I want to prove the version of Jordan's lemma which say's that:
Around a contour as shown below:
The integral: $$\int_{\Gamma} e^{az}f(z)dz$$ will go to $0$ if $$|f(z)|\le \frac{M}{R^\alpha}$$ for some positive $M$ and $\alpha$. Does anyone know how this can be proved?
A necessary condition is that $a >0$.
Assume that the semicircle is centered at the point $x_{0}$ on the real axis.
Then if $|f(z)|\le \frac{M}{R^\alpha}$ on the contour, we have
$$ \begin{align} \left|\int_{\Gamma} e^{a z} f(z) \, dz \right| &\le \frac{M}{R^{\alpha}} \int_{\pi/2}^{3 \pi /2} \left| e^{a(x_{0}+Re^{it})} i Re^{it} \, \right|dt \\ &= \frac{e^{ax_{0}}MR}{R^{\alpha}} \int_{\pi/2}^{3 \pi /2} e^{a R \cos \theta} \ d \theta \\ &= \frac{e^{ax_{0}}MR}{R^{\alpha}} \int_{0}^{\pi} e^{-a R \sin \phi} \ d \phi. \tag{1} \end{align} $$
Now the proof of Jordan's lemma can be used to show that the integral vanishes as $R \to \infty$.
$(1)$ Let $\phi=\theta - \frac{\pi}{2}$.