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.
 A: Consider the function $f(t)=e^{\large-\sqrt a|t|}$, then the Fourier transform of $f(t)$ is given by
$$
\begin{align}
F(x)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-ix t}\,dt\\
&=\int_{-\infty}^{\infty}e^{-\sqrt a|t|}e^{-ix t}\,dt\\
&=\int_{-\infty}^{0}e^{\sqrt at}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-\sqrt at}e^{-ix t}\,dt\\
 &=\lim_{u\to-\infty}\left. \frac{e^{(\sqrt a-ix)t}}{\sqrt a-ix} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(\sqrt a+ix)t}}{\sqrt a+ix} \right|_{0}^{t=v}\\
&=\frac{1}{\sqrt a-ix}+\frac{1}{\sqrt a+ix}\\
&=\frac{2\sqrt a}{x^2+a}.
\end{align}
$$
Next, the inverse Fourier transform of $F(x)$ is
$$
\begin{align}
f(t)=\mathcal{F}^{-1}[F(x)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(x)e^{ix t}\,dx\\
e^{-\sqrt a|t|}&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\sqrt a}{x^2+a}e^{ix t}\,dx\\
\frac{\pi e^{-\sqrt a|t|}}{\sqrt a}&=\int_{-\infty}^{\infty}\frac{e^{ix t}}{x^2+a}\,dx,\tag1
\end{align}
$$
where $(1)$ can be rewritten as
$$
\int_{0}^{\infty}\frac{e^{ix t}}{x^2+a}\,dx=\frac{\pi e^{-\sqrt at}}{2\sqrt a}.\tag2
$$
Now differentiate $(2)$ with respect to $a$ twice and with respect to $t$ five times, take the real part, and set $a=t=1$ yields
\begin{align}
\Re\left[\int_{0}^{\infty}\frac{\partial^2}{\partial a^2}\frac{\partial^5}{\partial t^5}\left(\frac{e^{ix t}}{x^2+a}\right)\,dx\right]_{t=1,\,a=1}&=\left.\frac{\partial^2}{\partial a^2}\frac{\partial^5}{\partial t^5}\left(\frac{\pi e^{-\sqrt at}}{2\sqrt a}\right)\right|_{t=1,\,a=1}\\
-2\int_{0}^{\infty}\frac{x^5\sin x}{(x^2+1)^3}\,dx&=-\frac{\pi}{4e}\\
\int_{0}^{\infty}\frac{x^5\sin x}{(x^2+1)^3}\,dx&=\large\color{blue}{\frac{\pi}{8e}}.
\end{align}
A: It is well known that $\displaystyle\int_0^\infty\frac{\sin x}{x}\,dx$ converges. Then
$$
\int_0^{\infty}\frac{x^5\sin x}{(1+x^2)^3}\,dx=\int_0^{\infty}\Bigl(\frac{x^5}{(1+x^2)^3}-\frac1x\Bigr)\sin x\,dx+\int_0^{\infty}\frac{\sin(x)}{x}.
$$
This implies that the integral converges, since the the first integral on the right hand side converges absolutely,
To compute the value use calculus of residues. Let 
$$
f(z)=\frac{z^5\,e^{iz}}{(1+z^2)^3}.
$$
$f$ is meromorphic on an open set contaning the upper half plane, with a pole of order $3$ at $z=i$. For $R>1$ let $C_R$ be the semicircle $z=R\,e^{it}$, $0\le t\le\pi$. Then
$$
\int_{-R}^R\frac{x^5e^{ix}}{(1+x^2)^3}\,dx+\int_{C_R}f(z)\,dz=2\,\pi\,i\,\text{Res}(f;i).
$$
Now you have to:


*

*Prove that $\lim_{R\to\infty}\int_{C_R}f(z)\,dz=0$

*Compute the residue.

A: An alternative to the obvious canonical answer given by Julián Aguirre:
it is easy to compute $$\int_{-\infty}^{\infty} \frac{e^{i b x}}{x^2+a}dx $$
for $a, b > 0$, because the integrand has simple poles only. Now differentiate with respect to $a$ twice and with respect to $b$ 5 times, and set $a = b = 1$.
The theorem you need to justify the differentiation can be found in your textbook.
A: I try to tackle the problem by differentiation under integral sign.
Using the famous result:
$$
\int_0^{\infty} \frac{\cos a x}{x^2+t^2} dx=\frac{\pi}{2 t} e^{-a t}
$$
where $a,t>0.$
we get the result of the integral
$$
\int_0^{\infty} \frac{\cos a x}{1+t x^2} d x=\frac{\pi}{2} \frac{1}{\sqrt{t} e^{\frac{a}{\sqrt t}}}
$$
Differentiating both sides w.r.t $t$ twice at $t=1$ yields
$$
\int_0^{\infty} \frac{x^4 \cos (a x)}{\left(1+x^2\right)^3} d x=\frac{\pi\left(3-5 a+a^2\right)}{16 e^a}
$$
Differentiating both sides w.r.t $a$ once yields
$$
\frac{\partial}{\partial a} \int_0^{\infty} \frac{x^4 \cos (a x)}{\left(1+x^2\right)^3} d x =\frac{\pi}{16} \frac{\partial}{\partial a}\left(\frac{3-5 a+a^2}{e^a}\right)
$$
Putting $a=1$ yields
$$
-\int_0^{\infty} \frac{x^5 \sin x}{\left(1+x^2\right)^3}=\frac{\pi}{16}\left(-\frac{2}{e}\right)
$$
Hence
$$\boxed{\int_0^{\infty} \frac{x^5 \sin x}{\left(1+x^2\right)^3}=\frac{\pi}{8e}}$$
A: 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
$$ I=\frac{\pi}{8e}.$$
