Jordan's Lemma - Proof of Jordan's Lemma

Proof of Jordan's Lemma

By definition of the complex line integral,

\begin{align} \int_{C_R} f(z)\, dz &=\int_0^\pi g(Re^{i\theta})\,e^{iaR(\cos\theta+i \sin\theta)}\,i Re^{i\theta}\,d\theta\\ &=R\int_0^\pi g(Re^{i\theta})\,e^{aR(i\cos\theta-\sin\theta)}\,ie^{i\theta}\,d\theta\,. \end{align}

Now the inequality

yields

\begin{align} I_R:=\biggl|\int_{C_R} f(z)\, dz\biggr| &\le R\int_0^\pi\bigl|g(Re^{i\theta})\,e^{aR(i\cos\theta-\sin\theta)}\,ie^{i\theta} \bigr|\,d\theta\\ &=R\int_0^\pi \bigl|g(Re^{i\theta})\bigr|\,e^{-aR\sin\theta}\,d\theta\,. \end{align}

Using MR as defined in (*) and the symmetry sin θ = sin(π – θ), we obtain

Since the graph of sin θ is concave on the interval θ ∈ , the graph of sin θ lies above the straight line connecting its endpoints, hence

for all θ ∈ , which further implies

I_R \le 2RM_R \int_0^{\pi/2} e^{-2aR\theta/\pi}\,d\theta =\frac{\pi}{a} (1-e^{-a R}) M_R\le\frac\pi{a}M_R\,.

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