Showing that a path in the complex plane converges so we can use the residue theorem

Question

I wanted to calculate the Fourier transform of $$f(t):=\frac{1}{t^2+6t+13} \tag{1}$$

so basically calculating

$$F[f](t)=\hat{f}(\omega)=\int_{-\infty}^\infty f(t) e^{-i\omega t}dt$$

and doing so by using the residue theorem. Now let's consider the path $\gamma_1=\gamma_{1,R}+\gamma_{1,I}$ along the upper half-circle with radius $R$.

Note that this question is only about the convergence of the non-real path, everything else is more or less clear.

$$\gamma_{1,R}: [0,1] \to \mathbb R, \quad t\mapsto 2Rt-R, \qquad \dot{\gamma}_{1,R}(t)=2R \tag{2}$$

$$\gamma_{1,I}: [0,1] \to \mathbb R, \quad t\mapsto Re^{\pi i t}, \qquad \dot{\gamma}_{1,I}(t)=\pi i Re^{\pi i t}$$

Now we have

$$\int_{\gamma_1}f(t)e^{-i\omega t}dt=\int_{\gamma_{1,R}}f(t)e^{-i\omega t}dt+\int_{\gamma_{1,I}}f(t)e^{-i\omega t}dt=2\pi i Res(f;t_1)\tag{3}$$

with $t_1$ being the singularity enclosed by $\gamma_1$.

Now we want to check the convergence of $\int_{\gamma_{1,I}}f(t)e^{-i\omega t}dt$:

$$\int_{\gamma_{1,I}}f(z)e^{-i\omega t}dz\sim\int_{\gamma_{1,I}}\frac{e^{-i\omega t}}{t^2}dt=\int_0^1\frac{e^{-i\omega Re^{\pi i t}}}{Re^{2\pi i t}}\pi i Re^{\pi i t}dt=\pi i\int_0^1\frac{e^{-i\omega Re^{\pi i t}}}{Re^{\pi i t}}dt \tag{4}$$

Now, I I'd like to somehow show that the last term in (4) converges to 0 for $R\to\infty$ but I can't really see how. I know there are probably other way to approach this, some lemmas and what not (you are welcome to mention it), but I'd like to find an estimate for it which shows me the convergence. :)

How would I proceed to show the convergence?

Answer 1

Suppose that $f$ is meromorphic and doesnt have poles in the real line, also suppose that $\lim_{|z|\to\infty} zf(z)=0$ and $\omega\le 0$. Then choosing the arc above the real line we have by the residue theorem that

$$\int_{-R}^R f(x)e^{-i\omega x}\, dx+i\int_0^\pi f(R e^{it})e^{-i\omega Re^{it}}Re^{i t}\, dt=2\pi i\sum_{z\in V}\operatorname{Res}(f e^{-i\omega\cdot},z)$$

where $V:=R\Bbb D\cap\{z\in\Bbb C:\Im(z)> 0\}$. But now note that

$$\left|\int_0^\pi f(R e^{it})e^{-i\omega Re^{it}}Re^{i t}\, dt\right|\le\pi\max_{t\in[0,\pi]} |f(Re^{it})Re^{\omega R\sin t}|\le\pi\max_{|z|=R}|zf(z)|$$

so the integral in the arc goes to zero as $R$ goes to infinity. For the case $\omega >0$ we choose the arc below the real line, with a similar result.