In [ ]:
using Plots, ComplexPhasePortrait, ApproxFun
gr();
Dr. Sheehan Olver
s.olver@imperial.ac.uk
Website: https://github.com/dlfivefifty/M3M6LectureNotes
This lecture we cover
Integrals on the real line are always to viewed as improper integrals:
$$ \int_{-\infty}^\infty f(x) \dx = \int_0^\infty f(x) \dx +\int_{-\infty}^0 f(x)\dx = \lim_{b \rightarrow \infty} \int_0^b f(x) \dx + \lim_{a \rightarrow -\infty} \int_a^0 f(x) \dx.$$Definition (Principal value integral on the real line) The (Cauchy) principal value integral on the real line is defined as $$ \infdashint f(x) \dx := \lim_{M\rightarrow \infty} \int_{-M}^M f(x) \dx $$
Proposition (Integability $\Rightarrow$ Prinipal value integrability) If $ \int_{-\infty}^\infty f(x) \dx < \infty$ then $$ \infdashint f(x) \dx = \int_{-\infty}^\infty f(x) \dx.$$
The real line doesn't have an inside and outside, rather an above and below, or left and right. Thus we get the following two versions of the Residue theorem:
Definition (Upper/lower half plane) Denote the upper/lower half plane by $$ {\mathbb H}^+ = \{z : \Re z > 0 \} \\ {\mathbb H}^- = \{z : \Re z < 0 \} $$ The closure is denoted $$ \bar{\mathbb H}^+ = {\mathbb H}^+ \cup {\mathbb R} \cup \{\infty\} \\ \bar{\mathbb H}^- = {\mathbb H}^- \cup {\mathbb R} \cup \{\infty\} $$
Theorem (Residue theorem on the real line) Suppose $f : \bar {\mathbb H}^+ \backslash \{z_1,\ldots,z_r \} \rightarrow {\mathbb C}$ is holomorphic in ${\mathbb H}^+ \backslash \{z_1,\ldots,z_r \}$, where $\Re z_k > 0$, and $\lim_{\epsilon \rightarrow 0} f(x + i \epsilon) = f(x)$ converges uniformly. If $$ \lim_{z \rightarrow \infty} z f(z) = 0 $$ uniformly for $z \in \bar {\mathbb H}^+$, then $$ \infdashint f(x) \dx = 2 \pi i \sum_{k=1}^r {\underset{z = z_k}{\rm Res}} \, f(z) $$ Similarly, if the equivalent conditions hold in the lower half plane for $f : \bar{\mathbb H}^- \backslash \{z_1,\ldots,z_r \} \rightarrow {\mathbb C}$ then $$ \infdashint f(x) \dx = -2 \pi i \sum_{k=1}^r {\underset{z = z_k}{\rm Res}} \, f(z) $$
Examples:
In [ ]:
f = x -> x^2/(x^4+1)
phaseplot(-3..3, -2..2, f)
This function has poles in the upper plane, but has sufficient decay that we can apply Residue theorem:
In [4]:
z₁,z₂,z₃,z₄ = exp(im*π/4), exp(3im*π/4), exp(5im*π/4), exp(7im*π/4)
res₁ = z₁^2 / ((z₁ - z₂)*(z₁ - z₃)*(z₁ - z₄) )
res₂ = z₂^2 / ((z₂ - z₁)*(z₂ - z₃)*(z₂ - z₄) )
2π*im*(res₁ + res₂), sum(Fun(f, Line()))
Out[4]:
We can also apply Resiude theorem in the lower-half plane, and we get the same result:
In [5]:
res₃ = z₃^2 / ((z₃ - z₁)*(z₃ - z₂)*(z₃ - z₄) )
res₄ = z₄^2 / ((z₄ - z₁)*(z₄ - z₃)*(z₄ - z₂) )
-2π*im*(res₃ + res₄), sum(Fun(f, Line()))
Out[5]:
An immediate consequence of the Residue theorem is Cauchy's integral formula on the real line:
Theorem (Cauchy's integral formula on the real line) Suppose $f : \bar {\mathbb H}^+ \rightarrow {\mathbb C}$ is holomorphic in ${\mathbb H}^+ $, and $\lim_{\epsilon \rightarrow 0} f(x + i \epsilon) = f(x)$ converges uniformly. If $$ \lim_{z \rightarrow \infty} f(z) = 0 $$ uniformly for $z \in \bar {\mathbb H}^+$, then $$ f(z) = {1 \over 2 \pi i} \infdashint {f(x) \over x - z} dx $$ for all $z \in {\mathbb H}^+$.
