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

1. 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
2. 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:

This function has poles in the upper plane, but has sufficient decay that we can apply Residue theorem:

We can also apply Resiude theorem in the lower-half plane, and we get the same result:

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:

Evaluating in lower half plane doesn't work b ecause it has a pole there:

But does for a function analytic in the lower half plane (with a minus sign):

It also works for functions with exponential decay in the upper-half plane:

This is difficult as a real integral as the integrand is very oscillatory:

An equivalent result holds in the negative real axis, but be careful:

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 $\sin \theta \geq {2 \theta \over \pi}$:

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:

However, it does converge conditionally: $$ \dashint_{\infty}^\infty f(x) \dx := \lim_{M\rightarrow \infty} \int_{-M}^M f(x) \dx $$ converges:

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.