Jordon's Lemma

Claim 1

If $zf(z)\rightarrow^{unif.} 0$ as $z\rightarrow\infty$, then

\[\lim_{R\rightarrow\infty} \int_{C(R,0,\pi)} f(z)dz = 0\]

where $C(R,0,\pi)$ is the top semicircle arc of radius $R$ centered at $0$, rotated ccw.

proof. Note that on the arc, $|z| = R$ so that $|zf(z)| = R |f(z)| \leq \kappa(R)$ so that $|f(z)| \leq \frac{\kappa(R)}{R}$.

Let $\epsilon > 0$. By definition of $\kappa$, take $S$ s.t. $\forall R > S. \kappa(R) < \frac{\epsilon}{\pi}$. Then, with ML inequality we have

\[|\int_{C(R,0,\pi)} f(z)dz| \leq \pi R \frac{\kappa(R)}{R} < \epsilon\]

Example 1

Compute $\int_{-\infty}^{\infty} (x^2 + 1)^{-1}dx$

For any $R > 0$, expanding $f$ to the complex domain, and consider

$$\int_{C} \frac{1}{z^2 + 1}dz = 2\pi i Res(f, i) = 2\pi i \frac{1}{2 i} = \pi $$ Also we have

\[\int_{C} f(z)dz = \int_{C^{line}_R} f(z)dz + \int_{C(R,0,\pi)}f(z)dz\]

Therefore, we can write

\[\lim_{R\rightarrow\infty} \int_{C} \frac{1}{z^2 + 1}dz = \lim_{R\rightarrow\infty} \int_{C^{line}_R} f(z)dz + \lim_{R\rightarrow\infty} \int_{C(R,0,\pi)}f(z)dz = \pi\]

Consider the latter term, because $zf(z) = \frac{z}{z^2+1}\rightarrow^{unif} 0$ as $z\rightarrow\infty$, so that $\lim_{R\rightarrow\infty} \int_{C(R,0,\pi)}f(z)dz = 0$ Note that

\[\lim_{R\rightarrow\infty} \int_{C^{line}_R} f(z)dz = \lim_{R\rightarrow\infty} \int_{-R}^R f(x)dx = \int_{-\infty}^\infty f(x)dx\]

so that

\[\pi = \int_{-\infty}^\infty \frac{1}{x^2 + 1}dx + 0\]

Jordon's Lemma

IF $f(z)\rightarrow^{unif} 0$ as $z\rightarrow \infty$ and $k > 0$ then

\[\lim_{R\rightarrow\infty} \int_{C(R, 0, \pi) } e^{ikz}f(z)dz = \lim_{R\rightarrow\infty} \int_{C(R, -\pi, 0) }e^{ikz}f(z)dz = 0\]

proof. parameterize $C(R,0,\pi)$ with $c(t):[0,\pi] \rightarrow \mathbb C, c(t)= Re^{it}$

\[\begin{align*} I_R&:=|\int_{C(R,0,\pi)} e^{ikz} f(z)dz|\\ &= |\int_0^\pi e^{ikRe^{it}} f(Re^{it}) iRe^{it}dt|\\ &= R|\int_0^\pi e^{kR(i\cos t- \sin t)} f(Re^{it}) ie^{it}dt|\\ &\leq R \int_0^\pi |e^{kR(i\cos t)}|e^{-kR\sin t}||ie^{it}| |f(Re^{it})| dt\\ &= R \int_0^\pi e^{-kR\sin t}|f(Re^{it})| dt\\ &\leq R \kappa(R) \int_0^\pi e^{-kR\sin t} dt\\ &= R \kappa(R) 2 \int_0^{\pi/2} e^{-kR\sin t} dt\\ &\leq R \kappa(R) 2 \int_0^{\pi/2} e^{-kR\frac{2t}{\pi}} dt &\sin t\geq \frac{2t}{\pi}. \forall t \in [0, \pi/2]\\ &= 2R\kappa(R) \frac{\pi (1-e^{-\pi R})}{2Rk}\\ &\leq \frac{\pi}{k} \kappa(R) \end{align*}\]

Since $\lim_{R\rightarrow\infty} \kappa(R) = 0, \lim_{R\rightarrow\infty} |I_R| = 0$

The proof for the second equation is similar

Example

$\int_{-\infty}^\infty \frac{\cos x}{x^2+1}$

First, expand to complex domain,

\[\frac{\cos z}{z^2+1} = \frac{1}{2}\frac{e^{iz}}{x^2+1} + \frac{1}{2}\frac{e^{-iz}}{x^2+1}\]

Let $f_1(z) = \frac{e^{iz}}{x^2+1}, f_2(z) = \frac{e^{-iz}}{x^2+1}$. We can compute $f_1$ on the top semicircle, and $f_2$ on the bottom semicircle.

\[\begin{align*} \lim_{R\rightarrow\infty}\oint_C f_1(z)dz &= \lim_{R\rightarrow\infty}\int_{C^{line}_R} f_1(z)dz + \lim_{R\rightarrow\infty}\int_{C(R, 0, \pi)} f_1(z)dz\\ Res(f_1, i) &= \int_{-\infty}^\infty \frac{e^{iz}}{x^2+1}dx + \lim_{R\rightarrow\infty}\int_{C(R, 0, \pi)} e^{iz}\frac{1}{x^2+1}dz\\ 2\pi i\lim_{z\rightarrow i} \frac{e^{iz}}{z+i} &= \int_{-\infty}^\infty \frac{e^{iz}}{x^2+1}dx + 0&\text{Jordon's Lemma}\\ \frac{\pi}{e} &= \int_{-\infty}^\infty \frac{e^{iz}}{x^2+1}dx \end{align*}\]

Simiarly,

\[\begin{align*} \lim_{R\rightarrow\infty}\oint_C f_2(z)dz &= \lim_{R\rightarrow\infty}\int_{-C^{line}_R} f_2(z)dz + \lim_{R\rightarrow\infty}\int_{C(R, -\pi, 0)} f_2(z)dz\\ Res(f_2, -i) &= -\int_{-\infty}^\infty \frac{e^{-iz}}{x^2+1}dx + \lim_{R\rightarrow\infty}\int_{C(R, -\pi, 0)} e^{-iz}\frac{1}{x^2+1}dz\\ 2\pi i\lim_{z\rightarrow -i} \frac{e^{-iz}}{z-i} &= -\int_{-\infty}^\infty \frac{e^{-iz}}{x^2+1}dx + 0&\text{Jordon's Lemma}\\ \frac{\pi}{e} &= \int_{-\infty}^\infty \frac{e^{-iz}}{x^2+1}dx \end{align*}\]

Therefore,

\[\frac{\cos z}{z^2+1} = \frac{1}{2}(2\frac{\pi}{e}) = \frac{\pi}{e}\]

Another convinient way is to notice

\[\int_{-\infty}^\infty \frac{e^{ix}}{x^2+1}dx =\int_{-\infty}^\infty \frac{\cos x}{x^2+1}dx + i\int_{-\infty}^\infty \frac{\sin x}{x^2+1}dx \]

Because $\frac{\sin x}{x^2+1}$ is odd around 0, $\int_{-\infty}^\infty \frac{\sin x}{x^2+1}dx = 0$ and we are left with

\[\frac{\pi}{e} = \int_{-\infty}^\infty \frac{e^{ix}}{x^2+1}dx =\int_{-\infty}^\infty \frac{\cos x}{x^2+1}dx\]