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QUESTION

Find the sine transform $F_s(\xi),\xi>0$ of
$$f(x)=\frac{x}{1+x^4}$$ by integrating $\ds\frac{ze^{i\xi
z}}{(1+z^4)}$ around a large semicircle.

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ANSWER

$$f(x)=\frac{x}{1+x^4},\ F_s(\xi)=\int_0^\infty
\frac{x}{1+x^4}\sin \xi x =\frac{1}{2}\int_{-\infty}^\infty\frac{x
e^{i\xi x}}{1+x^4}\,dx$$ For $\xi >0$ closed contour is in the
upper half plane.

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$\ds\textrm{Res}(\frac{ze^{i\xi z}}{1+z^4},z_0)=\frac{z_0e^{i\xi
z_0}}{4x_0^3}=-\frac{z_0^2}{4}e^{i\xi z_0}$
\begin{eqnarray*}
F_s(\xi)&=&\frac{1}{2}\textrm{Im}\ 2 \pi
i\left(-\frac{i}{4}e^{i\xi\frac{1+i}{\sqrt{2}}}-\frac{-i}{4}e^{i\xi\frac{-1+i}{\sqrt{2}}}\right)\\
&=&\frac{\pi}{2}\textrm{Re}\ e^{-\frac{\xi}{\sqrt{2}}}
\left(\frac{e^{i\xi}{\sqrt{2}}}{2i}-\frac{e^{-\frac{i\xi}{\sqrt{2}}}}{2i}-
\frac{e^{-\frac{i\xi}{\sqrt{2}}}}{2i}\right)\\
&=&\frac{\pi}{2}e^{-\frac{\xi}{\sqrt{2}}}\sin\frac{\xi}{\sqrt{2}}\textrm{
for }\xi>0
\end{eqnarray*}
$F_s(-\xi)=\overline{F_s(\xi)}=F_s(\xi)$

So for any $\xi,\ \
F_s(\xi)=\frac{\pi}{2}e^{-\frac{|\xi|}{\sqrt{2}}}\sin
\frac{|\xi|}{\sqrt{2}}$

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