 \pi i \operatorname{Res}\left(\frac{e^{-i z}}{z+i} ;-i\right) \
& =-2 \pi i \lim _{z \rightarrow-i} e^{-i z}=-2 \pi i e^{-1} .
\end{aligned}
$$
Consequently,
$$
I=\frac{1}{2 i}\left(I_1-I_2\right)=\frac{1}{2 i}\left(0+2 \pi i e^{-1}\right)=\frac{\pi}{e}
$$

Consider
$$
\begin{aligned}
\oint_C \frac{\log z}{z^2+4} d z= & \int_{-R}^{-\varepsilon} \frac{\log z}{z^2+4} d x+\int_{C_z} \frac{\log z}{z^2+4} d z \
& +\int_{\varepsilon}^R \frac{\ln x}{x^2+4} d x+\int_{C_R} \frac{\log z}{z^2+4} d z,
\end{aligned}
$$
It is necessary to show that the second integral vanishes as $\varepsilon \rightarrow 0$ and the fourth integral vanishes as $R \rightarrow \infty$. The integrand has a simple pole at $z=2 i$.
Now
$$
\begin{aligned}
& \int_{C_{\varepsilon}} \frac{\log z}{z^2+4} d z=O(\varepsilon \ln \varepsilon) \rightarrow 0 \text { as } \varepsilon \rightarrow 0 \
& \int_{C_R} \frac{\log z}{z^2+4} d z=O\left(\frac{\ln R}{R^2} \cdot R\right) \rightarrow 0 \text { as } R \rightarrow \infty .
\end{aligned}
$$
The sum of the first and third integrals is
$$
\begin{aligned}
& i \pi \int_{-\infty}^0 \frac{1}{x^2+4} d x+2 \int_0^{\infty} \frac{\ln x}{x^2+4} d x \
= & 2 \pi i \operatorname{Res}\left(\frac{\log z}{z^2+4}, 2 i\right) \
= & 2 \pi i \frac{\log 2 i}{4 i}=\frac{\pi}{2}\left(\ln 2+i \frac{\pi}{2}\right) .
\end{aligned}
$$
By equating the real parts, we obtain
$$
\int_0^{\infty} \frac{\ln x}{x^2+4} d x=\frac{\pi}{4} \ln 2
$$