I am trying to compute this integral: $$\int_{-\infty}^{\infty}\frac{x-\sin x}{x^3(x^2+16)}dx$$ In theory, I know how to do it. After some easy part (I will post it upon request) I stuck with the following: I will have to compute $$A=Res(\frac{z+ie^{iz}}{z^3(z^2+16)},0)$$ and $$A'=Res(\frac{z+ie^{iz}}{z^3(z^2+16)},4i).$$ I can compute $A'$ instantly. I know that $$A=\frac12\Bigg(\frac{z+e^{iz}}{(z^2+16)}\Bigg )''(0).$$ I can compute $A$ with CAS, and even manually and finally I can evaluate my integral. The answer is correct. The problem is that computation of the derivative takes huge amount of time. I can not perform it in class in reasonable time. I found this problem in a solution manual and there they evaluate very similar residue instantly, without computations: $$B=Res(\frac{z+i(e^{iz}-1)}{z^3(z^2+16)},0) =-\frac{i}{32}.$$ I completely understand when and how they get that $-1$ in enumerator, this is clear for me. So my questions:
- How do they compute $B$ instantly? Are they cheating, or they are smart?
- Is it possible to compute $A$ instantly?
- If they are not cheating and it is not possible to compute $A$ without long calculation of a derivative, then what makes the difference?
I can provide additional details about steps that are clear for me upon request.
Thanks again for your answers!
Update:
Ok, now I see something related to computation of $B$: $z+i(e^{iz}-1) = -\frac{i}{2}z^2+o(z)$ at $0$ so after you divide on $z^3(z^2+16)$ you can kinda see how did they get $-\frac{i}{32}$ and do it easily. Is my idea true? Can you please explain it in a rigourus way? I do not really understand what is going on with $z^2+16$. Thanks again!
Update 2: Ok, I figured out everything myself. It is just straightforward evaluation on residue by definition via Taylor series, without at derivatives. It works at $0$ very easily. $-1$ in enumerator was important trick that made it possible.
The singularity of $\dfrac{z - \sin z}{z^3 (z^2 + 16)}$ at $z = 0$ is removable. That should help.