what is my mistake in estimating CR integral by Jordan s lemma

Question

as part of evaluating this integral $\int_0^\infty \frac{x.\sin(x)}{x^2 +9} \,dx$, I reach to use Jordan Lemma to show : $$|\int_{-R}^R \frac{ze^{iz}}{z^2 +9} \,dz|\le\int_{-R}^R |\frac{ze^{iz}}{z^2 +9}| \,|dz|\le\frac{\pi.R^2}{R^2-9}$$ clearly as $R\rightarrow\infty, $ the integral above does not tend to zero! what is my mistake?

Answer 1

I'm not sure exactly where your difficulty lies, but Jordan's lemma applies to continuous functions $f(z)=e^{iaz}g(z)$ defined on the contour $C=C_R \cup [-R;R]$ where $C_R$ is the upper half circle of radius $R$. It then states that: $$|\int_{C_R}f(z)dz| \le \frac{\pi}{a}\max_{z\in C_R}{|g(z)|}$$

If you compute this upper bound you will indeed see that the integral tends to $0$ as $R\to \infty$.

At last, we have:

$$\int_Cf(z)dz=\int_{C_R}f(z)dz+\int_{-R}^Rf(z)dz$$

We conclude by taking the limit in R and using the residue theorem. Is it clearer now?

Answer 2

Let's define a function $ f $ on $ \mathbb{R} $ as follows : $$ \left(\forall t\in\mathbb{R}\right),\ f\left(t\right)=\int_{-\infty}^{+\infty}{\frac{\mathrm{e}^{\mathrm{i}xt}}{x^{2}+9}\,\mathrm{d}x} $$

Fixing $ t\in\mathbb{R} $ and $ R> 3 $, then integrating $ f_{t}:z\mapsto\frac{\mathrm{e}^{\mathrm{i}zt}}{z^{2}+9} $ on $ \gamma_{R}=\left[-R,R\right]\cup\mathscr{C}_{R} $, where $ \mathscr{C}_{R}=\left\lbrace z:\left|z\right|=R,\ \mathcal{Im}\left(z\right)\geq 0\right\rbrace $, will give us a closed form for our $ f \cdot $

We have : $$ 2\pi\mathrm{i}\,\mathrm{Res}\left(f_{t},3\mathrm{i}\right)=\oint_{\gamma_{R}}{f_{t}\left(z\right)\mathrm{d}z}=\int_{-R}^{R}{f_{t}\left(x\right)\mathrm{d}x}+\int_{\mathscr{C}_{R}}{f_{t}\left(z\right)\mathrm{d}z} $$

Since : $$ \left|\int_{\mathscr{C}_{R}}{f_{t}\left(z\right)\mathrm{d}z}\right|\leq\int_{\mathscr{C}_{R}}{\left|f_{t}\left(z\right)\right|\left|\mathrm{d}z\right|}\leq\frac{1}{R^{2}-9}\int_{\mathscr{C}_{R}}{\left|\mathrm{d}z\right|}=\frac{1}{R^{2}-9}\int_{0}^{\pi}{R\,\mathrm{d}\theta}=\frac{\pi R}{R^{2}-9}\underset{n\to +\infty}{\longrightarrow }0 $$

And : $$ 2\pi\mathrm{i}\,\mathrm{Res}\left(f_{t},3\mathrm{i}\right)=2\pi\mathrm{i}\lim_{z\to 3\mathrm{i}}{\left(z-3\mathrm{i}\right)f_{t}\left(z\right)}=2\pi\mathrm{i}\lim_{z\to 3\mathrm{i}}{\frac{\mathrm{e}^{\mathrm{i}zt}}{z+3\mathrm{i}}}=\frac{\pi\,\mathrm{e}^{-3t}}{3} $$

Tending $ R $ to infinity, we get : $$ \int_{-\infty}^{+\infty}{f_{t}\left(x\right)\mathrm{d}x}=\frac{\pi\,\mathrm{e}^{-3t}}{3} $$

And hence, for any real $ t $, we have : $$ \int_{-\infty}^{+\infty}{\frac{x\,\mathrm{e}^{\mathrm{i}xt}}{x^{2}+9}\,\mathrm{d}x}=-\mathrm{i}\int_{-\infty}^{+\infty}{\frac{\partial f_{t}}{\partial t}\left(t,x\right)\mathrm{d}x}=-\mathrm{i}f'\left(t\right)=\pi\mathrm{i}\,\mathrm{e}^{-3t} $$

Taking the imaginary part, we get that $ \forall t\in\mathbb{R} $ : $$ \fbox{$\begin{array}{rcl}\displaystyle\int_{-\infty}^{+\infty}{\frac{x\sin{\left(xt\right)}}{x^{2}+9}\,\mathrm{d}x}=\pi\,\mathrm{e}^{-3t}\end{array}$} $$