Find this integral $$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
my idea: $$I=2\int_{0}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
because $$\dfrac{x^3\sin{x}}{x^4+x^2+1}\approx\dfrac{\sin{x}}{x},x\to\infty$$ so $$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$ converges
then I can't,Thank you
Just like the comments given by @heropup and @Random Variable, this problem can be solved by using residue theorem and Jordan's lemma.
Note that the denominator of the integrand can be factorize by \begin{equation} z^4+z^2+1=(z^2-z+1)(z^2+z+1) \end{equation} It have four poles that \begin{equation} z_{1,2}=\pm\frac{1}{2}+\frac{\sqrt{3}}{2}i\\ z_{3,4}=\pm\frac{1}{2}-\frac{\sqrt{3}}{2}i\\ \end{equation} Only $z_1$ and $z_2$ are in upper half plane. Then, from residue theorem and Jordan's lemma, we have: \begin{equation} I=\int_{-\infty}^{+\infty} \frac{x^3\sin(x)}{x^4+x^2+1}dx\\ =\Im(2\pi i\sum_{k=1,2}\mathrm{Res}_{z=z_k}\frac{z^3e^{iz}}{z^4+z^2+1})\\ =-\frac{1}{3}e^{-\sqrt{3}/2}\sin(\frac{1}{2})\pi\sqrt{3}+e^{-\sqrt{3}/2}\cos(\frac{1}{2})\pi \end{equation} The approximation of the result is $0.7938888330$. You can check it by some numerical method.
May I confess that I sawer nicer integrals ?
Anyway, being patient and using Mhenni Benghorbal's suggestions plus a series of integrations by parts as well as a few changes of variables, the antiderivative is
$$\frac{1}{24} e^{-\sqrt[6]{-1}} \left(e^{\sqrt{3}} \left(\left(\sqrt{3}-3 i\right)
\left(\text{Ei}\left(i x+(-1)^{5/6}\right)-\text{Ei}\left(\frac{1}{2} \left(-2 i
x-\sqrt{3}+i\right)\right)\right)+\left(\sqrt{3}+3 i\right) e^i
\left(\text{Ei}\left(\frac{1}{2} \left(-2 i
x-\sqrt{3}-i\right)\right)-\text{Ei}\left(\frac{1}{2} \left(2 i
x-\sqrt{3}-i\right)\right)\right)\right)-\left(\sqrt{3}-3 i\right) e^i
\left(\text{Ei}\left(\frac{1}{2} \left(-2 i
x+\sqrt{3}-i\right)\right)-\text{Ei}\left(\frac{1}{2} \left(2 i
x+\sqrt{3}-i\right)\right)\right)+\left(\sqrt{3}+3 i\right)
\left(\text{Ei}\left(\frac{1}{2} \left(-2 i
x+\sqrt{3}+i\right)\right)-\text{Ei}\left(\frac{1}{2} \left(2 i
x+\sqrt{3}+i\right)\right)\right)\right)$$ Once integrated between $-\infty$ and $\infty$, again after some patience, I arrived to $$\frac{\left(-i+\sqrt{3}+\left(\sqrt{3}+i\right) e^i\right) e^{-\sqrt[6]{-1}} \pi }{2
\sqrt{3}}$$ that is to say $$-\frac{1}{3} e^{-\frac{\sqrt{3}}{2}} \pi \left(\sqrt{3} \sin
\left(\frac{1}{2}\right)-3 \cos \left(\frac{1}{2}\right)\right)$$ which reduces to $$\frac{2}{\sqrt{3}} e^{-\sqrt{3}/2} \pi \cos \frac{\pi+3}{6}$$ which corresponds to heropup's result.
Thanks to Mhenni Benghorbal and heropup for their suggestions and ideas.
Note that $$I(a) = \int_{-\infty}^\infty \frac{t\sin at} {t^2+1}dt = \pi e^{-a} $$
(derived from $I’’(a)=I(a)$, $I(0)= -I’(0) =\pi$). Then \begin{align} \int_{-\infty}^{\infty}\frac{x^3\sin x}{x^4+x^2+1}\> dx = &\frac12\int_{-\infty}^{\infty} \overset{x=\frac{\sqrt3t -1}2}{\frac{(x+1)\sin x}{x^2+x+1}} +\overset{x= \frac{\sqrt3t +1}2}{\frac{(x-1)\sin x}{x^2-x+1}} d x\\ =&\cos\frac1{2} \int_{-\infty}^{\infty} \frac{t\sin \frac {\sqrt3}{2}t}{t^2+1} dt - \frac1{\sqrt3}\sin\frac1{2} \int_{-\infty}^{\infty} \frac{\cos\frac {\sqrt3}{2}t}{t^2+1} dt \\ =& \cos\frac1{2} I(\frac {\sqrt3}{2}) +\frac1{\sqrt3}\sin\frac1{2} I’(\frac {\sqrt3}{2})\\ =&\pi \>e^{-\frac {\sqrt3}{2} }\left( \cos\frac12-\frac1{\sqrt3} \sin\frac12\right)\\ \end{align}