Examples Here is a simple example of $f(x) = {x^2 \over (x+ i)^3}$, which is analytic in the upper half plane:
In [7]:
f = x -> x^2/(x+im)^3
z = 2.0+2.0im
sum(Fun(x-> f(x)/(x - z), Line()))/(2π*im) - f(z)
Out[7]:
Evaluating in lower half plane doesn't work b ecause it has a pole there:
In [9]:
f = x -> x^2/(x+im)^3
z = 2.0-2.0im
sum(Fun(x-> f(x)/(x - z), Line()))/(2π*im) , f(z)
Out[9]:
But does for a function analytic in the lower half plane (with a minus sign):
In [10]:
f = x -> x^2/(x-im)^3
z = 2.0-2.0im
-sum(Fun(x-> f(x)/(x - z), Line()))/(2π*im) , f(z)
Out[10]:
It also works for functions with exponential decay in the upper-half plane:
In [11]:
f = x -> exp(im*x)/(x+im)
z = 2 + 2im
sum(Fun(x-> f(x)/(x - z), -500 .. 500))/(2π*im) - f(z)
Out[11]:
This is difficult as a real integral as the integrand is very oscillatory:
In [13]:
xx = -200:0.1:200
plot(xx,real.(f.(xx)))
Out[13]:
An equivalent result holds in the negative real axis, but be careful:
In [17]:
z = -2-im
f = x -> exp(im*x)/(x+im)
sum(Fun(x-> f(x)/(x - z), -500 .. 500))/(2π*im)
Out[17]:
In [15]:
z = -2-im
f = x -> exp(im*x)/(x-im)
sum(Fun(x-> f(x)/(x - z), -500 .. 500))/(2π*im), f(z)
Out[15]:
In [16]:
z = -2-im
f = x -> exp(-im*x)/(x-im)
-sum(Fun(x-> f(x)/(x - z), -500 .. 500))/(2π*im), f(z)
Out[16]:
We can use this to get sharper results than ML inequality:
Lemma (Jordan) Assume $\omega > 0$. If $g(z)$ is continuous in on the half circle $C_R = \{ R e^{i \theta} : 0 \leq \theta \leq \pi \}$ then $$\left| \int_{C_R} g(z) e^{i \omega z} dz \right| \leq {\pi \over \omega} M$$ where $M = \sup_{z \in C_R} |g(z)|$.
Sketch of proof We have $$\left| \int_{C_R} g(z) e^{i \omega z} dz \right| \leq R \int_0^\pi \left|g(R e^{i \theta}) e^{i \omega R e^{i \theta}}e^{i \theta}\right| d\theta \leq MR \int_0^\pi e^{- \omega R\sin \theta } d\theta = 2MR \int_0^{\pi\over 2} e^{- \omega R\sin \theta } d\theta $$ But we have $\sin \theta \geq {2 \theta \over \pi}$:
In [19]:
θ = range(0; stop=π/2, length=100)
plot(θ, sin.(θ); label="sin t")
plot!(θ, 2θ/π; label = "2t / pi")
Out[19]:
Hence $$\left| \int_{C_R} g(z) e^{i \omega z} dz \right| \leq 2MR \int_0^{\pi\over 2} e^{- {2\omega R\theta \over \pi} } d\theta = {\pi \over \omega} (1 - e^{-\omega R}) M \leq {\pi M \over \omega}. $$
In [20]:
f = x -> exp(im*x)*x/(x^2+1)
sum(abs.(Fun(f, 0 .. 2000)))
Out[20]:
However, it does converge conditionally: $$ \dashint_{\infty}^\infty f(x) \dx := \lim_{M\rightarrow \infty} \int_{-M}^M f(x) \dx $$ converges:
In [21]:
f = x -> exp(im*x)*x/(x^2+1)
sum(Fun(f, -30000 .. 30000))
Out[21]:
Thus we can construct a Residue theorem for calculating $$ \infdashint g(x) e^{i \omega x} \dx $$ provided that $g(z) \rightarrow 0$ and is analytic in the upper-half plane.
In [20]:
2π*im*exp(-1)*im/(im+im) # 2π*im* residue of g(z)exp(im*z) at z = im
Out[20]: