Is this the right way to use Jordan's Lemma to show that integration over the semi-circle does not contribute to the closed-contour integration for the integral $\displaystyle \int_{-\infty}^{\infty}\frac{x \sin (7x)}{(x^2-6x+18)} dx$
$$\text{Im}\left(\oint_C \frac{z e^{i7z}}{z^2-6z+18}dz\right) = \text{Im}\left(\oint_C \frac{Re^{ix} }{R^2e^{2ix}-6e^{ix}+18}dz\right) = \text{Im}\left(\oint_C \frac{R}{R^2e-6R+18}dz\right) $$
$\endgroup$ 11 Answer
$\begingroup$What does Jordan's lemma say?
If $f(z$) is analytic in the upper half-plane except for a finite number of poles in $\operatorname{Im} z \geq 0$, and if the maximum of $|f(z)|$ vanishes as $|z| \to \infty$ in the upper half-plane, then $$ I_{\Gamma} = \int_{\Gamma} e^{imz} f(z)\ dz \to 0 \quad \mbox{as} \quad R \to \infty \\ \mbox{where}\ \Gamma \ \mbox{is the contour over our semi-circle.} $$
Is our $f(z) = \frac{z}{z^2-6z+18}$ analytic in the upper half-plane, except for a finite number of poles? And do we have that the maximum of $|f(z)| \to 0$ as $|z| \to \infty$?
$\endgroup$
