5.8 KiB
Math416 Lecture 27
Continue on Application to evaluate \int_{-\infty}^\infty \frac{\cos x}{1+x^4}dx
Consider the function$f(z)=\frac{e^{iz}}{1+z^4}=\frac{\cos z+i\sin z}{1+z^4}$.
Our desired integral can be evaluated by \int_{-R}^R f(z)dz
To evaluate the singularity, z^4=-1 has four roots by the De Moivre's theorem.
z^4=-1=e^{i\pi+2k\pi i} for k=0,1,2,3.
So z=e^{i\theta} for \theta=\frac{\pi}{4}+\frac{k\pi}{2} for k=0,1,2,3.
So the singularities are z=e^{i\pi/4},e^{i3\pi/4},e^{i5\pi/4},e^{i7\pi/4}.
Only z=e^{i\pi/4},e^{i3\pi/4} are in the upper half plane.
So we can use the semi-circle contour to evaluate the integral. Name the path as \gamma.
\int_\gamma f(z)dz=2\pi i\left[\operatorname{Res}_{z=e^{i\pi/4}}(f)+\operatorname{Res}_{z=e^{i3\pi/4}}(f)\right].
The two poles are simple poles.
\operatorname{Res}_{z_0}(f)=\lim_{z\to z_0}(z-z_0)f(z).
So
\begin{aligned}
\operatorname{Res}_{z=e^{i\pi/4}}(f)&=\lim_{z\to e^{i\pi/4}}(z-e^{i\pi/4})\frac{e^{iz}}{1+z^4}\\
&=\frac{(z-e^{i\pi/4})e^{iz}}{(z-e^{i\pi/4})(z-e^{i3\pi/4})(z-e^{i5\pi/4})(z-e^{i7\pi/4})}\\
&=\frac{e^{ie^{i\pi/4}}}{(e^{i\pi/4}-e^{i3\pi/4})(e^{i\pi/4}-e^{i5\pi/4})(e^{i\pi/4}-e^{i7\pi/4})}
\end{aligned}
A short cut goes as follows:
We know p(z)=1+z^4 has four roots z_1,z_2,z_3,z_4.
\lim_{z\to z_0}\frac{(z-z_0)}{p(z)}=\frac{1}{p'(z_0)}
So
\operatorname{Res}_{z=e^{i\pi/4}}(f)=\frac{e^{ie^{i\pi/4}}}{4e^{i3\pi/4}}
Similarly,
\operatorname{Res}_{z=e^{i3\pi/4}}(f)=\frac{e^{ie^{i3\pi/4}}}{4e^{i\pi/4}}
So the sum of the residues is
\begin{aligned}
\operatorname{Res}_{z=e^{i\pi/4}}(f)+\operatorname{Res}_{z=e^{i3\pi/4}}(f)&=\frac{e^{ie^{i\pi/4}}}{4e^{i3\pi/4}}+\frac{e^{ie^{i3\pi/4}}}{4e^{i\pi/4}}\\
&=\frac{e^{\frac{i}{\sqrt{2}}} e^{-\frac{1}{\sqrt{2}}}}{4[-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}]}+\frac{e^{-\frac{i}{\sqrt{2}}}-e^{-\frac{1}{\sqrt{2}}}}{4[\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}]}\\
&=\frac{\pi\sqrt{2}}{2}e^{-\frac{1}{\sqrt{2}}}(\cos\frac{1}{\sqrt{2}}+\sin\frac{1}{\sqrt{2}})
\end{aligned}
For the semicircle part, we can bound our estimate by
\left|\int_{C_R}f(z)dz\right|\leq\pi R\max_{z\in C_R}|f(z)|\leq \pi \frac{1}{R^4}\to 0
as R\to\infty.
