Notebook

In [ ]:

using Plots, ComplexPhasePortrait, ApproxFun
gr();

M3M6: Methods of Mathematical Physics¶

$\def\dashint{{\int\!\!\!\!\!\!-\,}} \def\infdashint{\dashint_{\!\!\!-\infty}^{\,\infty}} \def\D{\,{\rm d}} \def\dx{\D x}$

Dr. Sheehan Olver

s.olver@imperial.ac.uk
Website: https://github.com/dlfivefifty/M3M6LectureNotes

Lecture 7: Integrals over the real line¶

This lecture we cover

Integrals over real lines
- Principal value integral
- Cauchy's integral formula and Residue theorem on the real line
- Jordan's lemma
- Application: Calculating Fourier tranforms of weakly decaying functions
Functions with branch cuts
- Logarithmic function

Integrals over the real line¶

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

$\infdashint f(x) \dx = \int_{-\infty}^\infty f(x) \dx.$

Residue theorem on the real line¶

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]:

(2.221441469079183 + 3.487868498008632e-16im, 2.221441469084968)

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]:

(2.221441469079183 + 5.231802747012948e-16im, 2.221441469084968)

Cauchy's integral formula on the real line¶

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]:

2.69638478211931e-13 - 3.032296636007459e-13im

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]:

(2.2331966377567175e-13 + 2.0218605967538834e-14im, 0.7040000000000001 - 0.128im)

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]:

(0.03277196176631425 + 0.16750113791564233im, 0.03277196176604461 + 0.1675011379153391im)

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]:

4.501316791527543e-9 + 5.93330299732131e-7im

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]:

-4.529525111429054e-9 + 5.865568293152012e-7im

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]:

(0.09196985577264773 - 0.09196926921581833im, 0.9007327639404081 + 0.3351305720620013im)

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]:

(-0.045354995402086956 - 0.1219015148055592im, -0.04535499089125899 - 0.12190092372837213im)

Jordan's lemma¶

The case of calculating

$\int_{-\infty}^\infty e^{i \omega x} g(x) dx$ is important because it is the Fourier transform of

$g(x)$ . Provided

$g$ is defined in the upper half plane and

$\omega > 0$ ,

$f(z) = e^{i \omega z} g(z)$ has exponential decay.

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

$\left| \int_{C_R} g(z) e^{i \omega z} dz \right| \leq 2MR \int_0^{\pi\over 2} e^{- {2\omega R\theta \over \pi} } d\theta = {\pi \over \omega} (1 - e^{-\omega R}) M \leq {\pi M \over \omega}.$

Application: Calculating Fourier integrals of weakly decaying functions¶

Why is this useful? We can use it to apply Residue theorem to We already know $O(z^{-2})$ decay gives us the integral via Residue theorem. And if we only have $z^{-1}$ decay our integral does not converge absolutely:

In [20]:

f = x -> exp(im*x)*x/(x^2+1)
sum(abs.(Fun(f, 0 .. 2000)))

Out[20]:

7.600902584542065

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]:

-6.776263578034403e-21 + 1.1557671135433842im

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]:

0.0 + 1.1557273497909217im