# Does improper integral $\int\limits_{1}^{\infty}\frac{x\sin x}{x^2+3x+3}\,dx$ converge?

Could you, please, give any reasons why this integral converges. I have tried a lot of different methods (e.g. Dirichlet's test) none of them helped me

$$\int\limits_1^\infty \frac{x\sin x}{x^2+3x+3}\,dx$$

• Is contour integration (complex) allowed? – imranfat May 28 '18 at 17:38
• @imranfat No, it isn’t – FoRRestDp May 28 '18 at 17:39
• Well, then the provided answer is also a little issue (it is not incorrect), because that assumes convergence of the integral involving $sinx/x$, so that needs to be addressed first – imranfat May 28 '18 at 17:40
• Wolfram gives a numerical result of it here – TheSimpliFire May 28 '18 at 17:49
• What was the trouble applying Dirichlet test? – A.Γ. May 28 '18 at 18:24

Hint: if you know that the integral of $\frac{\sin x}{x}$ is convergent, you can write $$\frac{\sin x}{x}=\frac{x^2+3x+3}{x^2+3x+3}\frac{\sin x}{x}=\frac{x\sin x}{x^2+3x+3}+\frac{3\sin x}{x^2+3x+3}+\frac{3\sin x}{(x^2+3x+3)x}$$ and combine the terms to express your integral of interest via integrals that are easier to prove convergence for.
P.S. If you don't then try integration by parts (integrate $\sin x$ and derivate the rational part).