Integral $\int_0^\infty \frac{x-\sin x}{x^3(x^2+4)} dx$
Solution 1:
Let $J(a)=\displaystyle\int_{0}^{\infty}\frac{ax-\sin ax}{x^3(x^2+1)}dx$. Differentiating by $a$ three times (which is admissible under the integral sign, because the resulting integrals converge uniformly for $a$ in any finite interval), we get $$J'''(a)=\displaystyle\int_{0}^{\infty}\frac{\cos ax}{x^2+1}dx=\frac{\pi}{2}e^{-|a|}.$$ With $J(0)=J'(0)=J''(0)=0$ this gives $J(a)=\dfrac{\pi}{2}\Big(1-|a|+\dfrac{a^2}{2}-e^{-|a|}\Big)\operatorname{sgn}(a)$.
The answer is $\dfrac{J(2)}{16}=\dfrac{\pi}{32}(1-e^{-2})$ as expected.