Residue calculus: Integrals go to zero

Question

Let f be a rational function without poles on the real axis and a nullpoint in $\infty$. Then we have: $\lim_{r \rightarrow \infty} \int_{-r}^r f(x) e^{ix} dx = 2 \pi i \sum_{Im(a)>0}\text{Res}_a (f(z)e^{iz})$.

Proof:

We choose a rectangle as the integration path: Let $r>0$ be large enough. Then by the residue theorem we know: $$\int_{-r}^r f(x) e^{ix} dx + \int_{r}^{r+ir} f(x) e^{ix} dx + \int_{r+ir}^{-r+ir} f(x) e^{ix} dx+\int_{-r+ir}^{-r} f(x) e^{ix} dx = 2 \pi i \sum_{Im(a)>0}\text{Res}_a (f(z)e^{iz})$$

Since $f(x) e^{ix} $ has a null at $\infty$, we know that $s_r = \sup_{|z|\geq r} |f(z)| $ exists for $r$ large enough and goes to $0$. But we can estimate the absolute values of the second and fourth integral term by $s_r \int_0^r e^{-t} dt \leq s_r$, and the third integral term by $2re^{-r}s_r$, so they go to $0$ for $r\rightarrow \infty$.

I do not understand the bold part. Why are those estimates true? I understand the rest, but I just don't see why these estimates are true.

Answer 1

I will first deal with$$\int_{r+ir}^{-r+ir}f(x)e^{ix}\,\mathrm dx.$$We have\begin{align}\left|\int_{r+ir}^{-r+ir}f(x)e^{ix}\,\mathrm dx\right|&\leqslant2re^{-r}\sup_{t\in[-1,1]}|f(tr+ir)\\&\leqslant2\sup_{t\in[-1,1]}|f(tr+ir)|,\end{align}since $(\forall x\in(0,\infty):xe^{-x}<1$. So, since $\lim_{z\to\infty}f(z)=0$, given $\varepsilon>0$ you have $\left|\int_{r+ir}^{-r+ir}f(x)e^{ix}\,\mathrm dx\right|<\varepsilon$ if $r$ is large enough.

Now, I will deal with$$\int_r^{r+ir}f(x)e^{ix}\,\mathrm dx.$$We have\begin{align}\left|\int_r^{r+ir}f(x)e^{ix}\,\mathrm dx\right|&\leqslant r\int_0^1\left|f(r+tir)e^{i(r+tir)}\right|\,\mathrm dt\\&\leqslant r\sup_{t\in[0,1]}\left|f(r+tir)\right|\int_0^1e^{-tr}\,\mathrm dt\\&=\sup_{t\in[0,1]}\left|f(r+tir)\right|(1-e^{-r})\\&\leqslant\sup_{t\in[0,1]}\left|f(r+tir)\right|.\end{align}So, once again, since $\lim_{z\to\infty}f(z)=0$, given $\varepsilon>0$ you have $\left|\int_r^{r+ir}f(x)e^{ix}\,\mathrm dx\right|<\varepsilon$ if $r$ is large enough.

The case of the remaining term is similar.

Answer 2

Perhaps a little simpler: The residue theorem implies

$$\int_{-r}^r f(x)e^{ixt}\,dx +\int_{\gamma_r}f(z)e^{iz}\,dz$$ $$ = 2 \pi i \sum_{\text {Im }(a)>0}\text{Res}_a (f(z)e^{iz}).$$

Here $\gamma_r(t) = re^{it}, 0\le t\le \pi.$ Now

$$|\int_{\gamma_r}f(z)e^{iz}\,dz| \le \int_0^{\pi}|f(re^{it}| re^{-r\sin t} dt$$

$$\le M_r\cdot 2\int_0^{\pi/2}re^{-r\sin t} dt $$

Here $M_r = \sup_{t\in [0,\pi]} |f(re^{it})|$ and we are given that $M_r\to 0.$ Since there is $c>0$ such that $\sin t \ge ct$ on $[0,\pi/2],$ the last integral is bounded as $r\to \infty.$ Hence the last expression $\to 0$ as desired.