We had the following theorem in class for a fourier transform $\widehat f$:

Let $\widehat f$ be the restriction of a $\mathbb C$ definied meromorphic function $F$. Let $F$ have a finite number of poles and let $z\cdot F(z)$ be bounded for big $z$. Then for $t>0$\begin{align*} f(t)=i\cdot\sum_{\Im(z)>0}\textrm{res}_z\left(F(z)e^{i zt}\right)\end{align*}

My proof: Consider $\gamma_r(s):=re^{i s}$: The residue theorem gives us for big $r$\begin{align*} \int_{\gamma_r}F(z)e^{i zt}\,d z+\int_{-r}^r\widehat f(\omega)e^{i \omega t}\,d \omega=2\pi i\cdot\sum_{\Im(z)>0}\textrm{res}_z\left(F(z)e^{i zt}\right) \tag{$\ast$}\end{align*} Now $|z\cdot F(z)|\leq M$ for $z\geq R$, so for $r\geq R$\begin{align*} \left|\int_{\gamma_r}F(z)e^{i zt}\,d z\right|&=\left|\int_0^\pi F(re^{i s})i re^{i s}\cdot e^{i \gamma_r(s)t}\,d s\right|\\ &\leq \int_0^\pi r\left|F\left(re^{i s}\right)\right|e^{-rt\sin s}\,d s\\ &\leq M\int_0^\pi e^{-rt\sin s}\,d s \end{align*} For $0\leq s\leq \frac{\pi}{2}$ we have $\sin s\geq\frac{2s}{\pi}$, and so \begin{align*} 2M\int_0^{\frac{\pi}{2}}e^{-rt\sin s}\,d s&\leq2M\int_0^{\frac{\pi}{2}}e^{-rt\frac{2s}{\pi}}\,d s\\ &=2M\left[\left(\frac{-\pi}{2rt}\right)\cdot e^{-rt\cdot\frac{2s}{\pi}}\right]_0^{\frac{\pi}{2}}\\ &=\frac{M\pi}{rt}\cdot\left(1-e^{-rt}\right)\\ &\xrightarrow[]{r\rightarrow\infty}0 \end{align*} So we get\begin{align*} f(t)=\frac1{2\pi}\lim_{r\rightarrow\infty}\int_{-r}^r\widehat f(\omega)e^{i\omega t}\,d\omega=i\cdot\sum_{\Im(z)>0}\textrm{res}_z\left(F(z)e^{i zt}\right) \end{align*}

So we had ($\ast$) be given in class, but I don't understand why this equation holds. How does it follow from the residue theorem?

And second why is $|e^{i\gamma_r(s)t}|=e^{-rt\sin s}$? (I did it with wolfram alpha because I didn't get any nice result)



1) It seems like you have assumed that $0 \leq s \leq \pi$ for $\gamma_r(s):=re^{i s}$, such that it is the arc of an upper half-circle. Together with the line $-r$ to $r$ this gives you a half-circle like in this picture, and for large $r$, all poles in the upper half plane will be inside it (since there are finitely many). Then it is just the residue theorem.

2) Remember that $\gamma_r(s):=re^{i s} = r \cos(s) + ir \sin(s)$, and that $|e^{ik}|$ is one when $k$ is real.

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