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QUESTION

Find the function whose Fourier transform is
$$\frac{1}{16+\xi^4}$$ by integrating
$\ds\frac{e^{izx}}{(16+z^4)}$ around a large semicircle.

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ANSWER
$$F(\xi)=\frac{1}{16+\xi^4};\ \ \
f(x)=\frac{1}{2\pi}\int_{-\infty}^\infty
\frac{e^{izx}}{16+z^4}\,dz$$ For $x>0$ we choose in the upper half
plane:

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Poles at $z=2e^\frac{i\pi}{4}=\sqrt{2}(1+i)$ and $z=2e^\frac{3
i\pi}{4}=\sqrt{2}(-1+i)$

(and two more poles in the lower half plane).\\
$\ds\textrm{Res}\left(\frac{e^{izx}}{16+z^4},z_0\right)=\frac{e^{iz_0x}}{4z_0^3}=\frac{z_0e^{iz_0x}}{-64}$
\begin{eqnarray*}
f(x)&=&\frac{1}{2\pi}2\pi
i\left(-\frac{1}{32}e^\frac{i\pi}{4}e^{i\sqrt{2}\left(1+i\right)x}-\frac{1}{32}e^\frac{3i\pi}{4}e^{i\sqrt{2}\left(-1+i\right)x}\right)\\
&=&-\frac{i}{32}e^{-\sqrt{2}x}\left(e^{i\sqrt{2}x+\frac{i\pi}{4}}-e^{-i\sqrt{2}x-\frac{i\pi}{4}}\right)\\
&=&-\frac{i}{32}e^{-\sqrt{2}x}\left(e^{i\sqrt{2}x+\frac{i\pi}{4}}+e^{-i\sqrt{2}x+\frac{3i\pi}{4}}\right)\\
&=&\frac{1}{16}e^{-\sqrt{2}x}\sin\left(\sqrt{2}x+\frac{\pi}{4}\right),\
x>0
\end{eqnarray*}
From the expression for $f(x), f(-x)=\overline{f(x)}=f(x)$, so for
any $x$, $$f(x)=\frac{1}{16}e^{-\sqrt{2}|x|}\sin
\left(\sqrt{2}|x|+\frac{\pi}{4}\right)$$

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