So
\int_{-\infty}^\infty\frac{\cos x}{1+x^4}dx=\frac{\pi\sqrt{2}}{2}e^{-\frac{1}{\sqrt{2}}}(\cos\frac{1}{\sqrt{2}}+\sin\frac{1}{\sqrt{2}})
Big idea of this course
f is holomorphic \iff f has complex derivative.
f is holomorphic \iff f satisfies Cauchy-Riemann equations \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} and \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}
f is holomorphic \iff f is analytic (is locally given by power series). The power series is integrable/differentiable term by term in the radius of convergence.
Laurent series
Similar to power series both with annulus of convergence.
f(z)=\sum_{n=-\infty}^\infty a_n(z-z_0)^n for z\in A(z_0,r,R).
Identity theorem: If f is holomorphic on a domain \Omega, it is uniquely determined by its values on any sets with a limit point in \Omega.
Cauchy's Theorem
\int_\gamma f(z)dz=0
If f is holomorphic on \Omega and \gamma is a closed path in \Omega and \gamma\cup \operatorname{int}\gamma\subset \Omega, then \int_\gamma f(z)dz=0.
Favorite estimate
\left|\int_\gamma f(z)dz\right|\leq \sup_{z\in\gamma}|f(z)|\cdot \operatorname{length}(\gamma)
Cauchy's Integral Formula
f(z_0)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-z_0}dz
where z_0\in \operatorname{int}\gamma and \gamma is a closed path.
Extension: If f is holomorphic on \Omega and z_0\in \Omega, then f is infinitely differentiable and
f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_\gamma \frac{f(z)}{(z-z_0)^{n+1}}dz
Residue theorem
If f is holomorphic on \Omega except for a finite number of isolated singularities z_1,z_2,\dots,z_p, and \Gamma is a curve inside \Omega that don't pass through any of the singularities (\Gamma\subset \Omega\setminus \{z_1,z_2,\dots,z_p\}), then
\int_\Gamma f(z)dz=2\pi i\sum_{z_i}\operatorname{ind}_{\Gamma}(z_i) \operatorname{res}_{z_i}(f)
Harmonic conjugate
Locally, always have harmonic conjugates.
Globally can do this iff domain is simply connected.
Schwarz-pick's Lemma:
If f maps D to D and f(0)=0, then |f(z)|\leq |z| for all z\in D. and |f'(0)|\leq 1.
For mobius map, f:D\to D holds, \varphi(f(z),f(w))=\varphi(z,w) for all z,w\in D.
\varphi(z,w)=\frac{z-w}{1-\overline{w}z}
Convergence
Types of convergence
Converge pointwise (Not very strong):
\forall x\in X, \lim_{n\to\infty}f_n(x)=f(x).
Or, \forall x\in X, \forall \epsilon>0, \exists N>0, \forall n\geq N \implies |f_n(x)-f(x)|<\epsilon.
Converge uniformly (Much better):
\forall \epsilon>0, \exists N>0, \forall n\geq N \implies \forall x\in X, |f_n(x)-f(x)|<\epsilon.
Converge locally uniformly (Strong):
\forall x\in X, \exists open x\in U, such that f_n\to f uniformly on U.
Converge uniformly on compact subsets (Good enough for local properties):
\forall compact K\subset X, f_n\to f uniformly on K.
Weierstrass' Theorem
If f_n\in O(\Omega) and f_n\to f locally uniformly, then f\in O(\Omega).
Cauchy-Hadamard's Theorem
For a power series, \sum_{n=0}^\infty a_n(z-z_0)^n, the radius of convergence is
R=\frac{1}{\limsup_{n\to\infty}|a_n|^{1/n}}
On B(z_0,R), the series converges locally uniformly and absolutely.
Argument and Logarithm
\arg z is any \theta such that z=re^{i\theta}.
\operatorname{Arg} z is the principal value of the argument, -\pi<\operatorname{Arg} z\leq \pi.
\log z is the principal value of the logarithm, \log z=\ln |z|+i\arg z.
\operatorname{Log} z is the set of all logarithms of z, \operatorname{Log} z=\{\log z+2k\pi i: k\in\mathbb{Z}\}.
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