<s>[INST] How do I evaluate the integral $\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$? I have no idea how to start, it looks like integration by parts won't work.
$$\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$$
If someone could shed some light on this I'd be very thankful.
 [/INST] Note
\begin{eqnarray}
I&=&\int_0^\infty \frac{x[(x^2+1)-1]^2\sin x}{(1+x^2)^3}dx\\
&=&\int_0^\infty \frac{x\sin x}{1+x^2}dx-2\int_0^\infty \frac{x\sin x}{(1+x^2)^2}dx+\int_0^\infty \frac{x\sin x}{(1+x^2)^3}dx.
\end{eqnarray}
From
$$ \int_0^\infty \frac{\cos(ax)}{b^2+x^2}dx=\frac{\pi}{2b}e^{-ab}, a>0, b>0, $$
we have
\begin{eqnarray}
\int_0^\infty \frac{x\sin(ax)}{b^2+x^2}dx&=&=-\frac{d}{da}\int_0^\infty \frac{\cos(ax)}{b^2+x^2}dt=\frac{\pi}{2e^{-ab}},\\
\int_0^\infty \frac{x\sin x}{(b^2+x^2)^2}dx&=&-\frac{1}{2b}\frac{d}{db}\int_0^\infty \frac{x\sin(ax)}{b^2+x^2}dt=\frac{a\pi}{4be^{-ab}},\\
\end{eqnarray}
and
\begin{eqnarray}
\frac{d^2}{db^2}\int_0^\infty \frac{x\sin (ax)}{b^2+x^2}dx&=&8b^2\int_0^\infty \frac{x\sin(ax)}{(b^2+x^2)^3}dx-2\int_0^\infty \frac{x\sin(ax)}{(b^2+x^2)^2}dx.
\end{eqnarray}
The latter one implies
\begin{eqnarray}
\int_0^\infty \frac{x\sin(ax)}{(b^2+x^2)^3}dx=\frac{a(ab+1)\pi}{16b^3e^{-ab}}.
\end{eqnarray}
Thus
$$ I=\frac{\pi}{8e}.$$
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