Jordan's Lemma

Jordan's lemma shows the value of the Integral

$$I \equiv \int_{-\infty}^\infty f(x)e^{iax}\,dx$$ (1)

along the Real Axis is 0 for ``nice'' functions which satisfy $\lim_{R\to\infty} \vert f(Re^{i\theta})\vert = 0$. This is established using a Contour Integral $I_R$ which satisfies

$$\lim_{R\to\infty} \vert I_R\vert \leq {\pi\over a} \lim_{R\to\infty} \epsilon = 0.$$ (2)

To derive the lemma, write

$$x \equiv Re^{i\theta} = R(\cos\theta+i\sin\theta)$$ (3)

$$dx = iRe^{i\theta}\,d\theta$$ (4)

and define the Contour Integral

$$I_R = \int^\pi_0 f(Re^{i\theta})e^{iaR\cos\theta-aR\sin\theta}iRe^{i\theta}\,d\theta$$ (5)

Then

$$\vert I_R\vert = R\int^\pi_0 \vert f(Re^{i\theta})\vert \,\vert e^{iaR\cos \theta}... ... \vert e^{-aR\sin \theta}\vert \,\vert i\vert\, \vert e^{i\theta}\vert\,d\theta$$

$$= R\int^\pi_0 \vert f(Re^{i\theta})\vert e^{-aR\sin \theta}\,d\theta$$

$$= 2R\int^{\pi/2}_0 \vert f(Re^{i\theta})\vert e^{-aR\sin \theta }\,d\theta.$$ (6)

Now, if $\lim_{R\to\infty} \vert f(Re^{i\theta})\vert = 0$, choose an $\epsilon$ such that $\vert f(Re^{i\theta})\vert \leq \epsilon$, so

$$\vert I_R\vert \leq 2R\epsilon \int^{\pi/2}_0 e^{-aR\sin\theta}\,d\theta.$$ (7)

But, for $\theta \in [0, \pi/2]$,

$${2\over\pi} \theta \leq \sin\theta,$$ (8)

so

$$\vert I_R\vert \leq 2R\epsilon \int^{\pi/2}_0 e^{-2aR\theta/\pi}\, d\theta$$

$$= 2\epsilon R {1-e^{-aR}\over {2aR\over\pi}} = {\pi\epsilon\over a}(1-e^{-aR}).$$ (9)

As long as $\lim_{R\to\infty} \vert f(z)\vert=0$, Jordan's lemma

$$\lim_{R\to \infty} \vert I_R\vert \leq {\pi\over a} \lim_{R\to\infty} \epsilon = 0$$ (10)

then follows.

See also Contour Integration

References

Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 406-408, 1985.